9 - 1 conservative vs. non-conservative forces the fundamental theorem of calculus can be written as...

9 - 1

Conservative vs. Non-conservative Forces

)()()(

12

2

1

xfxfdxdx

xdfx

x

The fundamental theorem of calculus can be written as follows:

Why is this important? Because from it you can derive the definition of conservative forces, which is broadly defined as:

0)(

ifi xxfxfdxdx

xdf

The circle is used to denote that the limits of integration occur on a closed path, and thus the positions for evaluating the integral (i.e. at the initial and final positions) occur at the same point.

9 - 2

)()( 21

2

1

2

1

xUxUdxdx

xdUFdx

x

x

x

x

12

2

1

2

1

2

1

mgxmgxdxdx

mgxdmgdxFdx

x

x

x

x

x

x

2x

What if the force was due to gravity?

Starting from the force equation, let’s re-write force as the derivative of a scalar function, U, and evaluate the integral as the derivative of U is integrated around a closed path.

1x3x

23

3

2

mgxmgxFdxx

x

31

1

3

mgxmgxFdxx

x

9 - 3

ififif xxmgmgxmgxxUxUU

Now let’s assume that the particle’s motion occurred around a closed path (meaning that it eventually returned to its starting position). Evaluating the integral

Let’s now define the scalar quantity U as the potential energy.

0312312

1

3

3

2

2

1

mgxmgxmgxmgxmgxmgxFdx

FdxFdxFdxFdxx

x

x

x

x

x

shows that the integral of gravitational force around a closed path is zero. In other words, one can say that the state at which one can find the particle under the action of gravity is independent the path it takes between two points. Only the change in the scalar quantity U is required to compute the change in state of its initial and final positions. (i.e.)

9 - 4

2222

2

1

2

1

2

1ififif xxkkxkxxUxUU

Now let’s assume that the particle’s motion around a closed path is due to the force by a spring…

The scalar quantity U is now called the potential energy of the spring.

02

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

23

21

22

23

21

22

21

22

2

1

3

3

2

2

1

2

1

2

1

2

1

kxkxkxkxkxkxFdx

FdxFdxFdxFdx

kxkxdxkxdx

dkxdxFdx

x

x

x

x

x

x

x

x

x

x

x

x

The integral of the force due to a spring around a closed path is zero, and thus a spring is a conservative force. If we were to evaluate the potential energy of a spring at two different points:

9 - 5

Now let’s generalize this to 3-dimensional vector fields. Assume the you have a force vector written in Cartesian coordinates.

zyxUzyxF

zz

zyxUy

y

zyxUx

x

zyxUzyxF

zFyFxFzyxF zyx

,,,,

ˆ,,

ˆ,,

ˆ,,

,,

ˆˆˆ,,

fi

r

r

r

r

rUrUrrUrrFf

i

f

i

where is denoted the gradient operator

In order to test this concept, let’s look at the 3-D gravitational field...

9 - 6

rmGmrU

rUr

mGmr

rmGmF

1

1ˆ

21

21221

The force of gravity can be written as the gradient of a scalar function:

Since the integral of the force over a closed path is zero, gravity can be defined as a conservative force even in three dimensions.

011

2121 ii r

mGmr

mGmrrUrrF

Which leads to an expression for the gravitational potential energy defined purely as a scalar of the inverse distance away from the source.

9 - 7

What are some examples of non-conservative forces?

DvxDxFf

1212

12

2

1

2

1

xxt

xxDxxDvdxDvdxxF

x

x

x

x

f

2121

21

1

2

1

2

xxt

xxDxxDvdxDvdxxF

x

x

x

x

f

Friction and any processes that causes dissipation of heat.

0

221

212

1

2

2

1

t

xxD

t

xxDdxxFdxxFdxxF

x

x

f

x

x

ff

For example, the energy dissipated in going from position 1 to position 2, assuming velocity is constant (this assumption does not matter, but just facilitates the derivation), is given by:

Likewise, the energy dissipated in going from position 2 back to position 1 is:

Thus, the total energy dissipated in a closed path is:

which is not zero! Thus friction is a non-conservative force (i.e. is path dependent).

9 - 8

What about kinetic energy?

TUrmrmrUrU

rmrmrmrrmdrdtdt

rdm

dtrrmdtdt

rdrmrdrmrdrF

rUrUdxrUrdrF

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

21

2221

21

22

2

21

2

1

2

1)()(

2

1

2

1

2

1

)()(

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

Starting from the previous derivation for potential energy

Thus the change in potential energy, U, due to movement in a conservative force from position 1 to position 2 is re-allocated as kinetic energy, T. Another way to say this is that the sum of potential energy and kinetic energy must always remain constant. (i.e.) This is a very powerful result because now we only have to look at the initial and final states instead of integrating through the entire path of motion.

.1122 constTUTU

9 - 10

Solution by energy method

Velocity is zero, so T=0 Potential energy is U=mgy

AB

ABAABBAB y

x

yxyyxxyx

dt

d 02222

22222ABBABA yxlyyxx

ffBiAiii TUmgymgyTU 0

Constraints: Bar has constant length

22

2

1

2

1

2ABA ymxmmgy

lmg

We must find the position at which the acceleration of slider B is zero. This may be an inflection point (i.e. where the velocity is also zero), it may be a point where the velocity is at a local maxima, or where the velocity is at a global maximum.

Initial conditions at time t=0

9 - 11

2

222

22

22

A

BBAAAB y

xxy

lgyy

lgx

A

ABA

AB

A

BB y

l

l

gyxy

lg

y

lx

y

xx

2

2

221

2

22

2

22

2

2

A

A

BB y

lg

y

xx

221

2

2

032

322

2

2

22

22

222

22

AAB

AB

AAAAAA

BBB

ylyxl

ygx

yyyyl

l

gy

l

l

gy

dt

dxxx

dt

d

Solving for the velocity of slider B:

So, what conditions leads to the acceleration of slider B equal to zero?

9 - 12

032 2 AA yly

0Ay

lyA 3

2

The condition that must be satisfied for the acceleration of slider B to equal zero is shown below.

There are two solutions. One of them is: , however this is an inflection point, where the velocity is changing its direction. The other solution is:

9

21lxB

54

8

54

8

2

22

22 gl

xgl

yl

l

gyx BA

AB

Due to the constraints of the length of the bar, the x-position is therefore:

And therefore the maximum velocity is given as:

9 - 13

r = 600 mm

C

A

B

200 g

O

Problem 1

A small 200-g collar C can slide on asemicircular rod which is made to rotateabout the vertical AB at the constant rate of6 rad/s. Determine the minimum requiredvalue of the coefficient of static frictionbetween the collar and the rod if the collaris not to slide when (a) = 90o, (b) = 75o,(c) = 45o. Indicate in each case thedirection of the impending motion.

9 - 14

Problem 1

r = 600 mm

C

A

B

200 g

O


1. Kinematics: Determine the acceleration of the particle.

2. Kinetics: Draw a free body diagram showing the appliedforces and an equivalent force diagram showing the vectorma or its components.

9 - 15

Problem 1

r = 600 mm

C

A

B

200 g

O


3. Apply Newton’s second law: The relationship between theforces acting on the particle, its mass and acceleration is givenby F = m a . The vectors F and a can be expressed in terms ofeither their rectangular components or their tangential and normalcomponents. Absolute acceleration (measured with respect toa newtonian frame of reference) should be used.

9 - 16

Problem 1 Solution

r = 600 mm

C

A

B

200 g

O

Kinematics.

r = 600 mm

C

A

B

O

an

r sin

an = (r sin) 2

an = (0.6 m) sin ( 6 rad/s )2

an = 21.6 sin m/s2

9 - 17

Kinetics; draw a free body diagram.

Problem 1 Solution

r = 600 mm

C

A

B

200 g

O

(0.2 kg)(9.81 m/s2)

O

N

F

man = (0.2) 21.6 sin = 4.32 sin N

=

9 - 18

Problem 1 Solution

Apply Newton’s second law.

(0.2 kg)(9.81 m/s2)

O

N

F

man = (0.2) 21.6 sin = 4.32 sin N

=

+ Ft = 0: F - 0.2 (9.81) sin = - 4.32 sin cos

F = 0.2 (9.81) sin - 4.32 sin cos

+ Fn = man: N - 0.2 (9.81) cos = 4.32 sin sin

N = 0.2 (9.81) cos + 4.32 sin2

F = N

For a given , the values of F , N , and can be determined

9 - 19

Problem 1 Solution

(0.2 kg)(9.81 m/s2)

O

N

F

man = (0.2) 21.6 sin = 4.32 sin N

=

Solution:

(a) = 90o, F = 1.962 N, N = 4.32 N, = 0.454 (down)

(b) = 75o, F = 0.815 N, N = 4.54 N, = 0.1796 (down)

(c) = 45o, F = -0.773 N, N = 3.55 N, = 0.218 (up)

9 - 20

b

b

r

O

A

B CD

E

Problem 2

Pin B weighs 4 oz and is free to slidein a horizontal plane along the rotatingarm OC and along the circular slot DEof radius b = 20 in. Neglecting frictionand assuming that = 15 rad/s and = 250 rad/s2 for the position = 20o,determine for that position (a) theradial and transverse components ofthe resultant force exerted on pin B,(b) the forces P and Q exerted on pinB, respectively, by rod OC and the wallof slot DE.

...

9 - 21

Problem 2

b

b

r

O

A

B CD

E

1. Kinematics: Examine the velocity and acceleration of theparticle. In polar coordinates:

v = r er + r e

a = (r - r 2 ) er + (r + 2 r ) e

.

.. .

.

... . r = r er

eer

Pin B weighs 4 oz and is free to slidein a horizontal plane along the rotatingarm OC and along the circular slot DEof radius b = 20 in. Neglecting frictionand assuming that = 15 rad/s and..

.

= 250 rad/s2 for the position = 20o, determine for that position(a) the radial and transverse components of the resultant forceexerted on pin B, (b) the forces P and Q exerted on pin B,respectively, by rod OC and the wall of slot DE.

9 - 22

Problem 2


.


b

b

r

O

A

B CD

E


9 - 23

Problem 2


.


b

b

r

O

A

B CD

E

3. Apply Newton’s second law: The relationship between theforces acting on the particle, its mass and acceleration is givenby F = m a . The vectors F and a can be expressed in terms ofeither their rectangular components or their radial and transversecomponents. With radial and transverse components:

Fr = m ar = m ( r - r 2 ) and F = m a = m ( r + 2 r ) .. . .. . .

9 - 24

Problem 2 Solution

Kinematics.b

b

r

O

A

B CD

E

b

r

O

A

B

b

r = 2 b cos

r = - 2 b sin

r = - 2 b sin - 2 b cos 2..

. .

...

= 20o

= 15 rad/s

= 250 rad/s2..

.

9 - 25

Problem 2 Solution

b

r

O

A

B

b

r = 2 b cos , r = - 2 b sin , r = - 2 b sin - 2 b cos 2... . ...

For: b = 20/12 ft, = 20o,

= 15 rad/s = 250 rad/s2

r = 2 (20/12 ft) cos 20o = 3.13 ft

r = - 2 (20/12 ft) sin 20o (15 rad/s) = - 17.1 ft/s

r = -2(20/12 ft) sin 20o (250 rad/s2 ) - 2(20/12 ft) cos 20o (15 rad/s)2

r = - 989.79 ft/s2

.

. ..

..

..

9 - 26

Problem 2 Solution

b

b

r

O

A

B CD

EKinetics; draw a free body diagram.

r

O A

B

r

O A

B

Fr

F

mar

ma

=

(a) Radial and transversecomponents of the resultant forceexerted on pin B.

9 - 27

Problem 2 Solution

r

O A

B

r

O A

B

Fr

F

mar

ma

=


+ Fr = mar: Fr = m ( r - r 2 )

Fr = [- 989.79 - ( 3.13 )(152 )] = -13.16 lb Fr = 13.16 lb

+ F = ma: F = m ( r + 2 r )

F = [(3.13)(250) + 2 (-17.1)(15)] = 2.1 lb

(4/16)32.2

.

.

.

..

(4/16)32.2

..

F = 2.10 lb

9 - 28

Problem 2 Solution

b

b

r

O

A

B CD

E

r

O A

B

r

O A

B

Fr

F

=Q

P

Fr = - Q cos

-13.16 = - Q cos 20o

Q = 14.00 lb 40o

F = - Q sin + P

2.10 = - 14.0 sin 20o + P

P = 6.89 lb 20o

9 - 29

400 mm100 mm

A B

Problem 3

A 250-g collar can slide on ahorizontal rod which is free to rotateabout a vertical shaft. The collar isinitially held at A by a cord attachedto the shaft and compresses a springof constant 6 N/m, which isundeformed when the collar islocated 500 mm from the shaft. As

the rod rotates at the rate o = 16 rad/s, the cord is cut and thecollar moves out along the rod. Neglecting friction and themass of the rod, determine for the position B of the collar (a) thetransverse component of the velocity of the collar, (b) the radialand transverse components of its acceleration, (c) theacceleration of the collar relative to the rod.

.

9 - 30

Problem 3400 mm100 mm

A B The collar is initially held at A by acord attached to the shaft andcompresses a spring. As the rodrotates the cord is cut and thecollar moves out along the rod to B.

1. Kinematics: Examine the velocity and acceleration of theparticle. In polar coordinates:

v = r er + r e

a = (r - r 2 ) er + (r + 2 r ) e

.

.. .

.

... . r = r er

eer

9 - 31



2. Angular momentum of a particle: Determine the particlevelocity at B using conservation of angular momentum. In polarcoordinates, the angular momentum HO of a particle about O isgiven by

HO = m r vThe rate of change of the angular momentum is equal to the sumof the moments about O of the forces acting on the particle.

MO = HO

If the sum of the moments is zero, the angular momentum isconserved and the velocities at A and B are related by m ( r v)A = m ( r v)B

.

9 - 32




9 - 33

Problem 3400 mm

100 mm

A B

4. Apply Newton’s second law: The relationship between theforces acting on the particle, its mass and acceleration is givenby F = m a . The vectors F and a can be expressed in terms ofeither their rectangular components or their radial and transversecomponents. Absolute acceleration (measured with respect toa Newtonian frame of reference) should be used.With radial and transverse components:

Fr = m ar = m ( r - r 2 ) and F = m a = m ( r + 2 r ) ..... ..

The collar is initially held at A by acord attached to the shaft andcompresses a spring. As the rodrotates the cord is cut and thecollar moves out along the rod to B.

9 - 34

Problem 3 Solution400 mm100 mm

A BKinematics.

A B

v

a

ar

.

v = r

ar = r - r 2

a = r + 2 r

...

....

.r

9 - 35


A B

Angular momentum of a particle.

A B

(v.

(v

rB = 0.4 m

rA = 0.1 m

m rA (v)A = m rB (v)B

since (v)A = rA

(v)B =

(v)B = (16 rad/s)

(v)B = 0.4 m/s

(rA)2

rB

( 0.1 m )2

0.4 m

(a) The transverse component of the velocity of the collar.

.

.

r

9 - 36


A B(b) The radial and transverse components of acceleration.

Kinetics; draw a freebody diagram.

F m ar

m a

=

Only radial force F (exerted by the

spring) is applied to the collar.

For r = 0.4 m:

F = k x = (6 N/m)(0.5 m - 0.4 m)

F = 0.6 N

+ Fr = mar: 0.6 N = (0.25 kg) ar

ar = 2.4 m/s2

+ F = ma: 0 = (0.25 kg) a

a = 0


9 - 37


A B(c) The acceleration of the collar relative to the rod.

Kinematics.

A B

v

a

ar

.

r

For r = 0.4 m:

v = r , =

= = 1 rad/s

vr

0.4 m/s0.4 m

ar = r - r 2

(2.4 m/s2) = r - (0.4m)(1 rad/s)2

r = 2.8 m/s2

The acceleration of the collar relative to the rod is r.

..

.

..

.

..

..

..