solutionsto supplementary problems - springer · d.gross•w.hauger j.schr¨oder •w.a.wall...

D. Gross • W. HaugerJ. Schroder • W.A. WallS. Govindjee

Engineering Mechanics 3Dynamics

Solutions toSupplementary

Problems

The numbers of the problems and the figures correspondto the numbers in the textbook Gross et al., EngineeringMechanics 3, Dynamics, 2nd Edition, Springer 2013

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1Chapter 1

Motion of a Point Mass

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2 1 Motion of a Point Mass

E1.17 Example 1.17 A point P moves on a given path from A to B

(Fig. 1.42). Its velocity v decreases linearly with arc-length s from

a value v0 at A to zero at B.

How long does it take P to reach point B?

l s

v

s

B

P

A

s = l

v

v0

Fig. 1.42

Solution First we write down the velocity v as a linear function

of the arc-length s:

v(s) = v0(1− s

l

).

Separation of variables

ds

dt= v → ds

v0(1− s

l

) = dt

and indeterminate integration lead to∫

ds

1− sl

= v0

∫dt → s = l

[1− C exp

(−v0t

l

)].

The integration constant C is determined from the initial condi-

tion:

s(0) = 0 : 0 = 1− C → s = l

[1− exp

(−v0t

l

)].

Point B is reached when s = l. This yields the corresponding time

tB :

s(tB) = l → tB →∞ .

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1 Motion of a Point Mass 3

E1.18Example 1.18 A radar screen tracks a rocket which rises vertically

with a constant acceleration

a (Fig. 1.43). The rocket is

launched at t = 0.

Determine the angular ve-

locity ϕ and the angular acce-

leration ϕ of the radar screen.

Calculate the maximum an-

gular velocity ϕ and the cor-

responding angle ϕ.

ϕ

lFig. 1.43

Solution Since the acceleration

a is constant, the velocity v

and the position x of the rocket

are

v = at+ v(0) ,

x = at2/2 + v(0)t+ x(0) .

xϕ

Applying the initial conditions yields

v(0) = 0 → v = at,

x(0) = 0 → x =at2

2.

The angle ϕ of the radar screen follows as

tanϕ =x

l→ ϕ(t) = arctan

(at2

2l

).

Differentiation leads to the angular velocity

ϕ(t) =at

l/

[1 +

(at2

2l

)2]

and the angular acceleration

ϕ(t) =

(a

l− 3a3t4

4l3

)/

[1 +

(at2

2l

)2]2.

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The time t∗ of the maximum angular velocity ϕmax is obtained

from

ϕ(t∗) = 0 → t∗ =

(4l2

3a2

)1/4

.

Thus,

ϕmax = ϕ(t∗) → ϕmax =

√3√3a

8l

and

ϕ(t∗) = arctan(1/√3) → ϕ(t∗) = 30 .

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E1.19Example 1.19 Two point masses P1 and P2 start at point A with

zero initial velocities and travel on a circular path. P1 moves with

a uniform tangential acceleration at1 and P2 moves with a given

uniform angular velocity ω2.

a) What value must at1 have

in order for the two masses

to meet at point B?

b) What is the angular veloci-

ty of P1 at B?

c) What are the normal acce-

lerations of the two masses

at B?

r

P2

P1

A B

Fig. 1.44

Solution The velocity and position of P1 follow from at1 = const =

v with the initial conditions s01 = 0 and v01 = 0 as

v1 = at1t , s1 =1

2at1t

2 .

Similarly, for point P2 we obtain from at2 = 0 with s02 = 0 and

v02 = rω2:

v2 = rω2 , s2 = rω2t .

a) Both points meet at time tB at point B. Thus,

πr =1

2at1t

2B , πr = rω2tB .

This leads to

tB =π

ω2→ at1 =

2πr

t2B=

2rω22

π.

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b) The tangential acceleration at1 and the time tB are now known.

Hence, we can calculate the angular velocity of P1 at B:

v1(tB) = at1tB = rω1(tB)

→ ω1(tB) =at1tBr

=2rω2

2π

πrω2= 2ω2 .

c) The normal accelerations at B follow from an = rω2:

an1 = rω21(tB) = 4rω2

2 , an2 = rω22 .

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E1.20Example 1.20 A child of massm jumps up and down on a trampo-

line in a periodic manner. The child’s jumping velocity (upwards

upon leaving the trampoline) is v0 and during the contact time

∆t the contact force F (t) has a triangular form.

Find the necessary contact force amplitude F0 and the jum-

ping period T0.

Fig. 1.45

F

t∆t T0

zF0

Solution The child is subjected

to its constant weight W = mg

and, during the contact with the

trampoline, to the contact for-

ce F (t). While the child is in the

air, the equation of motion is gi-

ven by

↑: mz = −mg.

F

mg

F

t20 tt1

T0

F0

Integration between t = t0 = 0 (end of a contact) and t = t1(beginning of a new contact) yields

mv1 −mv0 = −mgt1,

where v1 = v(t1) and v0 = v(0). Recall that the velocity at the

beginning of a vertical motion and the velocity at the end of a

free fall of a body are equal in magnitude: v1 = −v0 (note the

different signs). Therefore, we obtain

2v0 = gt1 → t1 = 2v0g.

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The jumping period T0 follows with ∆t = t2 − t1 as

T0 = t1 +∆t → T0 =2v0g

+∆t .

We now apply the Impulse Law between time t0 and time t2 (see

the figure):

↑ : mv2 −mv0 = −mgT0 +1

2F0(t2 − t1) .

In order to have a periodic process, the velocities v2 and v0 have

to coincide: v2 = v0. With this condition the Impulse Law yields

−mgT0 +1

2F0∆t = 0 → F0 =

2T0∆t

mg = 2( 2v0g∆t

+ 1)mg .

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E1.21Example 1.21 A car is travelling in a circular arc with radius R

and velocity v0 when it starts to brake.

If the tangential deceleration is

at(v) = −(a0+κv), where a0 and κ are

given constants, find the time to brake

tB, the stopping distance sB, and the

normal acceleration an during the bra-

king.

v0

R

s

Fig. 1.46

Solution The acceleration at is given as a function of the velocity:

at(v) = v = −(a0 + κv) .

We separate the variables and integrate (initial condition: v(0) = v0):

t(v) = −∫ v

v0

dv

a0 + κv=

1

κlna0 + κv0a0 + κv

.

The time tB when the car comes to a stop follows from the con-

dition v = 0:

tB = t(v = 0) =1

κln(1 +

κv0a0

).

Now we determine the inverse function of t(v):

eκt =a0 + κv0a0 + κv

→ v(t) =a0κ

[(1 +

κv0a0

)e−κt − 1

].

Integration with the initial condition s(0) = 0 leads to the position

s(t) =

∫ t

0

v(t)dt =a0κ2

[(1 +

κv0a0

)(1− e−κt

)− κt

].

The stopping distance sB is obtained for t = tB :

sB = s(tB) =a0κ2

[(1 +

κv0a0

)(1− 1

1 + κv0a0

)− ln

(1 +

κv0a0

)]

=a0κ2

[κv0a0− ln

(1 +

κv0a0

)].

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The normal acceleration an(t) during the breaking is found as

an =v2

R=

a20Rκ2

(1 +

κv0a0

)e−κt − 1

2

.

As a check on the correctness of the result we calculate an for

t = tB and obtain an = 0.

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E1.22Example 1.22 A point P moves on

a parabola y = b(x/a)2 from A to

B. Its position as a function of the

time is given by the angle ϕ(t) =

arctanω0 t (see Fig. 1.47).

Determine the magnitude of ve-

locity v(t) of point P . How much

time elapses until P reaches point

B? Calculate its velocity at B.

r

B

P

a xA

y

b

ϕ

Fig. 1.47

Solution The parabola y = b(x/a)2 and the angle ϕ(t) = arctanω0t

are given. The angle ϕ can also be expressed as ϕ = arctan y/x.

Thus, y/x = ω0t. Solving for x and y yields the position of

point P :

x(t) =a2

bω0t , y(t) =

a2

b(ω0t)

2 .

To obtain the velocity of P , we differentiate:

x(t) =a2

bω0 , y(t) = 2

a2

bω20t ,

v =√x2 + y2 → v(t) =

a2

bω0

√1 + 4ω2

0t2 .

Point B is reached at time tB when x = a:

x(tB) = a → tB =b

aω0.

Thus, the velocity at B is obtained as

vB = v(tB) → vB =a2

bω0

√

1 + 4b2

a2.

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E1.23 Example 1.23 A rod with length l rota-

tes about support A with angular po-

sition given by ϕ(t) = κ t2. A body

G slides along the rod with position

r(t) = l(1− κ t2).a) Find the magnitude of velocity and

acceleration of G when ϕ = 45.

b) At what angle ϕ does G hit the sup-

port?

Given: l = 2 m, κ = 0.2 s−2.

A

l

B

r G

ϕ(t)

Fig. 1.48

Solution a) First we calculate the time t = t1 that it takes the

rod to reach the position ϕ = ϕ1 = 45:

ϕ1 = π/4 = κt21 → t1 =√π/(4κ) = 1.98 s .

Then we determine the derivatives of the given functions r(t) and

ϕ(t):

r = l(1− κt2) , r = −2κlt , r = −2κl ,

ϕ = κt2 , ϕ = 2κt , ϕ = 2κ .

With t = t1 we obtain the velocity

vr = r = −2κlt1 = −1.58 m/s ,

vϕ = rϕ = l(1− κt21)2κt1 = 0.34 m/s ,

v =√v2r + v2ϕ = 1.62 m/s

v vrπ

4

vϕ

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and the acceleration

ar = r − rϕ2 = −2κl− l(1− κt21)4κ2t21= −1.07 m/s2 ,

aϕ = rϕ+ 2rϕ = l(1− κt21)2κ− 2 · 2κlt12κt1= −2.34 m/s

2,

a =√a2r + a2ϕ = 2.57 m/s2 .

ar aϕ

a

π

4

b) Body G reaches the support (r = 0) at time tE :

r = 0 = l(1− κt2E) → tE =√1/κ = 2.24 s .

This yields the corresponding angle ϕE :

ϕE = κt2E = 1 (= 57.3) .

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E1.24 Example 1.24 A mouse sits in a tower (with radius R) at point A

and a cat sits at the center 0.

If the mouse runs at a constant

velocity vM along the tower wall and

the cat chases it in an Archimedian

spiral r(ϕ) = Rϕ/π, what must the

cat’s constant velocity vC be in order

to catch the mouse just as the mou-

se reaches its escape hole H? At what

time does it catch the mouse?

H

R

0r

A

ϕ

Fig. 1.49

Solution We determine the components of the velocity of the cat

from the given equation r(ϕ) = Rϕ/π of the Archimedian spiral:

vr = r =dr

dϕ

dϕ

dt=R

πϕ and vϕ = rϕ =

R

πϕϕ .

Thus, the constant velocity vC can be written as

vC =√v2r + v2ϕ =

R

πϕ√1 + ϕ2 =

R

π

dϕ

dt

√1 + ϕ2 .

Separation of variables and integration lead to

vC

t∫

0

dt = vCt =R

π

ϕ∫

0

√1 + ϕ2 dϕ .

With the integral∫ √

1 + x2dx =1

2

[x√1 + x2 + arsinh x

]

this results in

vCt =R

2π

(ϕ√1 + ϕ2 + arcsinh ϕ

).

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The time T when both the cat and the mouse reach the hole

follows from the constant velocity vM of the mouse along the wall:

πR = vMT → T =πR

vM.

Introduction of T and ϕ(T ) = π into the expression for vCt yields

the velocity of the cat:

vC =vM2π2

(π√1 + π2 + arcsinh π

)= 0.62 vM .

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E1.25 Example 1.25 A soccer player kicks the ball (mass m) so that

it leaves the ground at an angle α0 with an initial velocity v0(Fig. 1.50). The air exerts a drag force Fd = k v on the ball; it

acts in the direction opposite to the velocity.

Determine the velocity v(t) of the ball. Calculate the horizon-

tal component vH of v when the ball reaches the teammate at a

distance l.

x

z

l

α0

v0

Fig. 1.50

Solution The equations of motion are

mx = −Fd cosα , mz = mg + Fd sinα .

We introduce the kinematic relations

x =dx

dt, z =

dz

dt

x

z

vFd

mg

α

and the drag force Fd = kv. Noting that the components of the

velocity are given by

x = v cosα , z = −v sinα

we obtain

mdx

dt= −kx , m

dz

dt= mg − kz .

Separation of variables gives

dx

x= − k

mdt ,

dz

z − mg

k

= − kmdt

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and integration yields

lnx

C1= − k

mt , ln

z − mg

kC2

= − kmt .

The constants of integration C1 and C2 can be determined from

the initial conditions. For t = 0 the left-hand sides of the equations

above have to be zero. Thus, the arguments of the ln-functions

have to be equal to one (numerator and denominator have to be

equal):

C1 = x(0) = v0 cosα0 , C2 = z(0)− mg

k= −v0 sinα0 −

mg

k.

This leads to the components of the velocity:

x(t) = v0 cosα0e−kt/m , z(t) =

mg

k−(mgk

+ v0 sinα0

)e−kt/m .

We now integrate the x-component to obtain the horizontal posi-

tion of the ball:

x(t) = −mkv0 cosα0e

−kt/m + C .

Exploiting the initial condition

x(0) = 0 → C =m

kv0 cosα0

yields

x(t) =m

kv0 cosα0

(1− e−kt/m

).

The time t∗ when the ball reaches the teammate at the distance

l is found to be

x(t∗) = l → t∗ =m

kln

mv0 cosα0

mv0 cosα0 − kl.

This leads to the corresponding horizontal component of the ve-

locity:

x(t∗) = v0 cosα0 −kl

m.

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E1.26 Example 1.26 A body with mass m

starts at a height h at time t = 0

with an initial horizontal velocity v0.

If the wind resistance can be

approximated by a horizontal force

H = c0mx2, at what time tB and

location xB does it hit the ground?

z

m

x

h

B

v0

Fig. 1.51

Solution The equations of

motion are given by

→ : mx = −mc0x2 , ↑ : mz = −mg .

We integrate and apply the

z

x

mg

H

initial conditions x(0) = 0, z(0) = h, x(0) = v0, z(0) = 0 to obtain

∫ x

v0

d ˙x˙x2

= −c0∫ t

0

dt → x =1

1v0 + c0t

,

∫ x

0dx =

∫ t

0

11v0 + c0t

dt → x =1

c0ln(1 + c0v0t) ,

z = −gt , z = −g2t2 + h .

The time tB when the body hits the ground follows from zB = 0:

tB =

√2h

g.

Thus, the location of point B is obtained as

xB = x(t = tB) =1

c0ln

(1 + c0v0

√2h

g

).

In the case of a vanishing wind resistance (c0 = 0) this result

reduces to

xB = limc0→0

1

c0ln

(1 + c0v0

√2h

g

)= v0

√2h

g.

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E1.27Example 1.27 A mass m slides on a

rotating frictionless and massless

rod S such that it is pressed against

a rough circular wall (coefficient of

friction µ).

If the mass starts in contact with

the wall at a velocity v0, how many

rotations will it take for its velocity

to drop to v0/10?

g

rS

m

µ

Fig. 1.52

Solution If we express the accelera-

tion vector in terms of the Serret-

Frenet frame, then the equations of

motion are written as

տ : mat = −R , ւ : man = N .R

at

an

N

Note that the weight of the mass does not influence the motion

since the motion takes place in a horizontal plane. If we use the ki-

nematic relations at = v, an = v2/r, and the friction law R = µN

we can write the first equation of motion in the form

mv = −µm v2

r.

Separation of variables and integration yield∫ v

v0

dv

v2= −

∫ t

0

µ

rdt → v(t) =

v0

1 +µv0r t

.

We integrate again to obtain the distance traveled:∫ s

0

ds = v0

∫ t

0

dt

1 + µv0r t→ s(t) =

r

µln (1 +

µv0r

t) .

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Now we calculate the time t1 that it takes for the velocity to drop

to v0/10:

v010

=v0

1 +µv0r t1

→ t1 =9r

µv0.

The corresponding value s(t1) is found as

s1 = s(t1) =r

µln (1 +

µv0r

t1) =r

µln 10 .

This yields the corresponding number of rotations:

n =s12πr

=ln 10

2πµ.

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E1.28Example 1.28 A car (mass m) is traveling with constant velocity

v along a banked circular curve (radius r, angle of slope α), see

Fig. 1.53. The coefficient of static friction µ0 between the tires of

the car and the surface of the road is given.

Fig. 1.53

r

m

α

µ0

Determine the region of the allowable velocities so that sliding

(down or up the slope) does not take place.

Solution We model the car as a point mass and express the ac-

celeration vector in terms of the Serret-Frenet frame. Then the

equation of motion in the direction of the normal vector is (see

the free-body diagram)

man = N sinα+H cosα .

v2/gr

µ0 = 0.3

π/60 π/3

2.0

1.5

1.0

0.5

αN

mg

H

M r

α

vr

M

eten

en

g

Since the car does not move in the vertical direction we can apply

the equilibrium condition

↑: 0 = N cosα−H sinα−mg .

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We now solve these two equations for the normal force N and the

force of static friction H :

H = man cosα−mg sinα ,N = man sinα+mg cosα .

The car does not slide if the condition of static friction

|H | ≤ µ0N

is satisfied. With an = v2/r this leads to the allowable region of

the velocity:

tanα− µ0

1 + µ0 tanα≤ v2

gr≤ tanα+ µ0

1− µ0 tanα.

This is displayed for µ = 0.3 as a function of the angle α in the

figure.

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E1.29Example 1.29 A car (mass m) has a velocity v0 at the beginning of

a curve (Fig. 1.54). Then it slows down with constant tangential

acceleration at = −a0. The coefficient of static friction between

the road and the tires is µ0.

Calculate the velocity v

of the car as a function of

the arc-length s. What is the

necessary radius of curvature

ρ(s) of the road so that the

car does not slide?

s

v0

m

Mρ(s)

Fig. 1.54

Solution We model the car as a point mass. Its velocity v can

be determined from the given constant tangential acceleration atthrough integration (initial condition v(0) = v0):

v = at = −a0 → v(t) = v0 − a0t .

The distance traveled follows from

s = v → s(t) = s0 + v0t− a0t2/2 .

With the initial condition s(0) = s0 = 0 we obtain

s(t) = v0t− a0t2/2 .

Elimination of the time t yields the velocity as a function of the

arc-length:

v(s) =√v20 − 2a0s .

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In order to determine the necessary radius of curvature ρ(s) we

write down the equations of motion:

mat = Ht → Ht = −ma0 ,

man = Hn → Hn = mv2

ρ,

0 = N −mg → N = mg .

The static friction forceH = (H2t +H

2n)

1/2 and the normal force N

have to satisfy the condition of static friction to avoid sliding of

the car:

|H | ≤ µ0N → a20 +v4

ρ2≤ µ2

0g2 .

Solving for ρ yields

ρ(s) ≥ v20 − 2a0s√µ20g

2 − a20.

This condition can not be satisfied if a0 > µ0g.

M

mg

Hn

Ht

et

Nen

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E1.30Example 1.30 A bowling ball (mass m) moves with constant ve-

locity v0 on the frictionless return of a bowling alley. It is lifted

on a circular path (radius r) to the height 2r at the end of the

return. The upper part of the circular path has a frictionless guide

of length r ϕG (Fig. 1.55).

Given the angle ϕG, determine the velocity v0 such that the

bowling ball reaches the upper level.

Fig. 1.55

g

v0 mr

ϕG

Solution The velocity v0 has to be large enough so that the bow-

ling ball reaches the upper level (height 2r) with a velocity v ≥ 0.

It is convenient to use the Conservation of Energy Law to deter-

mine the relation between v0 and v. With T0 = mv20/2, V0 = 0,

T1 = mv2/2 and V1 = 2mgr we obtain

T0 + V0 = T1 + V1 → mv20/2 + 0 = mv2/2 + 2mgr .

This yields a first requirement on v0:

v20 = v2 + 4gr → v20 ≥ 4gr .

In addition, the velocity v0 has to be large enough so that the

normal force between the bowling ball and the lower part of the

circular path does not become zero before the ball reaches the

guide of length rϕG. The necessary velocity v(ϕG) follows from

the equation of motion in the normal direction (note that ϕ is

measured from the upper level):

ց: mv2(ϕ)

r= N +mg cosϕ .

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The condition N ≥ 0 for ϕ = ϕG leads to

v2(ϕG) ≥ gr cosϕG .

In order to calculate the corresponding velocity v0 we use the Con-

servation of Energy Law between the lower level and the beginning

of the guide:

T0 + V0 = T2 + V2

→ mv20/2 + 0 = mv2(ϕG)/2 +mgr(1 + cosϕG) .

Thus, we obtain the second requirement

v20 ≥ (2 + 3 cosϕG)gr .

Both requirements are plotted in the figure, which shows that

the minimum velocity v0 is obtained as

v20 =

(2 + 3 cosϕG)gr for ϕG < ϕ∗

G = arccos2/3 ,

4gr for ϕG > ϕ∗

G .

2+3 cos ϕG

8

6

4

2

0.2π0 0.6π π

ϕG

v20/gr

0.4π 0.8π

ϕ∗

G

N

mg

ϕ

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E1.31Example 1.31 A point mass m is

subjected to a central force

F = mk2r, where k is a constant

and r is the distance of the mass

from the origin 0. At time t = 0

the mass is located at P0 and has

velocity components vx = v0 and

vy = 0.

Find the trajectory of the

mass.

α

y

F

m

x

r0

0

P0

r

v0

Fig. 1.56

Solution The equations of motion in the directions x and y are

→: mx = −F cosα ,

↑ : my = −F sinα .

With F = mk2r, x = r cosα and y = r sinα we obtain

x+ k2x = 0 , y + k2y = 0 .

Both equations are equivalent to the differential equation that de-

scribes undamped free vibrations of a point mass (see Chapter 5).

The solutions are

x = A cos kt+B sin kt , y = C cos kt+D sin kt .

The constants of integration are calculated from the initial condi-

tions:

x(0) = 0 → A = 0 , y(0) = r0 → C = r0 ,

x(0) = v0 → B =v0k, y(0) = 0 → D = 0 .

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We now have a parametric representation of the curve describing

the motion:

x =v0k

sin kt , y = r0 cos kt .

To eliminate the time, these equations are squared and then ad-

ded. Thus, we finally obtain

(x

v0/k

)2

+

(y

r0

)2

= 1 .

The point mass moves along an ellipse which reduces to a circle

for k = v0/r0.

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E1.32Example 1.32 A centrifuge with radius r rotates with constant an-

gular velocity ω0. A massm is to be placed at rest in the centrifuge

and accelerated within a time t1 to the angular velocity ω0.

What will be the needed (con-

stant) moment M acting on the

mass? What is the power P of this

moment?ω0

mr

Fig. 1.57

Solution The centrifuge rotates with a

constant angular velocity. Thus, the

driving torque M needed to accelera-

te the point mass is equal to the mo-

ment of the friction force R which acts

between the mass and the cylinder:

M = rR .

R

N

Nat

R

The equation of motion in the tangential direction for the mass is

↑ : mat = R ,

where at = rω. Thus,

mrω =M

r→ ω =

M

mr2.

Integration with the initial condition ω(0) = 0 yields

ω =M

r2mt .

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The condition ω(t1) = ω0 leads to the required moment:

M =r2mω0

t1.

The corresponding power is

P = M · ω0 =Mω0 =r2mω2

0

t1.

Note that the moment M has to be reduced to zero after time t1.

Otherwise the mass would be further accelerated.

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E1.33Example 1.33 A skier (mass m) has the velocity vA = v0 at point

A of the cross country course (Fig. 1.58). Although he tries hard

not to lose velocity skiing uphill, he reaches point B with only a

velocity vB = 2 v0/5. Skiing downhill between point B and the

finish C he again gains speed and reaches C with vC = 4 v0.

Between B and C assume that a constant friction force acts due

to the soft snow in this region; the drag force from the air on the

skier can be neglected.

Calculate the work done by the skier on the path from A to B

(here the friction force is negligible). Determine the coefficient of

kinetic friction between B and C.

finish

10h

B

3h

C

A

h

Fig. 1.58

Solution In order to calculate the work done by the skier on the

path from A to B we use the work-energy theorem:

TB + VB = TA + VA + U

→ mv2B/2 +mgh = mv2A/2 + 0 + U

→ U = m(gh− 21v20/50) .

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Now we apply the work-energy theorem between points B and C,

where the work of the friction force is given by −RlBC :

TC + VC = TB + VB + U

→ mv2C/2 + 0 = mv2B/2 + 3mgh−RlBC .

With the normal force

N = mg cosα → N = mg10h

lBC

and the law of kinetic friction

R = µN → R = µmg10h

lBC

mg

R

Nα

we obtain

µ =3

10− v2C − v2B

20gh→ µ ≈ 3

10− 4v20

5gh.

The result is valid only for µ ≥ 0.

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E1.34Example 1.34 A circular disk (radius R)

rotates with the constant angular veloci-

ty Ω. A point P moves along a straight

guide; its distance from the center of

the disk is given by ξ = R sinω t whe-

re ω = const (Fig. 1.59).

Determine the velocity and the acce-

leration of P .

P

R ξ

Ω

Fig. 1.59

Solution We use polar coordinates r, ϕ to

solve the problem. From Ω = const we find

ϕ = Ω = const → ϕ = Ωt , ϕ = 0 .

The position vector is written as

r = ξ er → r = R sinωt er .

eϕ ϕ

Prer

Differentiation yields the velocity vector

v = ξ er + ξϕ eϕ → v = Rω cosωt er +RΩ sinωt eϕ

and the acceleration vector

a = (ξ − ξϕ2) er + (ξϕ+ 2ξϕ) eϕ

→ a = −R(ω2 +Ω2) sinωt er + 2RωΩcosωt eϕ .

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The paths of point P are displayed for several values of Ω/ω in

the following figures.

Ω/ω = 0.5Ω/ω = 0.25

Ω/ω = 2.0Ω/ω = 1.0

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E1.35Example 1.35 A chain with length l and

mass m hangs over the edge of a friction-

less table by an amount e.

If the chain starts with zero initial

velocity, find the position of the end of

the chain as a function of time.

e

Fig. 1.60

Solution All the links of the chain have the same displacement,

velocity and acceleration. The corner only produces a change of

direction. We therefore consider the chain to be a single mass with

an applied force that depends on the length x of the overhanging

part. Thus, with a = x, the equation of

motion is

ma = mx

lg → x− g

lx = 0 .

This differential equation of second order

with constant coefficients has the soluti-

on

mx

lg

x

x(t) = A cosh

√g

lt+B sinh

√g

lt .

We calculate the integration constants from the initial conditions:

x(0) = 0 → B = 0

x(0) = e → A = e

→ x(t) = e cosh

√g

lt .

This solution is valid only for x ≤ l.We may also solve the problem by

making an imaginary section cut at

the corner. Then the equations of

motion for each part of the chain are

d

dt

(ml− xl

x)= S ,

d

dt

(mlxx)=m

lxg − S .

If we eliminate S we again obtain the

differential equation

x− g

lx = 0 .

x

S

S

mx− ll

mx

l

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E1.36 Example 1.36 A rotating source of

light throws a beam of light onto a

screen (Fig. 1.61). Point P on the

screen should move with constant

velocity v0.

Determine the required angu-

lar acceleration ϕ(ϕ) of the source

of light. Sketch ϕ(ϕ) and ϕ(ϕ).

x

v0

r0

ϕ

P

Fig. 1.61

Solution The position of point P is given by

x = r0 tanϕ .

We write down the inverse function

ϕ = arctanx

r0.

Differentiation with x = v = v0 = const yields

ϕ =1

1 +(xr0

)2x

r0=

v0r0r20 + x2

=v0

r0(1 + tan2 ϕ)=v0r0

cos2 ϕ ,

ϕ =v0r0

2 cosϕ(− sinϕ)ϕ = −2(v0r0

)2sinϕ cos3 ϕ .

v0r

ϕ

ϕ−π2

π

2

ϕ

ϕ−π2

π

2

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Note: the maximum value of ϕ, located at ϕ = ±30, is obtainedas

|ϕmax| =3

8

√3(v0r0

)2.

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E1.37 Example 1.37 A car travels with velocity v0 = 100 km/h. At time

t = 0, the driver fully applies the brakes. At that moment the car

starts sliding on a rough road (coefficient of kinetic friction µ).

Use the simplest model for the car (point mass) and calculate

the time t∗ and the distance x∗ until the car comes to a stop

a) on a dry road (µ = 0.8),

b) on a wet road (µ = 0.35).

Solution The equation of motion in the horizontal direction

← : ma = mx = −R ,

the equilibrium condition in the

vertical direction

↑ : 0 = N −mg

x

R

N

mg

and the law of friction

R = µN

lead to

a = −µg .

Integration with v(t=0) = v0 and x(t=0) = 0 yields

v(t) = v0 − µgt , x(t) = v0t−1

2µgt2 .

The time t∗ follows from the condition v = 0:

t∗ =v0µg

.

This leads to

x∗ = x(t∗) =v20µg− v20

2µg=

v202µg

.

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a) On a dry road (µ = 0.8) we obtain

t∗ =100 · 1000

3600 · 0.8 · 9.81 = 3.55 s ,

x∗ =

(100

3.6

)2 1

2 · 0.8 · 9.81 = 49 m .

b) On a wet road (µ = 0.35) we are led to

t∗ =100 · 1000

3600 · 0.35 · 9.81 = 8.1 s ,

x∗ =(1003.6

)2 1

2 · 0.35 · 9.81 = 112 m .

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E1.38 Example 1.38 In order to determine the

coefficient of restitution e experimental-

ly, a ball is dropped from height h0 onto

a horizontal rigid surface (Fig. 1.62). Af-

ter the ball has hit the surface 7 times, it

reaches only 20% of the original height

h0.

Calculate the coefficient of restituti-

on.

h0

h7=0.2h0

Fig. 1.62

Solution With the positive direction of

the velocity taken as upwards, the velo-

city of the ball immediately before the

first impact is given by

v = −√2gh0 .

The definition

e = −vv

h1h0

h2

h7

of the coefficient of restitution in terms of the velocities yields the

velocity v immediately after the impact:

v = −ev = e√2gh0 .

From the Conservation of Energy Law we obtain the height which

the body will reach:

1

2mv2 = mgh1 → h1 =

v2

2g= e2h0 .

Similarly, we obtain

hi = e2hi−1

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for the subsequent impacts. Thus,

h7 = e2h6 = e4h5 = . . . = e14h0

and

e =

(h7h0

)1/14

= 0.21/14 = 0.891 .

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2Chapter 2

Dynamics of Systems of PointMasses

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44 2 Dynamics of Systems of Point Masses

E2.10 Example 2.10 Two vehicles (masses m1 and m2, velocities v1 and

v2) crash head-on, see Fig. 2.20. After a plastic impact the vehicles

are entangled and slide with locked wheels a distance s to the right.

The coefficient of kinetic

friction between the wheels

and the road is µ.

Calculate v1 if v2 and s

are known.

v1 v2

m1 m2

Fig. 2.20

Solution The total momentum of the system remains unchanged

during the collision. We assume that positive velocities are direc-

ted to the right (note the direction of v2). We then obtain the

velocity v of the vehicles after the impact:

m1v1 −m2v2 = (m1 +m2)v → v =m1v1 −m2v2m1 +m2

.

In order to relate the distance s to the velocity v we apply the

work-energy theorem

T1 − T0 = U .

We insert the kinetic energies

T0 = (m1 +m2)v2/2 , T1 = 0 ,

the work of the friction force

U = −Rs

and Coulomb’s friction law

R = µN → R = µ(m1 +m2)g .

We then obtain

−1

2(m1 +m2)v

2 = −µ(m1 +m2)gs

→ v1 =m2

m1v2 +

(1 +

m2

m1

)√2µgs .

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2 Dynamics of Systems of Point Masses 45

E2.11Example 2.11 A block (mass m2) rests on a horizontal platform

(mass m1) which is also initially at rest (Fig. 2.21). A constant

force F accelerates the

platform (wheels rolling

without friction) which

causes the block to slide

on the rough surface of

the platform (coefficient of

kinetic friction µ).

m2F

m1

l

µ

Fig. 2.21

Determine the time t∗ that it takes the block to fall off the

platform.

Solution We separate the

block and the platform and

introduce the coordinates

x1 and x2 as shown in the

free-body diagram. Then

the equations of motion for

the two bodies are

① → : m1x1 = F −R ,

② → : m2x2 = R .

With the friction law R =

µN = µm2g, we obtain

N

1

m1g

F

BA

x2

x1

m2g

R

R

2

x1 =F − µm2g

m1, x2 = µg .

Now, we integrate twice. Using the initial conditions x1(0) =

x2(0) = 0 and x1(0) = x2(0) = 0 we obtain

x1 =F − µm2g

m1t , x2 = µgt ,

x1 =F − µm2g

m1

t2

2, x2 = µg

t2

2.

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The block falls off the platform if

x1 − x2 = l .

This yields

F − µm2g

m1

(t∗)2

2− µg (t

∗)2

2= l → t∗ =

√2lm1

F − µg(m1 +m2).

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E2.12Example 2.12 A railroad wagon (mass m1) has a velocity v1(Fig. 2.22). It collides with a wagon (mass m2) which is initi-

ally at rest. Both wagons roll without friction after the collision.

The second wagon is connected via a spring (spring constant k)

with a block (mass m3) that lies on a rough surface (coefficient of

static friction µ0).

Assume the impact to be plastic and determine the maximum

value of v1 so that the block stays at rest.

v1

m3

km1 m2 µ0

Fig. 2.22

Solution The total momentum of the system remains unchanged

during the impact. This yields the velocity v of the wagons after

the collision:

m1v1 = (m1 +m2)v → v =m1

m1 +m2v1 .

xm1 m2

m3 F

H

m3g

N

We assume that the block stays at rest. Then we can write down

the equilibrium conditions

→: H = F , ↑: N = m3g .

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The force of static friction H and the normal force N have to

satisfy the condition of static friction:

H ≤ µ0N → F ≤ µ0m3g .

With

F = kx → x ≤ µ0m3g

k

we obtain the maximum compression of the spring:

xmax =µ0m3g

k.

Now we apply the Conservation of Energy Law to determine v1max:

1

2(m1 +m2)v

2 =1

2kx2max → v1max =

µ0m3g

m1

√m1 +m2

k.

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E2.13Example 2.13 A point mass m1 strikes a point mass m2 which is

suspended from a string (length l, negli-

gible mass) as shown in Fig. 2.23. The

maximum force S∗ that the string can

sustain is given.

Assume an elastic impact and de-

termine the velocity v0 that causes the

string to break.

l

m1 m2

v0

Fig. 2.23

Solution First we formulate the con-

servation of linear momentum

m1v0 = m1v1 +m2v2

and of energy (elastic impact!):

m1v20/2 = m1v

21/2 +m2v

22/2 .

From these equations we can calcu-

late the velocity v2 of mass m2 im-

mediately after the impact:

S

m2g

ϕ

v2 =2m1

m1 +m2v0 .

Now, we write down the equation of motion

տ: m2an = S −m2g cosϕ .

With an = v22/l we obtain the force in the string:

S = m2v22/l +m2g cosϕ .

Gro

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The maximum force Smax in the string is found for ϕ = 0:

Smax = m2(v22/l + g) .

The string breaks if the maximum force Smax is larger than the

allowable force S∗:

Smax > S∗ .

This yields

v0 >m1 +m2

2m1

√l(S∗/m2 − g) .

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E2.14Example 2.14 A ball ① (mass m1) hits a second ball ② (mass

m2, velocity v2 = 0) with a velocity v1 as shown in Fig. 2.24.

Assume that the impact

is partially elastic (coeffi-

cient of restitution e) and

all surfaces are smooth.

Given: r2 = 3 r1, m2 =

4m1.

Determine the veloci-

ties of the balls after the

collision.

1v1

m2

m1

r22

r1

Fig. 2.24

Solution We introduce the auxiliary angle α. Since the surfaces

are smooth, the linear impulse F acts in the direction of the line

of impact and the Impulse Laws are given by

① → : m1(v1 − v1) = −F cosα ,

② ր : m2v2 = F .

Note that the weights of

the balls can be neglected

during impact and that

ball ① moves only horizon-

tally.

With the hypothesis

e = −v1x − v2xv1x − v2x

and

1α

N

F

F2

v1, v1α

αr2

r1

x

v2

r2−r1

v1x = v1 cosα , v2x = 0 , v1x = v1 cosα , v2x = v2

we obtain

v1 = v1

1− e m2

m1cos2 α

1 +m2

m1cos2 α

, v2 = v1(1 + e) cosα

1 +m2

m1cos2 α

.

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Introduction of m2 = 4m1 and

sinα =r2 − r1r1 + r2

=1

2→ cosα =

√1− sin2 α =

√3

2

yields

v1 =1− 3e

4v1 , v2 =

√3

8(1 + e) v1 .

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E2.15Example 2.15 A hunter (mass m1) sits in a boat (mass m2 =

2m1) which can move in the water without resistance. The boat

is initially at rest.

a) Determine the velocity vB1of the boat after the hunter fires a

bullet (mass m3 = m1/1000) with a velocity v0 = 500 m/s.

b) Find the direction of the velocity of the boat after a second

shot is fired at an angle of 45 with respect to the first one.

Solution a) We introduce a coordinate system and assume that

the bullet is fired in the direction of the negative x-axis. Since the

linear momentum before the firing of the bullet is zero, the total

linear momentum of the system after the firing (velocities positive

to the right) also has to be zero:

→ : (m1 +m2 −m3)vB1−m3v0 = 0 .

This yields the velocity vB1of the boat:

vB1=

m3

m1 +m2 −m3v0 =

1

1000(1 + 2− 1

1000

) 500 =500

2999

≈ 0.167m

s.

The algebraic sign shows that the boat moves in the direction of

the positive x-axis.

b) Now, we have to formu-

late the Impulse Laws in

the x- and y-direction:45

m3

v0

wB2

m1 +m2α

vB2

y

x

→ : (m1 +m2 −m3)vB1= (m1 +m2 − 2m3)vB2

−m3v0 cos 45 ,

↑ : 0 = (m1 +m2 − 2m3)wB2−m3v0 sin 45

.

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They lead to

vB2=

m1 +m2 −m3

m1 +m2 − 2m3vB1

+m3

m1 +m2 − 2m3v0 cos 45

≈ 0.285m

s,

wB2=

m3

m1 +m2 − 2m3v0 sin 45

≈ 0.118m

s,

→ tanα =wB2

vB2

= 0.414 → α = 22.5 .

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E2.16Example 2.16 A car ② goes into a skid on a wet road and co-

mes to a stop sideways across the road as shown in Fig. 2.25.

In spite of having fully applied the brakes

a distance s1 from car ②, a second car

① (sliding with the coefficient of kinetic

friction µ) collides with car ②. This cau-

ses car ② to slide an additional distance

s2. Assume a partially elastic central im-

pact. Given: m1 = 2m2, µ = 1/3, e =

0.2, s1 = 50m, s2 = 10m.

Determine the velocity v0 of car ① be-

fore the brakes were applied.

s1

2m2

v01m1

s2

Fig. 2.25

Solution The velocity v1 of car ① immediately before impact fol-

lows from the work-energy theorem:

1

2m1v

21 −

1

2m1v

20 = −(µm1g)s1 → v21 = v20 − 2µ gs1 .

The velocity v2 of car ② immediately after impact can be deter-

mined from the conservation of linear momentum (note v2 = 0)

m1v1 = m1v1 +m2v2

and the hypothesis

e =v2 − v1v1

.

We obtain

v2 =1 + e

1 +m2

m1

v1 .

Now we apply the work-energy theorem to the sliding of car ②:

−1

2m2v

22 = −(µm2g)s2 → v22 = 2µ gs2 .

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If we eliminate v1 and v2 we obtain

v20 =

(1 +m2

m1

1 + e

)2

2µ gs2 + 2µ gs1

=(1.51.2

)2· 23· 9.81 · 10 + 2

3· 9.81 · 50 = 429.19m2/s2

or

v0 = 20.7m/s → v0 = 74.6 km/h .

Note: the velocity of car ① at impact is v1 = 36.4 km/h.

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E2.17Example 2.17 Two cars (point masses m1 and m2) collide at an

intersection with velocities

v1 and v2 at an angle α

(Fig. 2.26). Assume a per-

fectly plastic collision.

Determine the magni-

tude and the direction of

the velocity immediately

after the impact. Calcula-

te the loss of energy during

the collision.

βm1

m2

α

v2

v1

Fig. 2.26

Solution Both cars have the

same velocity v after the

collision (plastic impact).

We write down the Impul-

se Laws in the x- and y-

directions: m2

α

v2

v1m1

y

x

m1+m2

β

v

→ : m1v1 +m2v2 cosα= (m1 +m2)v cosβ ,

↑ : m2v2 sinα = (m1 +m2)v sinβ .

If we square and then add these equations we obtain

v =1

m1 +m2

√(m1v1)2 + 2m1m2v1v2 cosα+ (m2v2)2 ,

whereas a division leads to

tanβ =m2v2 sinα

m1v1 +m2v2 cosα.

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The loss of mechanical energy during impact is given by the dif-

ference ∆T of the kinetic energies before and after impact:

∆T =m1v

21

2+m2v

22

2− (m1 +m2)v

2

2

=m1m2

2(m1 +m2)(v21 + v22 − 2v1v2 cosα).

Note that the loss of energy is a maximum if the cars collide head-

on (α = π). The energy ∆T causes the deformation of the cars.

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E2.18Example 2.18 A bullet (mass m)

has a velocity v0 (Fig. 2.27). An

explosion causes the bullet to

break into two parts (point mas-

ses m1 and m2). The directions

α1 and α2 of the two parts and

the velocity v1 immediately after

the explosion are given.

Calculate m1 and v2. Deter-

mine the trajectory of the center

of mass of the two parts.

v2m2

m

m1

v1

α2

y

xv0

α1 =π

2

Fig. 2.27

Solution Since no external forces are acting on the system, its

linear momentum does not change (linear momentum before the

explosion = linear momentum after the explosion). Component-

wise, the principle of conservation of linear momentum gives

→ : mv0 = m2v2 cosα2 ,

↑ : 0 = m1v1 −m2v2 sinα2 .

Thus, we have two equations for the two unknowns m1 and v2.

Solving with m = m1 +m2 yields

m1 = mv0v1

tanα2 , v2 =v0

cosα2 − v0v1

sinα2.

We measure the time t from the moment of the explosion. Then,

with |yi| = vit sinαi, the location of the center of mass is given by

yc =1

m(m1y1 +m2y2)

=1

m

mv0v1

tanα2 v1t+(m−m v0

v1tanα2

) (−v0 sinα2)

cosα2 − v0v1

sinα2t

= 0 .

The center of mass of the system continues to move on the original

straight line y = 0 (law of motion for the center of mass).

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E2.19 Example 2.19 A ball (point mass m1) is attached to a cable. It is

released from rest at a height h1 (Fig. 2.28). After falling to the

vertical position at A it collides with

a second ball (point mass m2 = 2m1)

which is also initially at rest. The co-

efficient of restitution is e = 0.8.

Determine the height h2 which the

first ball can reach after the collision

and the velocity of the second ball im-

mediately after impact.

m2

h1

m1

A

Fig. 2.28

Solution The velocities of the massesm1 andm2 before the impact

are

v1 =√2gh1 , v2 = 0 .

The Impulse Laws

① → : m1(v1 − v1) = −F ,

② → : m2v2 = +F ,

2v2v1, v1

1

Fm1 m2F

and the hypothesis

e = −v1 − v2v1

yield the velocities immediately after the impact:

v1 = v1m1 − em2

m1 +m2= v1

1− 1.6

1 + 2= −0.2

√2gh1 ,

v2 = v1(1 + e)m1

m1 +m2= v1 1.8

1

1 + 2= 0.6

√2gh1 .

The height h2 follows from

the conservation of energy

1

2m1v

21 = m1gh2

as

h2 =v212g

= 0.04 h1 .

h2m2v1 m1 v2

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E2.20Example 2.20 The rigid rod (negli-

gible mass) in Fig. 2.29 carries two

point masses. It is struck by an im-

pulsive force F at a distance a from

the support A.

Determine the angular veloci-

ty of the rod immediately after the

impact and the impulsive reaction

at A. Calculate a so that the reac-

tion force at A is zero.m

m

a

Fl

l

A

Fig. 2.29

Solution We introduce a coordinate

system. The y-component of the ve-

locity of the center of mass is zero

after the impact. Since the impulsi-

ve force F also has no y-component,

the impulsive reaction at A has on-

ly an x-component, i.e. Ay = 0. We

apply the principle of linear impulse

and momentum

y

3l/2

C

Ax

Ay

Fx

→: 2m(vc − vc) = Ax − F

and the principle of angular impulse and momentum (compare

Chapter 3)

y

C : 2

(l

2

)2

m(ω − ω) = Ax3

2l − F

(3

2l − a

).

Inserting the kinematic relation

vc = −3lω/2

and the velocities

vc = 0 , ω = 0

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before the impact we obtain

ω =F a

5l2m, Ax =

(1− 3a

5l

)F .

The reaction force at A is zero (Ax = 0) if the distance a is chosen

as

a = 5l/3 .

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E2.21Example 2.21 A wagon (weight W1 = m1g) hits a buffer (spring

constant k) with velocity v. The

collision causes a block (weight

W2 = m2g) to slide on the

rough surface (coefficient of sta-

tic friction µ0) of the wagon

(Fig. 2.30).

Determine a lower bound for

the wagon’s speed before the

collision.

k

µ0

W1

v

W2

Fig. 2.30

Solution We separate the wagon and the block. The orientation

of the friction force H is assumed arbitrarily in the free-body

diagram. Both bodies have the same acceleration (x1= x2= x) as

long as the block does not sli-

de. Therefore, the equations of

motion are

① ← : m1x = −H − F ,

② ← : m2x = H .

m2g

x

H

N

m1g

1

2

F

H

With F = k x we obtain

H = − m2

m1 +m2k x .

The maximum friction force Hmax is obtained at maximal com-

pression xmax of the spring which follows from the Conservation

of Energy Law:

1

2(m1 +m2)v

2 =1

2k x2max → xmax =

√m1 +m2

kv .

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The block is on the verge of slipping if the condition of “limiting

friction”

|Hmax| = µ0N = µ0m2g → k

m1 +m2xmax = µ0g

is satisfied. This yields the velocity that is necessary to cause

slippage of the block:

v = µ0g

√m1 +m2

k.

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E2.22Example 2.22 The system shown in Fig. 2.31 consists of two blocks

(massesm1 and m2), a spring (spring constant k), a massless rope

and two massless pulleys. Block 1 lies on a rough surface (coeffi-

cient of kinetic friction µ).

At the beginning of the

motion (t = 0), the spring

is unstretched and the po-

sition of block 1 is given

by x1 = 0.

Determine the velocity

x1 as a function of the po-

sition x1.

x2

x1

m2

µ

g

k m1

Fig. 2.31

Solution Since we want to find the velocity as a function of the

position, we use the work-energy theorem

T1 − T0 = U .

In moving from the initial position to an arbitrary position, the

gravitational force, the force in the spring and the friction force

perform work:

UW = m2gx2 , Uk = −1

2kx21 ,

UR = −Rx1 = −µNx1 = −µm1gx1 .

The kinetic energies are given by

T1 =1

2m1x

21 +

1

2m2x

22 , T0 = 0 .

With the kinematic relation

x2 = 2x1

the work-energy theorem yields

1

2m1x

21 +

1

2m24x

21 = m2gx2 −

1

2kx21 − µm1gx1

→ x1 =

√2

5(2− µ)gx1 −

1

5

k

mx21 .

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3Chapter 3

Dynamics of Rigid Bodies

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68 3 Dynamics of Rigid Bodies

E3.24 Example 3.24 Point A of the rod

in Fig. 3.47 moves with constant

velocity vA to the left.

Determine the velocity and

the acceleration of point B of the

rod (point of contact with the

step) as a function of the angle

ϕ. Find the path y(x) of the in-

stantaneous center of rotation.

x

eϕ

y

ϕ

A

B

vA

er

h

Fig. 3.47

Solution We use the coordinate system with the basis vectors erand eϕ as shown in the figure. Then, the velocity of point A is

given by

vA = vA cosϕer − vA sinϕeϕ .

The velocity vB of point B

points in the direction of the

rod (the rod does not lift off the

step). Thus,

vB = vBer .


eϕ

aϕAB

ϕ

aB

A

er

vB

vAB

arAB

vA

vB = vA + vAB where vAB = aϕeϕ

and with

a =h

sinϕ,

we obtain

vBer = vA cosϕer − vA sinϕeϕ +hϕ

sinϕeϕ

→ vB = vA cosϕ and ϕ =vAh

sin2 ϕ .

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3 Dynamics of Rigid Bodies 69

In order to determine the acceleration of point B we use the rela-

tion

aB = aA + arAB + a

ϕAB ,

where

aA = 0 , arAB = −aϕ2

er , aϕAB = aϕeϕ .

The angular acceleration of the rod is found by differentiating its

angular velocity ϕ:

ϕ =vAh(sin2 ϕ)˙ → ϕ = 2

v2Ah2

sin3 ϕ cosϕ .

Hence, we obtain

aB =v2Ah

sin2 ϕ(− sinϕer + 2 cosϕeϕ) .

The instantaneous center of rotation Π is given by the point of

intersection of two straight lines which are perpendicular to two

velocities. In the present example, the directions of the velocities

vA and vB are known. The-

refore, the instantaneous cen-

ter of rotation can easily be

constructed (see the figure).

Point Π has the coordinates

x = h cotϕ , y = h+ x cotϕ .

Elimination of the angle ϕ

yields the path of Π:

y =x2

h+ h .

y/h

path of Π

Π1.5

2.0

1.0

0.5

00.5-1.0 0 0.5 1.0

x/h

ϕ

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E3.25 Example 3.25 The wheel of a

crank drive rotates with

constant angular velocity ω

(Fig. 3.48).

Determine the velocity and

the acceleration of the pi-

ston P .

l

Pψ

rω ϕ

Fig. 3.48

Solution If we introduce the

auxiliary angle ψ, then the

position xP of the piston is

given by

xP = r cosϕ+ l cosψ .

Now we differentia-

te and obtain (note

ϕ = ω = const)

xP = −rω sinϕ− lψ sinψ ,

xP = −rω2 cosϕ− lψ sinψ − lψ2 cosψ .

P

lψ

x

y

rϕ = ωt

The quantities sinψ, cosψ, ψ and ψ are as yet unknown. They

follow from the condition that the piston can move only in the

horizontal direction. Therefore, its vertical displacement is zero:

yP = 0 = r sinϕ− l sinψ .

Differentiation yields

yP = 0 = rω cosϕ− lψ cosψ ,

yP = 0 = −rω2 sinϕ− lψ cosψ + lψ2 sinψ .

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Thus,

sinψ =r

lsinϕ , cosψ =

√1−

(rlsinϕ

)2,

ψ = ωr

l

cosϕ

cosψ, ψ = −ω2 r

l

sinϕ

cosψ+ ψ2 sinψ

cosψ.

Hence, we finally obtain

xP = −rωsinϕ+

r

l

sinϕ cosϕ

cosψ

,

xP = −rω2

cosϕ− r

l

[sin2 ϕ

cosψ− cos2 ϕ

cos3 ψ

].

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E3.26 Example 3.26 LinkMA of the mechanism in Fig. 3.49 rotates with

angular velocity ϕ(t).

Determine the velocities of points B and C, the angular velo-

city ω and the angular acceleration ω of the angled member ABC

at the instant shown.

A C

MB

z x

y

r

l

l

α

ϕ

Fig. 3.49

Solution With the given x, y, z-coordinate system, the velocities

of points A and B at the instant shown can be written in the form

vA = rϕ

− sinϕ

cosϕ

0

, vB = vB

− cosα

sinα

0

,

where vB is as yet unknown (note that the direction of vB is

known). The angular velocity of the angled member ABC is re-

presented by

ω =

0

0

ω

.

We now use the kinematic relation

vB = vA + ω × rAB

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which, with rAB = (0,−l, 0)T , in coordinates reads

vB

− cosα

sinα

0

= rϕ

− sinϕ

cosϕ

0

+

ωl

0

0

.

Solving yields

vB = rϕcosϕ

sinα, ω =

rϕ

l

(sinϕ− cosϕ

tanα

).

In an analogous way we obtain the velocity of point C:

vC = vA + ω × rAC → vC = rϕ

− sinϕ

sinϕ+ cosϕ− cosϕ

tanα0

.

In order to determine the angular acceleration ω = (0, 0, ω)T of

the angled member, we first write down the relation

aB = aA + ω × rAB + ω × (ω × rAB) ,

which in coordinates reads

aB

− cosα

sinα

0

= rϕ

− sinϕ

cosϕ

0

− rϕ

2

cosϕ

sinϕ

0

+

ωl

0

0

+

0

ω2l

0

.

Solving for aB and ω yields

aB =1

sinα(rϕ cosϕ− rϕ2 sinϕ+ lω2) ,

ω =1

l

[rϕ sinϕ+ rϕ2 cosϕ− 1

tanα(rϕ cosϕ− rϕ2 sinϕ+ lω2)

].

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E3.27 Example 3.27 Wheel ① rolls in

a gear mechanism without slip

along circle ②. The mechanism

is driven with a constant angu-

lar velocity Ω (Fig. 3.50).

Determine the magnitudes

of the velocity and the accele-

ration of point P on the wheel.

1

RΩ

ϕ

P

2

r

Fig. 3.50

Solution The motion of P is composed of the rotation of B about

A with

vB = RΩ (vB ⊥ AB) , aB = RΩ2 (aB ‖ AB)

and the rotation of P about B. In order to determine the angular

velocity ω of wheel ① we consider two positions of the wheel. The

figure shows that the relation

(R+ r)α = r(β + α)

→ Rα = rβ

for the arclengths holds. Diffe-

rentiating and using α = Ω and

β = ω yields

RΩ = rω → ω = ΩR

r.

R

A

r

C

B′

β

C ′

αB

α

Note that Ω is positive counterclockwise, whereas ω is positive

clockwise.

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We can now construct the velocity

diagram and the acceleration dia-

gram. Since the velocity diagram

is represented by an isosceles tri-

angle, we obtain

vP = 2ΩR sinϕ

2.

vP

ϕ = Rω

vPB = rω

vB = RΩ

Finally, the law of cosines yields

a2P = (Ω2R)2 +

(Ω2R2

r

)2

−2Ω2RΩ2R2

rcos(π − ϕ) ,

aP = Ω2R

√

1 +

(R

r

)2

+ 2R

rcosϕ .

anPB = rω2ϕ

aB = RΩ2

aP

=Ω2R2

r

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E3.28 Example 3.28 A disk 1 (mass m, radius r1) rests in a frictionless

support (ω0 = 0). A second disk

2 (mass m, radius r2) rotates

with the angular velocity ω2. It

is placed on disk 1 as shown in

Fig. 3.51. Due to friction, both

disks eventually rotate with the

same angular velocity ω.

Determine ω. Calculate the

change ∆T of the kinetic energy.

1

ω2

2

r2

r1

mm

Fig. 3.51

Solution Since there is no moment of external forces acting on the

system, the angular momentum is conserved during the process:

Θ2ω2 = (Θ1 +Θ2)ω .

With the mass moments of inertia

Θ1 = mr21/2 , Θ2 = mr22/2

we obtain the resulting angular velocity ω:

ω =r22

r21 + r22ω2 .

The change ∆T of the kinetic energy follows as

∆T =1

2(Θ1 +Θ2)ω

2 − 1

2Θ2ω

22 → ∆T = −1

4mω2

2

r21r22

r21 + r22.

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E3.29Example 3.29 The door (mass m, moment

of inertia ΘA) of a car is open (Fig. 3.52).

Its center of mass C has a distance b from

the frictionless hinges. The car starts to

move with the constant acceleration a0.

Determine the angular velocity of the

door when it slams shut.

AC

a0

b

m

Fig. 3.52

Solution First, we write down

the acceleration of the center

of mass C:

aC = aA + arAC + a

ϕAC ,

where the magnitudes of the

individual terms are given by

At

Cϕ

AnA

aA = a0 , arAC = bϕ2 , aϕAC = bϕ .

We then formulate the principle of linear momentum (in the t-

direction, see the free-body diagram)

ւ: m(bϕ− a0 cosϕ) = At

and the principle of angular momentum (about the center of mass

C):

y

C : ΘC ϕ = −Atb .

The mass moment of inertia ΘC follows from the parallel-axis

theorem:

ΘC = ΘA −mb2 .

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Eliminating the force At yields

ϕ =ma0b

ΘAcosϕ .

The angular velocity of the door can be obtained through integra-

tion:

1

2ϕ2(ϕ) =

ma0b

ΘAsinϕ .

Thus, for ϕ = π/2 (closed door) we find

ϕ(π/2) =

√2ma0b

ΘA.

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E3.30Example 3.30 A child (mass m) runs along the rim of a circular

platform (mass M , radius r) starting from point A (Fig. 3.53).

The platform is initially at rest; its support is frictionless.

Determine the angle

of rotation of the plat-

form when the child ar-

rives again at point A.

0m

r

M

A

Fig. 3.53

Solution We apply the principle of angular momentum: the time

rate of change of the angular momentum is equal to the moment of

the applied forces. Since there are no external applied forces acting

in the present example, the angular momentum is constant:

dL(0)

dt= 0 → L(0) = const

with

L(0) = Θ0ω +mr2(ω + ωrel) .

Here,

Θ0 =Mr2/2

is the mass moment of inertia of the platform, ω is the angular

velocity of the platform and ωrel is the angular velocity of the child

relative to the platform. Integration of the principle of angular

momentum with the initial condition

L(0)(0) = 0 → L(0) ≡ 0

yields∫ t

0

L(0)dt = 0 → Θ0ϕ+mr2(ϕ+ ϕrel) = 0 .

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The child arrives again at

point A when the relati-

ve angle attains the value

ϕrel = 2π. Solving for ϕ

yields

ϕ = − 2π

M

2m+ 1

.

The result is displayed in

a diagram.

ϕ

2π

π

0 2 4 6 8 10

M/m

ϕrel= 2π

|ϕ|ϕchild =ϕ+ϕrel

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E3.31Example 3.31 A homogeneous

triangular plate of weight W =

mg is suspended from three

strings with negligible mass.

Determine the acceleration of

the plate and the forces in the

strings just after string 3 is cut.

l3

21

2l

αα

W

Fig. 3.54

Solution The motion of the plate is a translation after string 3 is

cut. Therefore, all the points of the plate have the same accele-

ration (and the same velocity). We choose point A which rotates

about a fixed point to represent the motion of the plate. Its acce-

leration can be written in the Serret-Frenet frame as

a = aA = vet +v2

ρAen .

Its velocity is zero at the mo-

ment of the cut:

v = 0 → a = vet .

The principles of linear and an-

gular momentum are then given

by (see the free-body diagram)

etC

mg

A

2

3l

en

S2

α

1

3l

S1

ւ : mv = mg sinα ,

տ : 0 = S1 + S2 −mg cosα ,y

C : 0 =2

3lS1 cosα−

l

3S1 sinα−

l

3S2 sinα−

4

3S2 cosα .

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Solving these equations yields

v = g sinα,

S1 =mg

6(sinα+ 4 cosα), S2 =

mg

6(− sinα+ 2 cosα) .

Note that S2 = 0 for sinα = 2 cosα (i.e., for tanα = 2 →α = 63.4). In this case the line of action of S1 passes through the

center of mass C.

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E3.32Example 3.32 A sphere (mass

m1, radius r) and a cylindrical

wheel (mass m2, radius r) are

connected by two bars (mass

of each bar m3/2, length l).

They roll down a rough incli-

ned plane (with angle α) wi-

thout slipping (Fig. 3.55).

Find the acceleration of the

bars.α

r

r

l

m1

m3

m2

xϕ

Fig. 3.55

Solution The only external forces that act on the system are the

weights of the individual parts. Therefore, conservation of energy

T + V = const

leads to the solution. The kinetic energy is given by

T = (m1 +m2 +m3)x2/2 + (Θ1 +Θ2)ϕ

2/2

and the potential energy is

V = −(m1 +m2 +m3)gx sinα .

With the kinematic relation (rolling without slipping)

x = rϕ

and the mass moments of inertia

Θ1 = 2m1r2/5 , Θ2 = m2r

2/2

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we obtain

(7m1/10 + 3m2/4 +m3/2)x2 − (m1 +m2 +m3)gx sinα = const .

Differentiation finally yields the acceleration of the bars:

2(7m1/10 + 3m2/4 +m3/2)xx− (m1 +m2 +m3)gx sinα = 0

→ a = x =(m1 +m2 +m3) sinα

7m1/5 + 3m2/2 +m3g .

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E3.33Example 3.33 The cylindrical

shaft shown in Fig. 3.56 has

an inhomogeneous mass den-

sity given by ρ = ρ0(1 + αr).

Find the moments of iner-

tia Θx and Θy.

xRr

y

y z

l

Fig. 3.56

Solution We determine the

mass moment of inertia

Θx from

Θx =

∫(y2 + z2)dm =

∫r2dm .

r

y

dr

R

z

With

dm = ρ 2πr dr dx = 2πρ0(r + αr2)dr dx

we obtain

Θx = 2πρ0l∫0

R∫0

(r3 + αr4)dr dx = 2πρ0l(R4

4+ α

R5

5

)

=π

2ρ0lR

4(1 +

4

5αR).

The mass moment of inertia Θy = Θz (symmetry) follows from

Θy =

∫(z2 + x2)dm

with

dm = ρr dϕ dr dx

= ρ0(r + αr2) dϕ dr dx ,

z = r sinϕ .

r

rdϕz

yϕ

dϕ

dr

r sinϕ

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Using the symmetry we obtain

Θy = 4ρ0

l∫

0

R∫

0

π/2∫

0

(r2 sin2 ϕ+ x2)(r + αr2)dϕ dr dx

= πρ0

l∫

0

R∫

0

(r2 + 2x2)(r + αr2)dr dx

=πρ0R2l

[R2

4+l2

3+ α

(R3

5+

2

9Rl2

)].

Note: For α = 0 the results reduce with m = ρ0πR2l to the mass

moments of inertia of a homogeneous cylindrical shaft (note the

term due to the parallel-axis theorem):

Θx =1

2mR2 , Θy =

1

12m(3R2 + 4l2

).

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E3.34Example 3.34 Determine the

moment of inertia Θa of a ho-

mogeneous torus with a circular

cross section and mass m.

c

Ra

a

Fig. 3.57

Solution The mass moment of inertia Θa is determined from

Θa =

∫r2dm .

We can determine the geo-

metrical relations

r = R+ c sinϕ ,

dm = ρ 2πr2c cosϕdr

from the figure. With

dr = c cosϕdϕ

2c cosϕR c

r

ϕ

c sinϕ

a

a dr

we obtain

dm = 4πρc2(R + c sinϕ) cos2 ϕdϕ .

Thus,

Θa = 4πρc2+π/2∫

−π/2

(R+ c sinϕ)3 cos2 ϕdϕ

= 4πρc2+π/2∫

−π/2

[R3 cos2 ϕ + 3R2c sinϕ cos2 ϕ

+ 3Rc2 sin2 ϕ cos2 ϕ+ c3 sin3 ϕ cos2 ϕ]dϕ

= 4πρc2(π

2R3 +

3π

8Rc2

)= 2π2ρc2R

(R2 +

3

4c2).

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Note that the integrals of the odd functions over the even interval

are zero. Using m = 2π2ρc2R we finally get

Θa = m(R2 +

3

4c2).

This result reduces to Θa = mR2 in the case of a thin ring

(c≪ R).

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E3.35Example 3.35 A rope drum on a

rough surface is set into motion

by pulling the rope with a con-

stant force F0.

Determine the acceleration of

point C assuming that the drum

rolls (no slipping). What coeffi-

cient of static friction µ0 is ne-

cessary to ensure rolling? µ0

m,Θc

F0

Cr1

r2

α

Fig. 3.58

Solution The free-body diagram shows the forces that act on the

drum. We apply the principles of linear and angular momentum:

→ : mac = F0 cosα−H ,

↑ : 0 = N −mg + F0 sinα ,

y

C : Θcω = r1H − r2F0 .

In addition we have the kinema-

tic relation

vc = r1ω → vc = ac = r1ω

F0

α

r1r2vc

HN

ω

Cmg

between the velocity and the angular velocity (rolling without

slip). We solve these four equations to obtain the acceleration and

the forces:

ac =F0

m

cosα− r2r1

1 +Θcr21m

,

H = F0

Θc cosα

r21m+r2r1

1 +Θcr21m

, N = mg − F0 sinα .

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In order to ensure rolling, the condition of static friction H ≤ µ0N

has to be satisfied. This leads to

µ0 ≥

Θc cosα

r21m+r2r1(

mg

F0− sinα

)(1 +

Θcr21m

) .

Note: For cosα > r2/r1 we have ac > 0 (motion to the right),

whereas for cosα < r2/r1 we obtain ac < 0 (motion to the left).

In the case of F0 sinα > mg the drum lifts off the ground.

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E3.36Example 3.36 A homogeneous beam (massM , length l) is initially

in vertical position ① (Fig. 3.59). A small disturbance causes the

beam to rotate about the frictionless

support A (initial velocity equal to ze-

ro). In position ② it strikes a small

sphere (mass m, radius r ≪ l). As-

sume the impact to be elastic (e = 1).

Determine the angular velocities of

the beam immediately before and af-

ter the impact and the velocity of the

sphere after the impact.

l

m

M①

②

A

Fig. 3.59

Solution The angular velocity ω immediately before the impact

follows from the conservation of energy (ΘA =Ml2/3):

Mgl/2 = ΘAω2/2−Mgl/2 → ω =

√6g/l .

Since the impact is assumed to be elastic, the conservation of the

angular momentum

ΘAω = ΘAω +mlv

as well as the conservation of energy

ΘAω2/2 = ΘAω

2/2 +mv2/2

hold. Solving these two equations yields

ω =M − 3m

M + 3mω , v =

2lM

M + 3mω .

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E3.37 Example 3.37 A thin half-cylin-

drical shell of weight W = mg

rolls without sliding on a flat sur-

face (Fig. 3.60).

Determine the angular velocity

as a function of ϕ when the initial

condition ϕ(ϕ = 0) = 0 is given.

R

ϕ

W

Fig. 3.60

Solution The position of the center of mass and the mass moment

of inertia are given by

c =1

Rπ

∫ π

0

(R sinα)Rdα =2R

π,

Θc= Θ0 −mc2 = mR2 − 4

π2mR2

= mR2(1− 4

π2) .

Rdα

R sinα0

CR

y

xyccC

Rϕ c cosϕ

cα

ϕ

Thus,

xc = Rϕ+ c cosϕ = R(ϕ+2

πcosϕ) ,

yc = c sinϕ =2R

πsinϕ ,

xc = Rϕ(1− 2

πsinϕ) ,

yc =2R

πϕ cosϕ .

Since we want to find the angular velocity as a function of the

angle we apply the conservation of energy:

T + V = T0 + V0 .

With

T0 = 0 , V0 = 0 ,

T =1

2m(x2c + y2c) +

1

2Θcϕ

2 =1

2mR2ϕ2

(1− 4

πsinϕ+

4

π2sin2 ϕ

+4

π2cos2 ϕ

)+

1

2mR2ϕ2

(1− 4

π2

)= mR2ϕ2

(1− 2

πsinϕ

),

V = −mgyc = −2

πmgR sinϕGro

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we obtain

mR2ϕ2(1− 2

πsinϕ)− 2

πmgr sinϕ = 0

→ ϕ(ϕ) =

√2g sinϕ

R(π − 2 sinϕ).

Note that the angular velocity attains its maximum for ϕ = 90

(lowest position of the center of mass):

ϕmax =√2g/R(π − 2) .

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E3.38 Example 3.38 An elevator con-

sists of a cabin (weightW = mg)

which is connected through a ro-

pe (of negligible mass) with a ro-

pe drum and a band brake (coef-

ficient of dynamic friction µ).

Determine the necessary bra-

king force F such that a cabin

travelling downwards with velo-

city v0 stops after a distance h.

0r1

r2µ

l

v0W

FΘ0

Fig. 3.61

Solution First we determine the

brake momentum MB that acts

on the drum. To this end we se-

parate the drum and the lever.

Equilibrium at the lever

y

A : 0 = −2r2S2 + lF

→ S2 =Fl

2r2

and the formula S1 = S2e−µπ

for belt friction (see Volume 1,

Chapter 9) lead to

MB = r2S2 − r2S1 = (1− e−µπ)Fl

2.

r2

F

S1

S1

S2

S2

A

l

2r2

W

ω0

Now we apply the work-energy theorem

T1 − T0 = U .

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With the kinematic relations (constraints imposed by the rope)

v0 = r1ω0 , h = r1ϕh

the kinetic energies and the work are given by

T1 = 0 , T0 =1

2mv20 +

1

2Θ0ω

20 =

v202

(m+

Θ0

r21

),

U = mgh−∫ ϕh

0

MBdϕ = mgh− (1− e−µπ)Flh

2r1.

Thus, we obtain

−v20

2

(m+

Θ0

r21

)= mgh− (1− e−µπ)

Flh

2r1.

Solving for the force F yields

F =r1v

20m

lh(1− e−µπ)

(1 +

Θ0

mr21+

2gh

v20

).

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E3.39 Example 3.39 A homogeneous beam (mass m, length l) rotates

about frictionless support A (Fig. 3.62) until it hits support B.

The motion starts with zero

initial velocity in the vertical

position. The coefficient of re-

stitution e is given.

Calculate the impulsive

forces at A and B. Determi-

ne the distance a so that the

impulsive force at A vanishes.

Calculate the change of the

kinetic energy.

m

l

A

a

B

Fig. 3.62

Solution First we calculate the angular velocity of the beam im-

mediately before the impact with the aid of the conservation of

energy (ΘA = ml2/3):

mgl/2 = ΘAϕ2/2 → ϕ2 = 3g/l .

C

a

AH

AV Bx

y ϕ

We then apply the principles of impulse and momentum:

→: m(¯xC − xC) = AH ,

↑: m(¯yC − yC) = AV + B ,

y

A : ΘA( ¯ϕ− ϕ) = −Ba .

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Here, the velocity of the center of mass and the angular velocity

before the impact are given by

xC = 0 , yC = −√3gl/2 , ϕ =

√3g/l .

In order to obtain the velocities immediately after the impact we

use the hypothesis

e = −¯yByB

→ ¯ϕ = −eϕ ,

which leads to

¯xC = 0 , ¯yC = e√3gl/2 , ¯ϕ = −e

√3g/l .

Solving for the impulsive reactions yields

AH = 0 , AV =(1 + e)(3a− 2l)m

6a

√3gl ,

B =(1 + e)ml

3a

√3gl .

The impulsive force at A vanishes if

AV = 0 → a = 2l/3 .

The change of the kinetic energy is the difference ∆T = T1 − T0of the energies before and after impact:

∆T =1

2ΘA ˙ϕ2 − 1

2ΘAϕ

2 → ∆T = −(1− e2)mgl/2 .

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E3.40 Example 3.40 A homogeneous cir-

cular disk of weight W = mg is sus-

pended from a pin-supported bar (of

negligible mass). Initially the disk

rotates with the angular velocity ω0.

a) Determine the amplitude of os-

cillation of the pendulum, if the

bar suddenly prevents the disk

from rotating.

b) Calculate the energy loss ∆E.

lm

ω0

r

ϕ

Fig. 3.63

Solution a) First we determine the angular velocity ω of the bar

immediately after the rotation of the disk is prevented (note that

the bar is still in the vertical position). Since there are no external

moments acting with respect to A,

the angular momentum is conser-

ved. With

ΘB =1

2mr2 , ΘA =

1

2mr2 +ml2 =

1

2m(r2 + 2l2)

we obtain

ΘBω0 = ΘAω → ω =ΘB

ΘAω0 =

r2

r2 + 2l2ω0 .

The maximum value ϕ1 of the angle

ϕ (= amplitude of the oscillation)

can be calculated from the conser-

vation of energy:

T1+V1=T0+V0 .

If we choose the potential energy V0(initial position) to be equal to zero,

we can write

T1 = 0 ,

V1 = mgl(1− cosϕ1) ,

T0 =1

2ΘAω

2 =mr2ω2

0

4

r2

r2 + 2l2,

V0 = 0 .

ω

Diskblocked

B

A

ϕ1

l(1− cosϕ1)

position 0

position 1

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Then, conservation of energy leads to

cosϕ1 = 1− r2ω20

4gl

r2

r2 + 2l2.

b) The loss of energy ∆E is given by the difference ∆T of the

kinetic energies before and after the blocking:

∆E =ΘBω

20

2− ΘAω

2

2=mr2ω2

0

4− m

4(r2 + 2l2)

r4ω20

(r2 + 2l2)2

=mr2ω2

0

2

l2

r2 + 2l2.

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E3.41 Example 3.41 A homogeneous an-

gled bar of mass m is attached to a

shaft with negligible mass. The ro-

tation of the system is driven by the

moment M0.

Determine the angular accelera-

tion and the support reactions.

M0

m

2l

2l

2l

l

ξη

Fig. 3.64

Solution The following moments

and products of inertia with respect

to the body-fixed coordinate system

ξ, η, ζ are needed:

Θζ =2

3m

(2l)2

3+m

3(2l)2 =

20

9ml2 ,

Θξζ = −m3

2ll

2= −1

3ml2 ,

Θηζ = 0 .

ω, ω

AηAξ

Bξ Bη

M0

ξc

Cη

ξ

ζ

With the moments

Mξ = 2lBη − 2lAη , Mη = 2lAξ − 2lBξ , Mζ =M0

the principle of angular momentum in component form yields

y

ξ : 2l(Bη −Aη) = −ωml2

3→ Bη −Aη = −mlω

6,

y

η : 2l(Aξ −Bξ) = −ω2ml2

3→ Aξ −Bξ = −

mlω2

6,

y

ζ : M0 =20

9ml2ω → ω =

9M0

20ml2.

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In order to be able to calculate the support reactions, we now

have to formulate the principle of linear momentum. The center

of mass moves on a circle. With the distance ξc = 4l/3 from the

axis of rotation, we obtain the components of its acceleration as

acξ = −ξcω2 and acη = ξcω. Thus,

−mξc ω2 = Aξ +Bξ , m ξc ω = Aη +Bη

which leads to

Aξ = −3

4mlω2 , Aη =

27

80

M0

l,

Bξ = − 7

12mlω2 , Bη =

21

80

M0

l.

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E3.42 Example 3.42 A shaft (principal moments of inertia Θ1, Θ2, Θ3)

rotates with constant angular velocity ω0 about its longitudinal

axis. This axis undergoes a rotation

α(t) about the z-axis of the fixed in

space system x, y, z.

Calculate the moment which is

exerted by the bearings on the shaft

for

a) uniform rotation α = Ωt,

b) harmonic rotation α = α0 sinΩt.

2

3, zx

1

ω0

y

α(t)

Fig. 3.65

Solution We solve the problem with the aid of Euler’s equations

Θ1ω1 − (Θ2 −Θ3)ω2ω3 =M1 ,

Θ2ω2 − (Θ3 −Θ1)ω3ω1 =M2 ,

Θ3ω3 − (Θ1 −Θ2)ω1ω2 =M3 .

With

ω1 = ω0 , ω1 = 0 , ω2 = ω2 = 0 , ω3 = α , ω3 = α

we obtain

M1 = 0 , M2 = −(Θ3 −Θ1)ω0α , M3 = Θ3α

where α(t) represents an arbitrary rotation¡.

a) In the special case of a uniform rotation α = Ωt we get

α = Ω , α = 0 .

Thus,

M1 =M3 = 0 , M2 = (Θ1 −Θ3)ω0Ω .

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b) In the case of a harmonic rotation α = α0 sinΩt we find

α = α0ΩcosΩt , α = −α0Ω2 sinΩt

and

M1 = 0 , M2 = (Θ1 −Θ3)ω0Ωα0 cosΩt ,

M3 = −Θ3Ω2α0 sinΩt .

Note: Only a moment about the 2-axis acts in case a). It is caused

by two opposite support reactions of equal magnitude (= couple)

in the 3- and z-directions, respectively.

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E3.43 Example 3.43 A pin-supported rigid beam

(mass m, length l) is initially at rest. At

time t0 = 0 it starts to rotate due to an

applied constant moment M0.

Determine the stress resultants (inter-

nal forces and moments) as functions of x

for t > t0. Neglect gravitational effects.

ϕA

l

x

M0

Fig. 3.66

Solution First we determine the angular acceleration and the an-

gular velocity of the beam. The principle of angular momentum

with respect to A yields

x

A : ΘA ϕ =M0 → ϕ =M0

ΘA.

Integration with the initial condition ϕ(0) = 0 gives

ϕ(t) =M0

ΘAt .

Now we cut the beam at an arbitrary position and introduce the

bending moment M and the shear force V into the free-body dia-

gram (the normal force N will be considered later). The principle

of angular momentum (with respect to the center of mass C) and

the principle of linear momentum (in the y-direction) yield

x

C : ΘC ϕ = −M(x)− V (x)l − x2

,

տ : m yC = V (x) .

The acceleration yC follows from the ki-

nematics (circular motion):

yC = rC ϕ =l + x

2ϕ .

MV

xy

C1

2(l − x)

m, ΘC

rC = 1

2(l + x)

Thus, with m = (1 − x

l)m and ΘA =

ml2

3we obtain the shear

force:

V (x) = ml + x

2ϕ = m

l + x

2

M0

ΘA=

3

2

M0

l

[1−

(xl

)2 ].

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Introduction of the moment of inertia ΘC =1

12m (l − x)2 leads

to the bending moment:

M(x) = −V l − x2− ΘC ϕ

= −3

4M0

(1− x

l

)2 (1 +

x

l

)− ml2

12

(1− x

l

)3 M0

ΘA

= −M0

(1− x

l

)2 (1 +

1

2

x

l

).

The normal force can be determined from the equation of motion

in the x-direction:

ր : m xC = −N(x)

where xC = −rC ϕ2 is the centripetal

acceleration. This leads to

m

xy

N C

rC = 1

2(l + x)

N(x) = m rC ϕ2

= m (1− x

l)x+ l

2

(M0

ΘAt

)2

=9

2

M20 t

2

ml3

[1−

(xl

)2 ].

Note that the normal force increases with time t in contrast to

the bending moment and the shear force.

The stress resultant diagrams are presented below.

M0+3

2

M0

l

+9

2

M20 t

2

ml3

M(x)V (x)

N(x)

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E3.44 Example 3.44 The angled member

(weightW = mg) in Fig. 3.67 con-

sists of two homogeneous bars.

Derive the equation of motion

for the member’s center of mass. 2l

l

g

A

Fig. 3.67

Solution First, we locate the center

of mass of the angled member. With

the coordinate system as shown in

the figure we obtain

xc =

m

3

l

2m

=l

6,

yc =

m

32l+

2m

3l

m=

4

3l .

Thus, the distance of the center of

mass from point A is given by

x

y

1

3m

a

C

xc

yc

A

mg

ϕa

C

a sinϕ

2

3m

a =√y2c + x2c =

√65

6l .

The member rotates about a fixed axis that passes through A. We

introduce the angle ϕ and apply the principle of angular momen-

tum:

ΘA ϕ =MA .

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We measure ϕ from the position of equilibrium (C is vertically

below A). With the mass moment of inertia

ΘA =m

3

l2

12+

[(2l)2 +

(l

2

)2]

+2

3m

(2l)2

3=

7

3ml2

we obtain

x

A :7

3ml2ϕ = −mga sinϕ → ϕ+

√65

14

g

lsinϕ = 0 .

Note: In the case of small oscillations (ϕ ≪ 1, sinϕ ≈ ϕ) the

equation of motion reduces to the differential equation

ϕ+

√65

14

g

lϕ = 0

for harmonic vibrations.

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E3.45 Example 3.45 A bowling ball (massm) is placed on a rough surface

(coefficient of kinetic friction

µ = 0.3) with velocity v0 = 5 m/s

(Fig. 3.68). Initially, the ball does

not rotate.

What is the position xr of the

ball when it stops sliding? Calculate

the corresponding velocity vr.

v0

µ

Fig. 3.68

Solution When the ball is placed on the rough surface it slides.

RN

ω

x

mg

rC

The friction force R is opposed to the direction of the motion.

Thus, the equations of motion are given by

→ : mx = −R ,

↑ : 0 = N −mg → N = mg ,

y

C : Θcω = rR .

With Θc = 2mr2/5 and the law of friction R = µN = µmg, the

first and the third equation lead to (initial conditions v(t = 0) =

v0, x(t = 0) = 0, ω(t = 0) = 0)

v = x = v0 − µgt , x = v0t−1

2µg t2 , ω =

5µg

2rt .

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The ball rolls without sliding if the velocity of its center of mass

is given by

v = rω .

This condition leads to the corresponding time tr:

v0 − µgtr =5

2µ g tr → tr =

2v07µg

= 0.49 s .

Thus,

xr = x(tr) =12v2049µg

= 2.08 m ,

vr = v(tr) =5

7v0 = 3.57

m

s.

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E3.46 Example 3.46 The double pendulum in

Fig. 3.69 consists of two identical ho-

mogeneous bars (each mass m, length

l). It is struck by a linear impulse F at

point D.

Determine the distance d of point D

from the lower end of the pendulum so

that the angular velocity ω2 of the lower

bar is zero immediately after the im-

pact. Calculate the impulsive forces at

A and B.

A

B

l

ld

FD

g

Fig. 3.69

Solution We separate the two bars and

draw the free-body diagram. Note that

there is no linear impulse in the vertical

direction. The bars are at rest before the

impact. The principles of linear and an-

gular impulse and momentum are given

by

①x

A : ΘA ω1 = l B ,

② → : mv2 = F − B ,x

C2 : ΘC2ω2 =

l

2B −

(d− l

2

)F

where

F

B

B

A

ω2

v2

l ①

②

1

2l

d− 1

2lC2

ω1

ΘA =1

3ml2 , ΘC2

=1

12ml2 .

We desire the motion of bar ② to be a translation. Therefore,

ω2 = 0 , v2 = l ω1 .

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This leads to

ω1 =3

4

F

ml, B =

F

1 +ml2

ΘA

=F

4, d =

l

2

2 +ml2

ΘA

1 +ml2

ΘA

=5

8l .

The impulsive force at A follows from the principle of linear im-

pulse for bar ① with the kinematic relation v1 =1

2l ω1:

→ : mv1 = A+ B → A =F

8.

1

v1C1

B

ω1

A

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4Chapter 4

Principles of Mechanics

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114 4 Principles of Mechanics

E4.8 Example 4.8 A homogeneous disk (mass m, radius r) rolls without

slipping on a rough surface (Fig. 4.9). Its center of mass C is

connected with the wall by a spring (spring constant k).


using

a) Newton’s 2nd Law,

b) dynamic equilibrium conditions.

ϕ

k

mA

r

C

Fig. 4.9

Solution a) The free-body diagram shows the forces that act on

the disk.

x

FC

H

N

W ϕ

We use the coordinates x and ϕ. Then the principles of linear

and angular momentum yield

←: mxc = −F −H ,

↑: 0 = N −W → N =W ,

x

C : ΘCϕ = rH where ΘC = mr2/2 .

In addition, we have the kinematic relation

xc = rϕ → xc = rϕ ,

and the relation

F = kxc → F = krϕ

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4 Principles of Mechanics 115

for the force F in the spring. Solving these equations for the angle

ϕ yields

ϕ+ ω2ϕ = 0 where ω2 =2k

3m.

b) If we apply the dynamic equilibrium conditions, the inertial

force mxc and the pseudo moment ΘCϕ (both acting in the ne-

gative coordinate directions) have to be drawn into the free-body

diagram.

ΘCϕ

FC

H

N

B

mxcW

Dynamic moment equilibrium about point B then yields

y

B : ΘC ϕ+ rmxc + rF = 0 .

If we introduce the kinematic relation (see a)), the force F = krϕ

and ΘC = mr2/2 we again obtain

ϕ+ ω2ϕ = 0 , ω2 =2c

3m.

Note that we may choose point B to be the reference point for

the moment equilibrium. This is advantageous since then the lever

arm of the unknown force of static friction H is zero. The mass

moment of inertia must be ΘC (not ΘB!).

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E4.9 Example 4.9 A cylinder (mass

m, radius r) rolls without slip-

ping on a circular path (radius

R); see Fig. 4.10.

Derive the equation of moti-

on using dynamic equilibrium

conditions.

ϕ

Rr

m

Fig. 4.10

Solution We isolate the cylinder and introduce the coordinates

ϕ and ψ (angle of rotation of the cylinder). With the tangential

acceleration at = (R − r)ϕ (in the positive ϕ-direction) of the

center of mass C and the normal acceleration an = (R − r)ϕ2

(directed towards the center of the circular path), the inertial

forcesmat (opposite to at) andman (opposite to an) can be drawn

on the free-body diagram.

m(R − r)ϕ

m(R − r)ϕ2ϕ

ϕ

C

ψ

B H

mg N

ΘCψ

The pseudo moment ΘCψ acts in the negative ψ-direction. Mo-

ment equilibrium about point B yields the equation of motion:

y

B : ΘC ψ +m(R − r)ϕ+mgr sinϕ = 0 where ΘC = mr2/2 .

The system has one degree of freedom. Therefore a relation exists

between the two coordinates:

vC = (R− r)ϕ = rψ → ψ = (R/r − 1)ϕ .

This leads to

ϕ+2g

3(R− r) sinϕ = 0 .

Note that the moment equilibrium may be established with re-

spect to point B. This is advantageous since the lever arms of

the unknown force of static friction H and of the inertial force

m(R − r)ϕ2 are zero. The mass moment of inertia, however, has

to be taken with respect to the center of mass C.

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E4.10Example 4.10 Two blocks of weights

W1 = m1g and W2 = m2g are

suspended by a pin-supported ro-

pe drum (moment of inertia ΘA) as

shown in Fig. 4.11.

Determine the angular accelerati-

on of the drum and the force in ro-

pe ① using dynamic equilibrium con-

ditions. Neglect the mass of the ro-

pes.

ΘA

m2

1 2

m1

r1A

r2

ϕ

Fig. 4.11

Solution We first introduce the co-

ordinates x1 and x2 describing the

motion of the blocks. The inerti-

al forces −mixi point in the ne-

gative xi-directions (see the free-

body diagram). In addition, we ha-

ve to consider the pseudo moment

−ΘAϕ which acts in the negative ϕ-

direction. Moment equilibrium about

point A then yields

y

A : −r1m1(g + x1) + r2m2(g − x2)−ΘAϕ = 0 .

m2x2

m1g

m1x1

m2g

ϕ

A

x2

x1

ΘAϕ

Using the kinematic relations

x1 = r1ϕ → x1 = r1ϕ ,

x2 = r2ϕ → x2 = r2ϕ

we obtain the angular acceleration of the drum:

ϕ =r2m2 − r1m1

r21m1 + r22m2 +ΘAg .

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In order to determine the force in rope ① we cut the rope. Force

equilibrium (see the free-body diagram) yields

↑ : S1 −m1g −m1x1 = 0

m1g

S1

x1

m1x1

or

S1 = m1(g + r1ϕ) = m1gr2(r1 + r2)m2 +ΘA

r21m1 + r22m2 +ΘA.

Note: For r2m2 > r1m1 the drum rotates clockwise, for r2m2 <

r1m1 it rotates counterclockwise. In the special case r2m2 = r1m1

the system is in static equilibrium (ϕ = 0).

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E4.11Example 4.11 An angled arm (mass

m) rotates with constant angular ve-

locity Ω about point 0 (Fig. 4.12).

Calculate the bending moment,

shear force and normal force as func-

tions of position using dynamic equi-

librium conditions.

b

a

0

Ω

Fig. 4.12

Solution We introduce two coordinate systems xi, zi. Then, we

make a cut at the arbitrary position x1. The acceleration of the

mass element dm at the position s (distance from the left end of

the arm) is given by an = rΩ2 (pointing towards point 0; note

that at = 0). Therefore, this element is subjected to the inertial

force dmrΩ2 (see the free-body diagram).

ϕ Ω

s dsr

V

M

Ω

N

V (x2 = 0)

x2

z2M(x2 = 0)

N(x2 = 0)

ds

s

N(x1 = b)

M

NV

dmrΩ2 dm(a− s)Ω2

V (x1 = b)

M(x1 = b)

z1

x1

With

dm =m

a+ bds = µds ,

where µ = m/(a + b) is the mass per unit length, and with the

geometrical relations

cosϕ = (b − s)/r , sinϕ = a/r

we can determine the stress resultants through integration (note

that M ′ = V ).

Normal force (0 ≤ x1 ≤ b):

N(x1) =

∫rΩ2 cosϕdm =

∫ x1

0

Ω2(b − s)µds

= µΩ2[bs− s2/2]|x1

0 → N(x1) = µΩ2(bx1 − x21/2) .

Shear force (0 ≤ x1 ≤ b):

V (x1) =

∫rΩ2 sinϕdm =

∫ x1

0

Ω2aµds → V (x1) = µΩ2ax1 .

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Bending moment (0 ≤ x1 ≤ b):

M(x1) =

∫ x1

0

V (s)ds → M(x1) = µΩ2ax21/2 .

The matching conditions at the corner (x1 = b, x2 = 0) are given

by

N0 = N(x2 = 0) = V (x1 = b) = µΩ2ab ,

V0 = V (x2 = 0) = −N(x1 = b) = −µΩ2b2/2 ,

M0 =M(x2 = 0) =M(x1 = b) = µΩ2ab2/2 .

Now we make a cut at the position x2. The mass element dm at

the position s is subjected to the inertial force dm(a− s)Ω2. This

leads to the following stress resultants:

Normal force, shear force, bending moment (0 ≤ x2 ≤ a):

N(x2) = N0 +∫ x2

0 µΩ2(a− s)ds

→ N(x2) = µΩ2(ab+ ax2 − x22/2) ,

V (x2) = V0 → V (x2) = −µΩ2b2/2 ,

M(x2) =M0 + x2V0 → M(x2) = µΩ2b2(a− x2)/2 .

+

+

+

−

+ +N0 V0M0

N0| V0 | VN M

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E4.12Example 4.12 A wheel (weight W1 = m1g, moment of inertia ΘA)

on an inclined plane is connected to a

block (weight W2 = m2g) by a rope

which is guided over an ideal pulley

(Fig. 4.13). The wheel rolls on the pla-

ne without slipping.

Determine the acceleration of the

block applying d’Alembert’s principle.

Neglect the masses of the rope and the

pulley.

A

m1,ΘA

r

α

m2

Fig. 4.13

Solution Since the constraint forces (force in the rope, static

friction force) need not be determined, it is advantageous to apply

d’Alembert’s principle. The motion

is described by the coordinates xiand ϕ. The inertial forces mixi and

the pseudo moment ΘAϕ (acting in

the directions opposite to the chosen

positive coordinate directions) are

shown in the figure. D’Alembert’s

principle (principle of virtual work)

requires that the virtual work of all

forces vanishes:

δU + δUI = 0

ΘAϕ

α

m1g

ϕ

Ax2

m2g

x1

m2x2

m1x1

→ −m1x1δx1 −m1g sinαδx1 −ΘAϕδϕ+m2gδx2 −m2x2δx2 = 0 .

With the kinematic relations

x1 = x2 = rϕ = x →

δx1 = δx2 = rδϕ = δx

x1 = x2 = rϕ = x

we obtain[−m1x−m1g sinα−

ΘA

r2x+m2g −m2x

]δx = 0 .

Since δx 6= 0, the expression in the brackets must vanish. Thus,

x = x2 = gm2 −m1 sinα

m1 +m2 +ΘA

r2

.

Note that x < 0 for m1 sinα > m2. In this case, the wheel rolls

down the inclined plane.

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E4.13 Example 4.13 Two drums are connected by a rope and carry blocks

of weights m1 g and m2 g

(Fig. 4.14). Drum ① is dri-

ven by the moment M0.

Determine the accele-

ration of block ② using

d’Alembert’s principle.

Neglect the mass of the

ropes. m1

m2

ΘA ΘB

r2

B1 r1

M0

r2

A

2

Fig. 4.14

Solution We introduce the

inertial forces mixi and

the pseudo moments ΘAϕ1,

ΘBϕ2. They act in the

directions opposite to the

chosen positive coordina-

te directions. D’Alembert’s

principle requires

δU + δUI = 0 ,m1g

m1x1

B

M0

ϕ1

ΘAϕ1

x1

x2

m2g

m2x2

A

ΘBϕ2ϕ2

which leads to

−m1(g + x1)δx1 +m2(g − x2)δx2

+M0δϕ1 −ΘAϕ1δϕ1 −ΘBϕ2δϕ2 = 0 .

With the kinematic relations

x1 = r1ϕ1

x2 = r2ϕ2

ϕ1 = ϕ2

→ϕ1 = ϕ2 =

x2r2

, x1 =r1r2x2 ,

δϕ1 = δϕ2 =δx2r2

, δx1 =r1r2δx2

Gro

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auge

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we obtain

−m1

(g +

r1r2x2)r1r2

+m2(g − x2)

+M0

r2− ΘA

r22x2 −

ΘB

r22x2

δx2 = 0 .

Since δx2 6= 0, the expression in the curly brackets must vanish.

Thus, the acceleration of block ② is

x2 = g

1− m1r1m2r2

+M0

r2m2g

1 +m1

m2

(r1r2

)2+

ΘA

m2r22+

ΘB

m2r22

.

Gro

ss, H

auge

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E4.14 Example 4.14 The system shown in Fig. 4.15 consists of a block

(massM), a homogeneous disk (massm, radius r) and two springs

(spring constant k). The block moves on a frictionless surface; the

disk rolls without slipping

on the block. A force F (t)

acts on the block.

Derive the equations of mo-

tion using Lagrange’s for-

malism.

kM

m

rk

no friction

F (t)

Fig. 4.15

Solution The system has

two degrees of freedom. We

choose the displacement x

of the block and the angle

of rotation ϕ of the disk to

F (t)

ϕC

x

be the generalized coordinates.

The two springs are unstressed for x = 0 and ϕ = 0. The kinetic

energy and the potential energy of the springs, respectively, are

T =Mx2/2 + ΘCϕ2/2 +mv2C/2 ,

V = kx2/2 + k(rϕ)2/2 .

With ΘC = mr2/2 and the kinematic relation

vC = x− ϕ

we obtain the Lagrangian

L = T − V

→ L =Mx2/2 +mr2ϕ2/4 +m(x− rϕ)2/2− kx2/2− kr2ϕ2/2 .

Since the force F (t) is not given from a potential, we have to apply

the Lagrangian equations in the form

d

dt

(∂L

∂x

)− ∂L

∂x= Qx ,

d

dt

(∂L

∂ϕ

)− ∂L

∂ϕ= Qϕ .

The generalized forces Qx and Qϕ follow from the virtual work

δU of the force F (t):

δU = Qxδx+Qϕδϕ = F (t)δx → Qx = F (t) , Qϕ = 0 .

Gro

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To set up the equations of motion, the following derivatives

have to be calculated:

∂L

∂x=Mx+m(x− rϕ) ,

d

dt

(∂L

∂x

)=Mx+m(x− rϕ) ,

∂L

∂ϕ= mr2ϕ/2−mr(x − rϕ) ,

d

dt

(∂L

∂ϕ

)= mr2ϕ/2−mr(x − rϕ) ,

∂L

∂x= −kx , ∂L

∂ϕ= −kr2ϕ .

Thus, we obtain

(M +m)x−mrϕ+ kx = F (t) ,

−mx+3

2mrϕ + krϕ = 0 .

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E4.15 Example 4.15 Fig. 4.16 shows two blocks of mass m1 and m2

which can glide on a friction-

less surface. They are cou-

pled by springs (stiffnesses

k1, k2, k3).

Derive the equations of

motion using the Lagrange

formalism.

k1m1 m2

k3 k2

Fig. 4.16

Solution The system is conservative; it has two degrees of freedom.

We introduce the two coordinates x1 and x2 which describe the

positions of the two blocks.

They are measured from the

equilibrium positions of the

blocks. The kinetic and the

potential energy, respective-

ly, are given by

x1 x2

k1m1 m2

k3 k2

T =1

2m1x

21 +

1

2m2x

22 ,

V =1

2k1x

21 +

1

2k2x

22 +

1

2k3(x2 − x1)2 .

Thus, the Lagrangian of the system is

L = T − V =1

2m1x

21 +

1

2m2x

22 −

1

2k1x

21 −

1

2k2x

22 −

1

2k3(x2 − x1)2 .

To set up the Lagrange equations

d

dt

( ∂L∂x1

)− ∂L

∂x1= 0 ,

d

dt

( ∂L∂x2

)− ∂L

∂x2= 0

the following derivatives must be calculated:

∂L

∂x1= m1x1 ,

d

dt

( ∂L∂x1

)= m1x1 ,

∂L

∂x1= −k1x1 + k3(x2 − x1) ,

∂L

∂x2= m2x2 ,

d

dt

( ∂L∂x2

)= m2x2 ,

∂L

∂x2= −k2x2 − k3(x2 − x1) .

Gro

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Hence, we obtain

m1x1 + k1x1 − k3(x2 − x1) = 0

→ m1x1 + (k1 + k3)x1 − k3x2 = 0 ,

m2x2 + k2x2 + k3(x2 − x1) = 0

→ m2x2 + (k2 + k3)x2 − k3x1 = 0 .

Note: The two coupled differential equations describe the coupled

free vibrations of the two blocks. In the special case of k3 = 0 the

system is decoupled and we obtain two independent equations of

motion for two systems, each with one degree of freedom.

Gro

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E4.16 Example 4.16 Two simple pen-

dulums (each mass m, length

l) are connected by a spring

(spring constant k, unstretched

length b) as shown in Fig. 4.17.

Derive the equations of mo-

tion using the Lagrange forma-

lism.

g

l

m

mk

l

Fig. 4.17

Solution The system has two

degrees of freedom. We choose

the generalized coordinates ϕ1

and ϕ2 as shown in the figure.

With the kinetic and the poten-

tial energy, respectively,

2l sin ϕ2

0

ϕ1

ϕ2

m

mk

T = ml2(ϕ1 − ϕ2)2/2 +ml2(ϕ1 + ϕ2)

2/2

→ T = ml2(ϕ21 + ϕ2

2) ,

V = −mgl cos(ϕ1 − ϕ2)−mgl cos(ϕ1 + ϕ2) + k(2l sinϕ2 − b)2/2

→ V = −2mgl cosϕ1 cosϕ2 + k(2l sinϕ2 − b)2/2

(zero level: point 0 and unstressed spring) the Lagrangian becomes

L = T − V

→ L = ml2(ϕ21 + ϕ2

2) + 2mgl cosϕ1 cosϕ2 − k(2l sinϕ2 − b)2/2 .

To set up the Lagrange equations

d

dt

(∂L

∂ϕ1

)− ∂L

∂ϕ1= 0 ,

d

dt

(∂L

∂ϕ2

)− ∂L

∂ϕ2= 0

Gro

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the following derivatives are needed:

∂L

∂ϕ1= 2ml2ϕ1 ,

d

dt

(∂L

∂ϕ1

)= 2ml2ϕ1 ,

∂L

∂ϕ1= −2mgl sinϕ1 cosϕ2 ,

∂L

∂ϕ2= 2ml2ϕ2 ,

d

dt

(∂L

∂ϕ2

)= 2ml2ϕ2 ,

∂L

∂ϕ2= −2mgl cosϕ1 sinϕ2 − k(2l sinϕ2 − b)2l cosϕ2 .

This leads to the equations of motion:

lϕ1 + g sinϕ1 cosϕ2 = 0 ,

mlϕ2 +mg cosϕ1 sinϕ2 + k(2l sinϕ2 − b) cosϕ2 = 0 .

Gro

ss, H

auge

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E4.17 Example 4.17 A disk (weight m2g,

moment of inertia Θ2) glides along

a frictionless homogeneous bar of

weight m1g (Fig. 4.18).

Find the equations of motion

using the Lagrange formalism.

x

ϕ m1

l

m2,Θ2

Fig. 4.18

Solution The system is conservative;

it has two degrees of freedom. Its po-

sition is uniquely determined by the

distance x of point C1 from pin 0 and

by the angle ϕ (generalized coordina-

tes). With the kinetic and the poten-

tial energy, respectively,

0l

2cosϕ

x

x cosϕ

ϕC2

C1

T =1

2

(m1l2

3

)ϕ2 +

12m2[(xϕ)

2 + x2] +1

2Θ2ϕ

22

=1

2

[m1l2

3+m2x

2 + Θ2

]ϕ2 +

1

2m2x

2 ,

V = −m1gl

2cosϕ−m2gx cosϕ = −

[m1

l

2+m2x

]g cosϕ

(zero level at 0) and the Lagrangian L = T − V , we can write

down the Lagrange equations

d

dt

(∂L∂ϕ

)− ∂L

∂ϕ= 0 ,

d

dt

(∂L∂x

)− ∂L

∂x= 0 .

To this end, the following derivatives are needed:

∂L

∂ϕ=(m1l

2

3+m2x

2 +Θ2

)ϕ ,

∂L

∂ϕ= −

(m1

l

2+m2x

)g sinϕ ,

d

dt

(∂L∂ϕ

)=(m1l

2

3+m2x

2 +Θ2

)ϕ+ 2m2xxϕ ,

∂L

∂x= m2x ,

d

dt

(∂L∂x

)= m2x ,

∂L

∂x= m2xϕ

2 +m2g cosϕ .

This leads to the coupled equations of motion:

(m1l2

3+m2x

2 +Θ2

)ϕ+ 2m2xxϕ+

(m1

l

2+m2x

)g sinϕ = 0 ,

m2x−m2xϕ2 −m2g cosϕ = 0 → x− xϕ2 − g cosϕ = 0 .

Gro

ss, H

auge

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E4.18Example 4.18 A thin half-cylindri-

cal shell of weight W = mg rolls

without sliding on a flat surface

(Fig. 4.19).



mr

Fig. 4.19

Solution The system is conservative;

it has one degree of freedom. We in-

troduce the coordinate ϕ as shown

in the figure. With the distance a =

2r/π of the center of mass C from

the center M of the shell we have

M

ϕa

x

y

C

rϕ

ΘC = ΘM − a2m = r2m− a2m= (1− 4/π2)mr2 ,

xc = rϕ− a sinϕ → xc = rϕ − aϕ cosϕ ,

yc = a cosϕ → yc = −aϕ sinϕ .

Thus, the potential energy V , the kinetic energy T , the Lagrangian

and the pertinent derivatives are

V = mga(1− cosϕ) =2

πmgr(1− cosϕ) ,

T =1

2m(x2c + y2c) +

1

2ΘC ϕ

2 =1

2mr2ϕ2

[(1− 2

πcosϕ

)2

+( 2πsinϕ

)2+(1− 4

π2

)]= mr2ϕ2

(1− 2

πcosϕ

),

L = T − V = mr[rϕ2

(1− 2

πcosϕ

)− 2

πg(1− cosϕ)

],

∂L

∂ϕ= mr

[2rϕ

(1− 2

πcosϕ

)],

d

dt

(∂L∂ϕ

)= mr

[2rϕ

(1− 2

πcosϕ

)+

4

πrϕ2 sinϕ

],

∂L

∂ϕ= mr

[ 2πrϕ2 sinϕ− 2

πg sinϕ

].

Gro

ss, H

auge

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er, W

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Introduction into

d

dt

(∂L∂ϕ

)− ∂L

∂ϕ= 0

yields the equation of motion:

ϕ(π − 2 cosϕ) + ϕ2 sinϕ+g

rsinϕ = 0 .

Gro

ss, H

auge

r, Sc

hröd

er, W

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E4.19Example 4.19 A block (mass m1)

can move horizontally on a smooth

surface (Fig. 4.20). A simple pendu-

lum (mass m2) is connected to the

block by a pin.

Find the equations of motion


m1

l

m2

g

Fig. 4.20

Solution The system is conserva-

tive; it has two degrees of free-

dom. We use the generalized coor-

dinates x and ϕ as shown in the

figure. With the zero-level of the

potential of the force m2g chosen

at the height of the mass m1, we

have

l cosϕϕ

m2

x

m1x

l sinϕ

y

V = −m2gl cosϕ ,

T =1

2m1x

2 +1

2m2[(x+ lϕ cosϕ)2 + (lϕ sinϕ)2] ,

L = T − V =1

2(m1 +m2)x

2 +m2lxϕ cosϕ+1

2m2l

2ϕ2 +m2gl cosϕ .

Introduction of the derivatives

∂L

∂ϕ= m2lx cosϕ+m2l

2ϕ ,∂L

∂ϕ= −m2lxϕ sinϕ−m2gl sinϕ ,

d

dt

(∂L∂ϕ

)= m2lx cosϕ−m2lxϕ sinϕ+m2l

2ϕ ,

∂L

∂x= (m1 +m2)x +m2lϕ cosϕ ,

∂L

∂x= 0 ,

d

dt

(∂L∂x

)= (m1 +m2)x+m2lϕ cosϕ−m2lϕ

2 sinϕ

Gro

ss, H

auge

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into the Lagrange equations

d

dt

(∂L∂ϕ

)− ∂L

∂ϕ= 0 ,

d

dt

(∂L∂x

)− ∂L

∂x= 0

yields the coupled equations of motion

x cosϕ+ lϕ+ g sinϕ = 0 ,

(m1 +m2)x+m2lϕ cosϕ−m2lϕ2 sinϕ = 0 .

Gro

ss, H

auge

r, Sc

hröd

er, W

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5Chapter 5

Vibrations

Gro

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136 5 Vibrations

E5.11 Example 5.11 The system in Fig 5.30

consists of three bars and a beam

(with negligible masses) and a block

(mass m).

Determine the circular frequency

of the free vertical vibrations. l l

l

EA

EA

m

EA

EI

Fig. 5.30

Solution The system consisting of

the truss and the beam is equiva-

lent to a system consisting of two

springs in parallel (both springs

undergo the same elongation when

the block is displaced). To deter-

mine the spring constant kB of the

beam, we subject the beam to the

force FB = 1 which acts at the

location of the block. This force

produces the deflection (see Engi-

neering Mechanics 2: Mechanics of

Materials, Table 4.3)

FB=1

wB

③

wT

FT =1

②

①

wB =1 · (2l)348EI

.

Thus, the spring constant is given by

kB =1

wB=

48 EI

(2l)3=

6 EI

l3.

In order to find the spring constant kT of the truss, we apply the

force FT = 1 at bar ①. This force causes the displacement (see

Engineering Mechanics 2: Mechanics of Materials, Section 6.3)

wT =∑ S2

i liEA

.

Gro

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5 Vibrations 137

With

S1 = 1 , S2 = S3 = −1

2

√2 , l1 = l , l2 = l3 =

√2 l

we obtain

wT =1

EA

[12 · l + 2 ·

(√22

)2√2 l]= (1 +

√2)

l

EA

→ kT =1

wT=

EA

(1 +√2)l

.

Now, we replace the two springs in parallel by an equivalent single

spring. Its spring constant k∗ is given by

k∗ = kB + kT =6 EI

l3+

EA

(1 +√2)l

.

Thus, the eigenfrequency is

ω =

√k∗

m=

1

l

√1

ml

(6 EI +

EAl2

1 +√2

).

Gro

ss, H

auge

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138 5 Vibrations

E5.12 Example 5.12 The system in Fig. 5.31 consists

of a homogeneous drum (mass M , radius r), a

block (mass m), a spring (spring constant k)

and a string (with negligible mass). The sup-

port of the drum is frictionless. Assume that

there is no slip between the string and the

drum.

Determine the natural frequency of the sys-

tem.k

Mr

m

Fig. 5.31

Solution We first draw the free-body dia-

gram. The position of the block is given by

the coordinate x, measured from the posi-

tion with an unstrechted spring. Next, we

write down the equations of motion for the

block

↓: mx = mg − S1

and for the drum

y

B : Mr2ϕ/2 = S1r − S2r .

In addition, we need the kinematic relation

x = rϕ

and the equation

B

S2 S1

ϕ

S1S2

mg

x

S2 = kx

for the restoring force in the spring. Solving yields the differential

equation for harmonic oscillations:

x+2k

M + 2mx =

2mg

M + 2m.

Thus, the natural frequency is

ω =

√2k

M + 2m.G

ross

, Hau

ger,

Schr

öder

, Wal

l, G

ovid

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5 Vibrations 139

E5.13Example 5.13 Two drums rotate in opposite directions as shown in

Fig. 5.32. They support a homogeneous board of weight W (mass

m). The coefficient of kinetic

friction between the drums

and the board is µ.

Show that the board un-

dergoes a harmonic vibrati-

on and determine the natural

frequency.

x

a

Ω Ω

W

µ

Fig. 5.32

Solution We separate the various parts of the system. The free-

body diagram shows the forces acting if the board is displaced by

an amount x (the friction forces act on the drums in the opposite

directions of the rotations).

W

R1

R2 R1

R2

a/2

a/2

N1N2

x

Thus, the equation of motion in the x-direction of the board is

given by

mx = R2 −R1 .

The normal forces follow from force equilibrium in the vertical

direction and from moment equilibrium:

N1 =W

a

2+ x

a, N2 =W

a

2− xa

.

With the law of kinetic friction R = µN , we obtain

R2 −R1 = µ(N2 −N1) = −µmg2x

a.

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140 5 Vibrations

The equation of motion

mx = −µmg 2x

a

thus leads to the differential equation for harmonic vibrations:

x+ 2µg

ax = 0 .

The natural frequency is given by

ω =

√2µ

g

a.

Note that the natural frequency does not depend on the angular

velocity Ω of the drums.

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E5.14Example 5.14 A homogeneous bar (weight W = mg, length l) is

submerged in a viscous fluid and undergoes vibrations about point

A (Fig. 5.33). The drag force Fd acting

on every point of the bar is proportio-

nal to the local velocity (proportiona-

lity factor β).

a) Derive the equation of motion.

Assume small amplitudes and neglect

the buoyancy. b) Calculate the value

β = β∗ for critical damping.

ϕ

A

l

Fig. 5.33

Solution a) We consider an arbitrary

element of length dx of the bar. It is

subjected to the drag force

dFd = β v(x) dx = βxϕ dx .

We restrict ourselves to small amplitu-

des (sinϕ ≈ ϕ). In this case the prin-

ciple of angular momentum yields

ϕ

A

dx

x

mg

l

2sinϕ≈ lϕ

2

dFd

x

A : ΘAϕ = −mg l2ϕ−

l∫

0

βx2ϕdx .

Evaluation of the integral and introduction of ΘA = ml2/3 lead

to the equation of motion

ϕ+βl

mϕ+

3

2

g

lϕ = 0 → ϕ+ 2ξϕ+ ω2ϕ = 0

where

ξ =βl

2m, ω2 =

3g

2l.

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142 5 Vibrations

b) Critical damping is characterized by

ξ = ω or ζ = 1 .

Thus,

β∗l

2m=

√3

2

g

l→ β∗ =

m

l

√6g

l.

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5 Vibrations 143

E5.15Example 5.15 The pendulum of a clock consists of a homogeneous

rod (mass m, length l) and a homoge-

neous disk (mass M , radius r) whose

center is located at a distance a from

point A (Fig. 5.34). Assume small am-

plitudes and determine the natural fre-

quency of the corresponding oscillati-

ons. Choose m = M and r ≪ a and

calculate the ratio a/l which yields the

maximum eigenfrequency.

A

M

m

2r

la

Fig. 5.34

Solution We introduce the angle ϕ

as shown and apply the principle

of angular momentum:

x

A : ΘAϕ = −mgl/2 sinϕ−Mga sinϕ

where

ϕ

Mg

mg

A

ΘA = ml2/3 +M(r2 + 2a2)/2 .

If we assume small amplitudes (sinϕ ≈ ϕ), we can linearize the

equation of motion:

ϕ+(ml + 2Ma)g

2ΘAϕ = 0 .

Hence, the eigenfrequency is obtained as

ω =

√(ml + 2Ma)g

2ΘA.

In the special case of m = M and r ≪ a the natural frequency

simplifies to

ω =

√3l+ 6a

2l2 + 6a2g .

The ratio a/l which yields the maximum eigenfrequency is found

by setting the derivative dω/da equal to zero:

dω

da= 0 → a/l =

1

2

(√7

3− 1

).G

ross

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ger,

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144 5 Vibrations

E5.16 Example 5.16 The system in Fig 5.35

consists of a homogeneous pulley

(mass M , radius r), a block (mass m)

and a spring (spring constant k).

Determine the equation of motion

for the block and its solution for the

initial conditions x(0) = 0, v(0) = v0.

Neglect the mass of the string and

any lateral motion.

M

A

m

r

k

Fig. 5.35

Solution We separate the pulley and

the block and measure the displace-

ments x of the block and xA of point A

from the position of equilibrium. With

this choice, we do not have to consider

the weights Mg and mg in the free-

body diagram. Thus, the equations of

motion are

① ↓ : mx = −S1 ,

② ↓ : MxA = S1 + S2 − kxA ,

x

A : ΘAϕ = rS1 − rS2 .

If we use the kinematic relations (Π=:

instantaneous center of rotation, see

the figure)

x

kxA2

xAΠ

ϕ

1

S2S1

x

xA

xA =x

2, 2rϕ = x → xA =

x

2, ϕ =

x

2r

and ΘA =Mr2/2 we can solve the equations of motion for x and

obtain

x+k

4m+3

2M

x = 0 → ω =

√√√√k

4m+3

2M

.

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5 Vibrations 145

The general solution of this differential equation is given by

x(t) = A cosωt+B sinωt .

The initial conditions

x(0) = 0 → A = 0 , x(0) = v0 → B =v0ω

lead to

x(t) =v0ω

sinωt .

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E5.17 Example 5.17 A wheel (mass m, radius r) rolls without slipping

on a circular path (Fig. 5.36). The mass of the rod (length l) can

be neglected; the joints are

frictionless.

Derive the equation of

motion and determine the

natural frequency of small

oscillations. ϕ

g

m

r

l

A

Fig. 5.36

Solution We apply conserva-

tion of energy

T + V = const

to derive the equation of mo-

tion. The kinetic energy of

the rolling wheel is given by

(see the figure)

Cψ

vc

ϕ

T = mv2c/2 + ΘCψ2/2 ,

the potential energy is (zero-level at the height of point A)

V = −mgl cosϕ .

With the mass moment of inertia ΘC = mr2/2 and the kinematic

relations

vc = lϕ , vc = rψ → lϕ = rψ

the kinetic energy can be written as

T = 3ml2ϕ2/4 .

Introduction into the expression for conservation of energy gives

3lϕ2/4− g cosϕ = const .


3

2lϕϕ+ gϕ sinϕ = 0 → ϕ+

2g

3lsinϕ = 0 .G

ross

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5 Vibrations 147

If we restrict ourselves to small amplitudes (sinϕ ≈ ϕ) this equa-tion reduces to

ϕ+2g

3lϕ = 0 .

Thus, the natural frequency is obtained as

ω =

√2g

3l.

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148 5 Vibrations

E5.18 Example 5.18 The simple pendulum in Fig. 5.37 is attached to a

spring (spring constant k) and a dashpot (damping coefficient d).

a) Determine the maximum value of

the damping coefficient d so that

the system undergoes vibrations.

Assume small amplitudes.

b) Find the damping ratio ζ so that

the amplitude is reduced to 1/10

of its initial value after 10 full cy-

cles. Calculate the corresponding

period Td.

a

ad

m

k

Fig. 5.37

Solution a) The equation of motion

follows from the principle of angular

momentum (rotation about point A;

small amplitudes: sinϕ ≈ ϕ, cosϕ ≈ 1):

x

A : ΘAϕ = −Fda− Fk2a−mg2aϕ .

With mg

Fk

Fd

ϕ

A

ΘA = m (2a)2 , Fd = da ϕ , Fk = k2aϕ

we obtain

ϕ+d

4mϕ+

( km

+g

2a

)ϕ = 0 → ϕ+ 2ξϕ+ ω2ϕ = 0 ,

where

ξ =d

8m, ω2 =

k

m+

g

2a.

Gro

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5 Vibrations 149

In order to have oscillations, the system must be underdamped:

ζ < 1 → ξ < ω → d

8m<

√k

m+

g

2a

→ d < 8

√km+

gm2

2a.

b) The necessary damping ratio follows with xn+10 = xn/10 from

the logarithmic decrement:

102πζ√1− ζ2

= lnxn

xn+10= ln 10

→ ζ =

√√√√1

( 20π

ln 10

)2+ 1

= 0.037 .

This leads to the period

Td =2π

ω√1− ζ2

≈ 2π

ω= 2π

√2am

2ak + gm.

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150 5 Vibrations

E5.19 Example 5.19 The structure in Fig. 5.38 consists of an elastic beam

(flexural rigidity EI, axial rigidity EA→∞, negligible mass) and

three rigid bars (with negligible masses). The block (mass m) is

suspended from a spring (spring constant k).

Determine the eigenfrequency of the vertical oscillations of

the block.

k

a

a a a

mFig. 5.38

Solution We reduce the structure to an

equivalent system of a spring (spring

constant k∗, see the figure) and a mass.

To this end we first replace the beam

and the bars by a spring with the spring

constant kB. We can determine kB if we

subject the beam to a force F which acts

at the free end. This force produces the

deflection

m

k∗

w =Fa3

EI

(see Engineering Mechanics 2: Mechanics of Materials, Example

6.22). The spring constant is obtained from

kB =F

w→ kB =

EI

a3.

The displacement of the block is the sum of the elongations of the

springs with spring constants k and kB. Therefore, the beam/bars

and the given spring act as springs in series. Thus, the spring

constant k∗ of the equivalent system follows from

1

k∗=

1

kB+

1

k→ k∗ =

kEI

ka3 + EI.

This yields the eigenfrequency

ω =

√k∗

m→ ω =

√kEI

(ka3 + EI)m.

Gro

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5 Vibrations 151

E5.20Example 5.20 A rod (length l, with negligible mass) is elastically

supported at point A (Fig. 5.39). The rotational spring (spring

constant kT ) is unstretched for ϕ = 0. The rod carries a point

mass m at its free end.

Derive the equation of motion.

Determine the spring constant so

that ϕ = π/6 is an equilibrium posi-

tion. Calculate the natural frequen-

cy of small oscillations about this

equilibrium position.

g

m

kT

l

ϕ

A

Fig. 5.39

Solution To derive the equation of

motion we apply the principle of an-

gular momentum:

y

A : ΘAϕ = mgl cosϕ−MA .

With the mass moment of inertia

mg

MA

ϕ

ΘA = ml2 and the restoring moment MA = kTϕ we obtain

ϕ =g

lcosϕ− kT

ml2ϕ .

Since ϕ = ϕ0 = π/6 is a position of equilibrium, the conditi-

on ϕ(π/6) = 0 leads to the required spring constant (note that

cos(π/6) =√3/2):

√3

2

g

l− π

6

kTml2

= 0 → kT =3√3

πmgl .

Now we consider small oscillations about this position of equili-

brium. We assume that

ϕ = ϕ0 + ψ with |ψ| ≪ 1 .

Introduction into the equation of motion yields

ψ =g

lcos(ϕ0 + ψ)− kT

ml2(ϕ0 + ψ) .

We use the trigonometric relation

cos(ϕ0 + ψ) = cosϕ0 cosψ − sinϕ0 sinψ

→ cos(ϕ0 + ψ) =

√3

2cosψ − 1

2sinψG

ross

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ger,

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152 5 Vibrations

and linearize (cosψ ≈ 1, sinψ ≈ ψ) to obtain

ψ = −1

2

g

lψ − kT

ml2ψ +

√3

2

g

l− π

6

kTml2

→ ψ +

(1

2+

3√3

π

)g

lψ = 0 .

Thus, the natural frequency is given by

ω =

√√√√(1

2+

3√3

π

)g

l.

Gro

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5 Vibrations 153

E5.21Example 5.21 A single story frame

consists of two rigid columns (with

negligible masses), a rigid beam of

mass m and a spring-dashpot sys-

tem as shown in Fig. 5.40. The

ground is forced to vibration by an

earthquake; the acceleration uE =

b0cosΩt is known from measure-

ments.

Determine the maximum ampli-

tude of the steady state vibrations.

d

uE

m

k

45

Fig. 5.40

Assume that the system is underdamped and that the vibrations

have small amplitudes.

Solution We assume that the

amplitudes of the vibrations

are small. Then the elongati-

on of the diagonal is obtained

as

∆ =

√2

2(x− uE) .

The elongation produces the

force

uE

F

xm

x− uE

∆

45

F = k∆+ d ∆

in the spring-dashpot system (see the figure). The equation of

motion of the beam is given by

→: mx = −F√2

2

→ mx+d

2(x− uE) +

k

2(x− uE) = 0 .

Thus, the relative displacement y = x− uE is described by

my +d

2y +

k

2y = mb0 cosΩt

→ 1

ω2y +

2ζ

ωy + y = y0 cosΩt ,

where

ω2 =k

2m, ζ =

d

2

√1

2km, y0 =

2mb0k

.

Gro

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154 5 Vibrations

The maximum amplitude A is obtained for η = Ω/ω ≈ 1 (reso-

nance!). In the case of small damping (ζ ≪ 1) we obtain

A = y0Vmax ≈y02ζ

= 2√2b0d

√m3

k.

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5 Vibrations 155

E5.22Example 5.22 The undamped system in Fig. 5.41 consists of a

block (mass m = 4 kg) and a spring (spring constant k = 1

N/m). The block is subjected to a force F (t). The initial conditions

x(0) = x0 = 1 m, x(0) = 0 and the response

x(t) = x0

[cos

t

2t0+ 20

(1− cos

t− T2t0

)〈t− T 〉0

]

to the excitation are given. Here,

t0 = 1 s, T = 5 s and 〈t− T 〉0 = 0

for t < T and 〈t − T 〉0 = 1 for

t > T .

Calculate the force F (t).

F (t)

x(t)

mk

Fig. 5.41

Solution First we calculate the force F for t < T . Then 〈t−T 〉0 = 0

and the response is given by

x(t) = x0 cost

2t0.

The unknown force follows from the equation of motion:

F = mx+ kx .

With

x = − x02t0

sint

2t0and x = − x0

4t20cos

t

2t0

we obtain

F (t) =

(−mx0

4t20+ kx0

)cos

t

2t0.

Inserting the given numerical values of the parameters leads to

F (t) ≡ 0 for t < T .

Gro

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156 5 Vibrations

Now we consider the case t > T . Then 〈t − T 〉0 = 1 and the

response follows as

x(t) = x0

[cos

t

2t0+ 20

(1− cos

t− T2t0

)]

→ x =x02t0

[− sin

t

2t0+ 20 sin

t− T2t0

]

→ x =x04t20

[− cos

t

2t0+ 20 cos

t− T2t0

].

Introduction into the equation of motion yields

F (t) =

(−mx0

4t20+ kx0

)cos

t

2t0

+

(20mx04t20

− 20kx0

)cos

t− T2t0

+ 20kx0

→ F (t) ≡ 20 N for t > T .

105

20

10

30

t[s]

F (t) [N]

Gro

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5 Vibrations 157

E5.23Example 5.23 A simplified model of a car (mass m) is given by

a spring-mass system (Fig. 5.42). The car drives with constant

velocity v0 over an uneven surface in the form of a sine function

(amplitude U0, wavelength L).

a) Derive the equation of moti-

on and determine the forcing

frequency Ω.

b) Find the amplitude of the ver-

tical vibrations as a function

of the velocity v0.

c) Calculate the critical velocity

vc (resonance!).

x

L

cU0

v0

m

Fig. 5.42

Solution a) We denote the verti-

cal displacement of the car by x,

the uneven surface is described

by u. Then Newton’s law reads

↑ : mx = −k (x − u) .

With the position of the car, s =

v0t, in the horizontal direction

we obtain

LU0

u

s

m

x

u

u = U0 cos2πs

L= U0 cos

2πv0t

L= U0 cosΩt .

Thus,

mx+ k x = k U0 cosΩt with Ω =2πv0L

.

b) We assume the solution of the equation of motion to be of

the form of the right-hand side: x = x0 cosΩt. This leads to the

amplitude of the steady state vibrations:

x0 =U0

1− Ω2

ω2

=U0

1− 4π2v20L2

m

k

where ω2 = k/m.

c) Resonance occurs for Ω = ω:

4π2v2cL2

m

k= 1 → vc =

L

2π

√k

m.

Gro

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158 5 Vibrations

E5.24 Example 5.24 A homogeneous wheel (mass m) is attached to a

spring (spring constant k). The

wheel rolls without slipping on

a rough surface which moves

according to the function u =

u0cosΩt (Fig. 5.43).

a) Determine the amplitude of

the steady state vibrations.

x

k C

m

r

uµ0

Fig. 5.43

b) Calculate the coefficient µ0 of static friction which is necessary

to prevent slipping.

Solution The equations of mo-

tion for the wheel are given by

↑ : 0 = N −mg , (a)

→ : mx = −k x+H , (b)

y

C : ΘC ϕ = −r H . (c)


x

ϕr

H

kx

C

mg

N

x = u+ rϕ → x = u+ rϕ = −u0 Ω2 cosΩt+ rϕ

and ΘC = mr2/2 we obtain from (b) and (c) the differential equa-

tion for forced vibrations:

x+2

3

k

mx = −1

3u0Ω

2 cosΩt .

a) We assume the solution to be of the form of the right-hand

side: x = x0 cosΩt. This leads to the amplitude of the steady state

vibrations:

|x0| =u0

3

∣∣∣∣2

3

k

mΩ2− 1

∣∣∣∣.

b) The condition |H |max ≤ µ0N for static friction and (a), (b)

yield the required coefficient of static friction:

µ0 ≥|H |max

N=

∣∣∣∣x+k

mx

∣∣∣∣max

g=

∣∣∣∣−x0Ω2 +k

mx0

∣∣∣∣g

→ µ0 ≥u0Ω

2

3g

∣∣∣∣∣∣∣

k

mΩ2− 1

2

3

k

mΩ2− 1

∣∣∣∣∣∣∣.

Gro

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5 Vibrations 159

E5.25Example 5.25 A small homogeneous disk (mass m, radius r) is at-

tached to a large homogeneous disk (mass M , radius R) as shown

in Fig. 5.44. The torsion spring

(spring constant kT ) is unstret-

ched in the position shown.

Determine the eigenfrequency

of the oscillations. Assume small

amplitudes.

M AR

a

m

kT

r

g

Fig. 5.44

Solution We apply the principle

of angular momentum to derive

the equation of motion:

x

A : ΘAϕ = −kTϕ−mga sinϕ .

In the case of small amplitudes

(sinϕ ≈ ϕ) this equation reduces

to

mg

kTϕ

A

ϕ

ϕ+ ω2ϕ = 0 with ω2 =kT +mga

ΘA.

Inserting the mass moment of inertia

ΘA =MR2

2+

[mr2

2+ma2

]

we can write the eigenfrequency in the form

ω =

√√√√√kT +mga

1

2MR2 +m(

r2

2+ a2)

.

Note that the problem can also be solved with the aid of the

conservation of energy (we choose V = 0 for ϕ = 0):

T + V = const → 1

2ΘAϕ

2 +1

2kTϕ

2 +mga(1− cosϕ) = const.


ΘAϕϕ+ kTϕϕ+mga sinϕ ϕ = 0 .

With sinϕ ≈ ϕ and ϕ 6= 0 we obtain the same result as above.Gro

ss, H

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160 5 Vibrations

E5.26 Example 5.26 The systems ➀

and ➁ in Fig. 5.45 consist of

two beams (negligible masses,

flexural rigidity EI), a spring

(spring constant k) and a box

(mass m).

Determine the spring con-

stants k∗ of the equivalent

springs for the two systems.

a2a

EIEIk

m

EIEI

m

k

1

2

Fig. 5.45

Solution We reduce both sys-

tems to the equivalent sim-

ple systems of a mass and a

spring.

1l , EI

w

In system ①, the three “springs” are attached to the mass. The-

refore, they undergo the same deflection when the box is displaced:

they act as springs in parallel. The equivalent spring constant k∗

is the sum of the individual spring constants:

k∗ =∑

ki .

We obtain the spring constants kL and kR of the right and the

left beam, respectively, if we subject the cantilevers to a force 1

at their free ends. The corresponding deflections are

w =1 · l33EI

(see Engineering Mechanics 2: Mechanics of Materials, Section

4.5) which leads to

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5 Vibrations 161

kL =1

wL=

3EI

(2a)3, kR =

1

wR=

3EI

a3.

Thus,

k∗ = kL + kR + k =27

8

EI

a3+ k =

27EI + 8ka3

8a3.

m m m

kR k

kk

k∗kL

Now we consider system ②. Here, the two beams act as springs

in parallel with equivalent spring constant k. This then acts in

series with given spring (spring constant k). Hence,

k = kL + kR =27

8

EI

a3,

1

k∗=

1

k+

1

k=

8a3

27EI+

1

k

→ k∗ =27EIk

27EI + 8ka3=

27EI

8a3 + 27EI

k

.

Note that the stiffness of system ② is smaller than the one of

system ①. Therefore it vibrates with a smaller frequency.

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162 5 Vibrations

E5.27 Example 5.27 A homogeneous wheel (mass m, moment of inertia

ΘC , radius r) rolls without slipping on a rough beam (mass M).

The beam moves without

friction on roller supports

(Fig. 5.46).

Determine the natural fre-

quency of the system.

k C

m,ΘC

r

M

Fig. 5.46

Solution We separate the wheel

and the beam and introduce the

coordinates x1, x2 and ϕ (see

the figure). The coordinates are

measured from the position of

equilibrium. Then the equati-

ons of motion are

① → : mx1 = −kx1 −H ,

y

C : ΘC ϕ = r H ,

② → : M x2 = H .

If we use the kinematic relation

1

2 BA

x1

x2

H

N

ϕr

H

mg

Ckx1

x1 = x2 + rϕ → x1 = x2 + rϕ → x1 = x2 + rϕ

and solve for x1 we obtain

x1 +k

m+M

1 +Mr2/ΘC

x1 = 0 .

Thus, the natural frequency is given by

ω =

√√√√√k

m+M

1 +Mr2/ΘC

.

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auge

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6Chapter 6

Non-Inertial Reference Frames

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164 6 Non-Inertial Reference Frames

E6.5 Example 6.5 Point A of the simple

pendulum (mass m, length l) in

Fig. 6.8 moves with a constant ac-

celeration a0 to the right.

Derive the equation of motion.

g

a0A

lϕ

m

Fig. 6.8

Solution We introduce the ξ, η-co-

ordinate system as shown in the fi-

gure. It is a translating coordinate

system with point A as the origin.

The equation of motion in the mo-

ving system is

mar = F + Ff .

The (real) force F acting at the

mass is given by

F = −S sinϕeξ + (S cosϕ−mg)eη

and the fictitious force Ff is

W =mg

eη

S

η

Aeξ

ϕξ

m

Ff = −maf = −ma0eξ .

Note that the Coriolis force is zero since ω = 0.

The components of the relative acceleration ar follow from the

coordinates of the point mass in the moving system through dif-

ferentiation:

ξ = l sinϕ, η = − l cosϕ,ξ = lϕ cosϕ, η = lϕ sinϕ,

ξ = lϕ cosϕ− lϕ2 sinϕ, η = lϕ sinϕ+ lϕ2 cosϕ .

Gro

ss, H

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6 Non-Inertial Reference Frames 165

This yields the relative acceleration

ar = ξ eξ + η eη = (lϕ cosϕ− lϕ2 sinϕ)eξ

+(lϕ sinϕ+ lϕ2 cosϕ)eη .

Introduction into the equation of motion leads to the components

of the equation of motion in the direction of the axes ξ and η:

m (lϕ cosϕ− lϕ2 sinϕ) = −S sinϕ−ma0 ,

m (lϕ sinϕ+ lϕ2 cosϕ) = S cosϕ−mg.

These are two equations for the unknowns ϕ and S. Solving for ϕ

yields the equation of motion

lϕ+ g sinϕ+ a0 cosϕ = 0 .

Note that the position ϕ0 = − arctana0/g is obtained for ϕ = 0.

The pendulum oscillates about this position for ϕ 6= 0.

Gro

ss, H

auge

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E6.6 Example 6.6 The two disks in

Fig. 6.9 rotate with constant

angular velocities Ω and ω

about their respective axes.

Determine the absolute ac-

celeration of point P at the in-

stant shown.

ϕ

Ω

Pr

ω

a

z

xy

Fig. 6.9

Solution We describe the mo-

tion of point P in a coordina-

te system x, y, z which is fixed

to the large disk. The absolute

acceleration of P is

aP = af + ar + ac ,

where the acceleration of the

reference frame and the relati-

ve acceleration are given by

ϕPr

z

y

x

af =

0

−(a+ r cosϕ)Ω2

0

, ar =

0

−rω2 cosϕ

−rω2 sinϕ

.

We also write the angular velocity of the reference frame and the

relative velocity as column vectors:

Ω =

0

0

Ω

, vr =

0

−rω sinϕ

rω cosϕ

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ss, H

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and we calculate the Coriolis acceleration:

ac = 2Ω× vr → ac =

2rωΩ sinϕ

0

0

.

Combining yields

aP =

2rωΩ sinϕ

−(a+ r cosϕ)Ω2 − rω2 cosϕ

−rω2 sinϕ

.

Gro

ss, H

auge

r, Sc

hröd

er, W

all,

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idje

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E6.7 Example 6.7 A horizontal circular platform (radius r) rotates with

constant angular velocity Ω (Fig. 6.10). A block (massm) is locked

in a frictionless slot at a distance a

from the center of the platform. At

time t = 0 the block is released.

Determine the velocity vr of the

block relative to the platform when

it reaches the rim of the platform.

r a

m

Ω

Fig. 6.10

Solution We describe the motion of

the block in a coordinate system x, y

which is fixed to the platform. The

absolute acceleration of the block is

given by

a

y x

aB = af + ar + ac.

Here, the acceleration of the reference frame, the relative accele-

ration and the Coriolis acceleration are

af =

[−xΩ2

0

], ar =

[x

0

], ac =

[0

2Ωx

].

Thus, the absolute acceleration becomes

aB =

[x− Ω2x

2Ωx

].

The equation of motion for the block is

maB = F ,

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where the force F (which is exerted from the slot on the block) is

F =

[0

Fy

].

We now write down the x-component of the equation of motion:

m(x − Ω2x) = 0 → x− Ω2x = 0 .

The general solution of this differential equation is given by

x(t) = A coshΩt+B sinhΩt .

With the initial conditions

x(0) = a → A = a ,

x(0) = 0 → B = 0

we obtain

x(t) = a coshΩt .

When the block reaches the rim of the platform, the condition

x(tR) = r → coshΩtR = r/a

is satisfied. Thus the relative velocity is (note that cosh2 x −sinh2 x = 1)

x(tR) = aΩ sinhΩtR → x(tR) = Ω√r2 − a2 .

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E6.8 Example 6.8 A simple pendulum is

attached to point 0 of a circular disk

(Fig. 6.11). The disk rotates with

a constant angular velocity Ω; the

pendulum oscillates in the horizon-

tal plane.

Determine the circular frequency

of the oscillations. Assume small

amplitudes and neglect the weight

of the mass.

l mr

0

Ω

Fig. 6.11

Solution We introduce a rotating ξ, η, ζ-coordinate system. Then

Ω = Ωeζ , a0 = r0 = −rΩ2eξ ,

Ω = 0 , r0P = l cosϕ eξ + l sinϕ eη .

The relative velocity can be ex-

pressed by the relative angular ve-

locity ϕ∗ (∗: time derivative rela-

tive to the moving frame):

vr

r0P

ϕr0

P

ξ

Ωη

vr = lϕ∗ → vr = −lϕ∗ sinϕ eξ + lϕ∗ cosϕ eη .

Thus, the fictitious forces F f and F c are

F f = − ma0 −mΩ× (Ω× r0P ) = m(rΩ2 + lΩ2 cosϕ)eξ

+ mΩ2l sinϕ eη ,

F c = − 2mΩ× vr = 2mΩlϕ∗(eξ cosϕ+ eη sinϕ) .

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ss, H

auge

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With the tangential relative acceleration art = lϕ∗∗ the equa-

tion of motion in the tangential direction is obtained as

տ: mlϕ∗∗ = m(lΩ2 + 2lϕ∗Ω) sinϕ cosϕ

− m[rΩ2 + (lΩ2 + 2lϕ∗Ω) cosϕ] sinϕ

= − mrΩ2 sinϕ .

We assume small amplitudes

(sinϕ ≈ ϕ). This yields

ϕ∗∗ +rΩ2

lϕ = 0 .

ξ

m

art

ϕS

mΩ2l sinϕ

2mlϕ∗Ω sinϕ

η2mlϕ∗Ωcosϕ

m(rΩ2+lΩ2 cosϕ)

Hence, the circular frequency of the oscillations is

ω =√r/l Ω .

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E6.9 Example 6.9 A drum rotates with

angular velocity ω about point B

(Fig. 6.12). Pin C is fixed to the

drum; it moves in the slot of link

AD.

Determine the angular velocity

ωAD of link AD and the velocity

vr of the pin relative to the link at

the instant shown. ω

A

C

B

D

4 l

3 l

Fig. 6.12

Solution We use the rotating co-

ordinate system x, y as shown in

the figure. The (absolute) velocity

of pin C is given bya

β

A

C

y

3 ℓ

x

4 ℓ

vC = 3lω

[cosβ

− sinβ

].

With the geometrical relations

a =√16l2 + 9l2 = 5l , sinβ = 3/5 , cosβ = 4/5

we can write

vC =3lω

5

[4

−3

].

The velocity of the reference frame at point C and the velocity of

pin C relative to the moving frame are

vf = βl

[0

5

], vr = vr

[1

0

].

With

vC = vf + vr → 3lω

5

[4

−3

]= βl

[0

5

]+ vr

[1

0

],

we finally obtain

ωAD = β = −9ω

25, vr =

12

5ωl

[1

0

].G

ross

, Hau

ger,

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öder

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E6.10Example 6.10 A point P moves along a circular path (radius r)

on a platform with a constant relative velocity vr (Fig. 6.13). The

platform rotates with a con-

stant angular velocity ω about

point A. The eccentricity e is gi-

ven.

Determine the relative, fixed

frame-, Coriolis, and absolute

accelerations of P .

0

vr

A e

ω rP

Fig. 6.13

Solution We introduce the coor-

dinate system ξ, η, ζ as shown in

the figure. Its origin is located at

the center 0 of the platform; it

rotates with the platform. Thus

point P undergoes a circular

motion relative to this system.

With the magnitude ar = v2r/r

of the relative acceleration and

its direction (from P to 0) we

can write

ξϕ

η

0

vr

A e

ω r0P

P

ar = −v2r

r(eξ cosϕ+ eη sinϕ) .

With

ω = ωeζ , ω = 0 , r0P = eξr cosϕ+ eηr sinϕ ,

vr = vr(−eξ sinϕ+ eη cosϕ) , r0 = a0 = −e ω2eξ

we obtain

af = a0 + ω × (ω × r0P )

= −e ω2eξ + rω2[eζ × (eζ × eξ cosϕ) + eζ × (eζ × eη sinϕ)]

= −(e+ r cosϕ)ω2eξ − rω2 sinϕeη ,

ac = 2ω × vr = 2ωvr[eζ × (−eξ sinϕ) + eζ × eη cosϕ]

= −2ωvr(eξ cosϕ+ eη sinϕ) .Gro

ss, H

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Thus, the absolute acceleration is found as

a = af + ar + ac

= −[eω2 + (rω2 +v2rr

+ 2ωvr) cosϕ]eξ − [rω2 +v2rr

+ 2ωvr] sinϕ eη

= −[eω2 + r(ω +vrr)2 cosϕ]eξ − r(ω +

vrr)2 sinϕ eη .

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E6.11Example 6.11 A circular ring (ra-

dius r) rotates with constant an-

gular velocity Ω about the x-

axis (Fig. 6.14). A point mass m

moves without friction inside the

ring.

Derive the equations of moti-

on and determine the equilibrium

positions of the point mass relati-

ve to the ring.

Ω

g

m

x

yz

r

ϕ

Fig. 6.14

Solution We describe the motion of the point mass in the x, y, z-

coordinate system (see the figure) which rotates with the ring.

The absolute acceleration a of the point mass is given by

a = af + ar + ac ,

where the acceleration of the reference frame and the relative ac-

celeration are

af =

0

−rΩ2 sinϕ

0

, ar =

−rϕ2 cosϕ− rϕ sinϕ

−rϕ2 sinϕ+ rϕ cosϕ

0

.

With the angular velocity of the ring and the relative velocity

Ω =

Ω

0

0

, vr =

−rϕ sinϕ

rϕ cosϕ

0

we can calculate the Coriolis acceleration

ac = 2Ω× vr → ac =

0

0

2rΩϕ cosϕ

.

The equation of motion is given by

ma = W +N ,Gro

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auge

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where

W =

−mg0

0

is the weight of the point mass and

N =

Nx

Ny

Nz

with Ny/Nx = tanϕ

is the force exerted from the ring on the point mass. Now, we can

write down the components of the equation of motion:

−m(rϕ2 cosϕ+ rϕ sinϕ) = −mg +Nx ,

−m(rϕ2 sinϕ− rϕ cosϕ+ rΩ2 sinϕ) = Nx tanϕ ,

2mrΩϕ cosϕ = Nz .

Positions of equilibrium relative to the ring are characterized by

ϕ = 0 , ϕ = 0 →(rΩ2 +

g

cosϕ

)sinϕ = 0 .

This yields

ϕ1 = 0 , ϕ2 = π , ϕ3,4 = π ± arccosg

rΩ2.

The positions ϕ3,4 exist only for Ω2 > g/r.

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E6.12Example 6.12 A point P moves

on a square plate along a cir-

cular path (radius r) with a

constant relative velocity vr.

The plate moves horizontally

with the constant acceleration

a0 (Fig. 6.15).

Determine the magnitude of

the absolute acceleration of P .

a0

vrP

r

Fig. 6.15

Solution The components of the

relative velocity in the moving

reference frame ξ, η are

vrξ = ξ∗ = −vr sinϕ ,

vrη = η∗ = vr cosϕξ0

y

x

fixed

η

ϕ

r

vrP

(∗: time derivative relative to the moving frame). Differentiation

with respect to the moving frame leads to the components of the

relative acceleration (note: rϕ∗ = vr ):

arξ = ξ∗∗ = −vrϕ∗ cosϕ = −v2r

rcosϕ ,

arη = η∗∗ = −vrϕ∗ sinϕ = −v2r

rsinϕ .

Since the reference frame undergoes a translation, the absolute

acceleration is given by

ax = a0 + arξ = a0 −v2rr

cosϕ ,

ay = arη = −v2r

rsinϕ .

It has the magnitude

a =√a2x + a2y =

√a20 +

v4rr2− 2a0

v2rr

cosϕ .

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E6.13 Example 6.13 A crane starts to move from rest with a constant

acceleration b0 along a straight track. At the same time, the jib

begins to rotate with constant angu-

lar velocity ω, and the trolley on the

jib begins to move towards point 0

with constant relative acceleration bc(Fig. 6.16). The initial positions of the

jib and of the trolley are given by ϕ0

and s0.

Determine the absolute velocity and

the absolute acceleration of the trolley

as functions of the time t.

0

ϕ

s

b0

bc

b0

ω

bc

Fig. 6.16

Solution We use the fixed coordinate system x, y, z, where the

x-axis coincides with the track. In ad-

dition, we introduce the rotating coor-

dinate system ξ, η, ζ, where the ξ-axis

rotates with the jib. Then, the gene-

ral equations for the absolute velocity

and the absolute acceleration are

r0P

ζ

ηy

xz

0

P

ω

ξ

ϕ

v = v0 + ω × r0P + vr ,

a = a0 + ω × r0P + ω × (ω × r0P ) + 2ω × vr + ar .

We measure the time t from the beginning of the motion. Then,

making use of the given accelerations, angular velocity and initial

conditions, we obtain

a0 = b0ex → v0 = b0t ex ,

ω = ωeζ , ω = 0 ,

ar = −bceξ → vr = −bct eξ → r0P = (− 12bct

2 + s0)eξ ,

and

ω × r0P = ω(− 12bct

2 + s0)eη , 2ω × vr = −2ω bct eη ,ω × (ω × r0P ) = −ω2(− 1

2bct2 + s0)eξ .

Gro

ss, H

auge

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all,

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With the relation ex = eξ cosϕ−eη sinϕ, where ϕ = ϕ0 +ω t, we

finally obtain

v = [b0t cosϕ− bct]eξ + [−b0t sinϕ+ ω(− 12 bct

2 + s0)]eη ,

a = [b0 cosϕ− ω2(− 12bct

2 + s0)− bc]eξ − [b0 sinϕ+ 2ω bct]eη .

Gro

ss, H

auge

r, Sc

hröd

er, W

all,

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idje

e

Engi

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amics

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013

solutionsto supplementary problems - springer · d.gross•w.hauger j.schr¨oder •w.a.wall...

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