Rotational Motion

Rotation

So far we have only discussed linear motion, or a circular orbit for the case of centripetal acceleration.

But often objects rotate, we would like to be able to describe this motion too.

And many complex motions, like a diver or a gymnast, can be described as a combination of linear and rotational motion.

Angular QuantitiesTo describe rotational motion, it makes more sense to think in polar coordinates. Thus are coordinates go from x and y to r and θ. ~r

~✓~x

~y

In pure rotation, only the θ coordinate changes. We can define an angular velocity ω and angular acceleration, α

~↵ =d~!

dt~! =

d~✓

dt

The physical length traveled by something in rotational motion is l=Rθ, so

Note in this case ω must be in units or radians per time. For example rad/s.

Likewise the tangential acceleration can be expressed in terms of the angular acceleration for constant R.

and α should have units of rad/s2.

v =dl

dt=

d(R✓)

dt= R

d✓

dt= R!

atan =d2l

dt2=

d2(R✓)

dt2= R

d2✓

dt2= R↵

Example 10-3A carousel is initially at rest. At t=0 it is given a constant angular acceleration α= 0.060rad/s2, which increases its angular velocity for 8.0s. At t=8.0s, determine the magnitude of the following quantities: a) the angular velocity of the carousel; b) the linear velocity of a child located 2.5m from center; c) the tangential (linear) acceleration of the child; d) the centripetal acceleration of the child; and e) the total acceleration of the child.known

α = 0.060 rad/s2

t = 8.0sR = 2.5m

unknownω = ?v = ?

atan = ?aR = ?

! =

Z↵dt = ↵t+ !0

= (0.060rad/s2)(8.0s) = 0.48rad/s

v = R! = (2.5m)(0.48rad/s) = 1.2m/s

atan = R↵

aR =v2

R= !2R

a =q

a2tan + a2R

= (2.5m)(0.060rad/s2) = 0.15m/s2

= (0.48rad/s)2(2.5m) = 0.58m/s2

=p

(0.15m/s2)2 + (0.58m/s2)2 = 0.60m/s2

Interstellar (2014)

ω = 68 rpm = 68 × 2 π /60s =7.1 rad/s

What is the angular velocity of the space station?

What is the angular acceleration of the spacecraft?

It looks like it takes about 20s for them to reach the same angular velocity, so

α = Δω / Δt = 7.1 rad/s / 20s = 0.36 rad/s2

Constant Angular AccelerationIn the case of constant angular acceleration we can integrate just like for constant linear acceleration and get equations of motion for that special case.

constant linear acceleration constant angular accelerationv = v0 + at ! = !0 + ↵t

x = x0 + v0t+1

2at2 ✓ = ✓0 + !0t+

1

2↵t2

v2 = v20 + 2a(x� x0) !2 = !20 + 2↵(✓ � ✓0)

v̄ =1

2(v + v0) !̄ =

1

2(! + !0)

Notice these have to be the same since angular velocity and acceleration are defined the same way as linear velocity and acceleration.

TorqueAngular force is called torque and is represented by the Greek letter τ.

Torque depends on the force and the distance that force is applied from the axis of rotation.

Furthermore, torque depends on the component of that force which is perpendicular to the radius vector (applying a force parallel to the radius vector doesn’t rotate an object).

Torque is given by the equation:

⌧ = ~R⇥ ~F = RF sin ✓

Example 10-7Two thin disk shaped wheels of radii RA = 30cm and RB = 50cm are attached to each other on an axle that passes through the center of each. Calculate the net torque on this compound wheel due to the two forces shown, each with magnitude 50N.

RA

RB

FA

FB

θ = 30°⌧A = RAFA

= (0.30m)(50N)� (0.50m)(50N)(0.866) = �6.7 m ·N

6.7 m.N in the clockwise direction

⌧B = RBFB sin 60

⌧ = RAFA �RBFB sin 60

The Dark Knight (2008)

How much force (torque) must be applied to the truck to flip it?

mg

Fmass = 15000kg

Looking up properties of a 18 wheeler.

length = 16m

height = 4m

In order for the truck to flip the torque from the force F must be greater than the torque from the weight.

⌧ = RF sin ✓ ⌧F =1

2hF ⌧W =

1

2lmg

⌧F > ⌧W =>1

2hF >

1

24hmg => F > 4mg = 4(15000kg)(9.8m/s2) = 588, 000N

That’s a tremendous amount of force and if you watch the clip carefully you’ll see they don’t rotate the truck from the force of the wires, but there is an explosion in the middle of the truck that flips in over. Since the explosion occurs at half the length its force only need to be mg to overcome gravity.

Hellboy (2004)

Moment of InertiaWe now have torque and angular acceleration so we can try to express Newton’s 2nd law for rotational dynamics.

What we need though is the rotational equivalent to mass. We can start by just considering a small ball on a thin string going around in a circle.

So we see that the quantity that acts like mass, called the moment of inertia is equal to mR2. If we now imagine that an extended object is made up of many small elements of mass mi,

F = matan = mR↵ multiply both sides by R ⌧ = RF = mR2↵

X

i

⌧i = (X

i

miR2i )↵ or ⌧ = I↵ where I =

X

i

miR2i

I =

Z⇢(R)R2dV

or

Moment of InertiaYou can work out the

moment of inertia using calculus, but of course

most common shapes have been worked out long ago. The book has any we will

use worked out for you and on the test I’ll give you any

formula you might need (There is no point in memorizing these).

Example 10-11A uniform rod of mass M and length l can pivot freely (i.e. we ignore friction) about a hinge or pin attached to the case of a large machine. The rod is held horizontally and then released. At the moment of release (when you are no longer exerting a force holding it up) determine a) the angular acceleration of the rod, and b) the linear acceleration on the tip of the rod. Assume the force of gravity acts on the center of mass of the rod.

mg

⌧ = RF sin ✓ =1

2lmg

⌧ = I↵ => ↵ = ⌧/I

I =1

3ml2

=12 lmg13ml2

=3

2

g

l

atan = l↵ =3

2g

Example 10-12a) Show that the moment of inertia of a uniform hallow cylinder of inner radius R1 and outer radius R2, and mass M is I=1/2 M(R2

1 +R22),

if the rotation is through the center along the axis of symmetry. b) Obtain the moment of inertia for the for a solid cylinder.

lets consider a small ring of width dr, then for this ring

dm = ⇢(2⇡R)hdR dI = R2dm

I =

ZR2dm =

Z R2

R1

2⇡⇢R3dR

= 2⇡⇢h1

4R4

��R2

R1=

⇡⇢h

2(R4

2 �R41)

M = ⇢V = ⇢⇡(R22 �R2

1)h

I =⇡⇢h

2(R4

2 �R41) =

⇡⇢h

2(R2

2 �R21)(R

22 +R2

1) =1

2M(R2

2 +R21)

if R1 = 0 I = 1/2 M R2

Rotational Kinetic EnergyA rotating object must have kinetic energy just like an object with linear motion.

Let’s consider a bunch of points rotating around a common axis. Their kinetic energy would be:

KE =X 1

2miv

2i =

X 1

2miR

2i!

2i =

1

2

X(miR

2i )!

2 =1

2I!2

So the formula for kinetic energy is what we would guess, just changing velocity to angular velocity and mass to the moment of inertia

KE =1

2I!2

Example 10-15A uniform rod of mass M and is pivoted on a frictionless hinge at one end. The rod is held at rest horizontally and then released. Determine the angular velocity of the rod when it reaches the vertical position and the speed of the rod’s tip at this moment.

l/2

�U = Mgh =1

2Mgl

�KE =1

2I!2

I =1

3Ml2

1

2

1

3Ml2!2 =

1

2Mgl => !2 =

3g

l=> ! =

r3g

l

v = R! = l

r3g

l=

p3gl

Rotational and Translational Motion

We have discussed rotational motion, but many things both move linearly and rotate. The most common example of this are things that roll.

Rolling with out slipping means that v=Rω. In this case the amount something rotates, is directly related to how far it goes.

If slipping occurs, then these two quantities don’t need to be proportional.

Example 10-16What will be the speed of a solid sphere of mass M and radius r0 when it reaches the bottom of an incline if it starts from rest at a vertical height of H and rolls without slipping? Compare to the case of an object sliding down with no rotation.

Hv = R!

E =1

2mv2 +

1

2Iw2 +mgy

Ei = mgy = MgH

Ef =1

2mv2 +

1

2I!2

I =2

5Mr20

MgH =1

2Mv

2 +1

2

�25Mr

20

��v2

r20

�

gH = (1

2+

1

5)v2 => v =

r10

7gH In the case of no rotation we instead have

MgH =1

2Mv

2 v =p2gH

Example 10-19String is wrapped around a uniform solid cylinder (something like a yo-yo) of mass M and radius R, and the cylinder starts falling from rest. As the cylinder falls, find a) its acceleration and b) the tension in the string.

mg

FT

XF = Ma Mg � FT = Ma

X⌧ = I↵ FTR =

1

2MR2↵

I =1

2MR2

because there is no slipping a=Rα,

FTR =1

2MR2(

a

R) =

1

2MRa => FT =

1

2Ma

Mg � 1

2Ma = Ma => Mg =

3

2Ma => a =

2

3g

FT =1

2Ma =

1

3Mg

The Princess Bride (1987)

θ

v = R!

R= 0.5mh= 1oom

Let’s assume no friction, then energy is conserved.

mgh =1

2mv2 +

1

2I!2

m = 80kg

=1

2mv2 +

1

2(1

2mR2)(v/R)2

Let’s assume rolling without slipping.

I =1

2mR2

=3

4mv2

=> v2 =4

3gh => v =

r4

3gh =

r4

3(9.8m/s2)(100m) = 36m/s

Clearly this doesn’t happen. There are nonconservative forces so that at the end they have zero velocity.

Angular MomentumThe last property we have to add to our list of angular properties is angular momentum, which gets the symbol L.

Since p=mv it makes sense that L=Iω.

We saw before that Newton’s second law could be written ΣF = dp/dt. So in angular coordinates we would write:

And just like before, if there are no outside torques we expect the angular momentum to be conserved.

X⌧ =

dL

dt

Conservation of Angular Momentum

If there are no external torques on a system, which is very often the case, then the angular momentum should be conserved.

dL

dt= 0 L = I! = constant

Example 11-2A simple clutch consists of two cylindrical plates that can be pressed together to connect two sections of an axle, as needed, in a piece of machinery. The two plates have masses MA= 6.0kg and MB= 9.0kg with equal radii R0 = 0.60m. They are initially separated. Plate MA is accelerated from rest to an angular velocity ω1 = 7.2rad/s and in a time Δt=2.0s. Calculate a) the angular momentum of plate MA, b) the torque required to have accelerated MA from rest to ω1. Next, plate MB initially at rest but free to rotate without friction is placed in firm contact with freely rotation plate MA and the two plates rotate at constant angular velocity ω2, which is considerably less than ω1. Why does this happen and what is ω2?

LA = IA!1 = (1

2MAR

20)w1

⌧ =dL

dt assume constant or average ⌧ =�L

�t=

7.8kg ·m2/s� 0

2.0s= 3.9m ·N

conservation of angular momentum IA!1 = (IA + IB)!2

!2 =IA

IA + IB!1 =

MA

MA +MB!1 =

6.0kg

15.0kg7.2rad/s = 2.9rad/s

=1

2(6.0kg)(0.60m)2(7.2rad/s) = 7.8 kg ·m2/s

Angular MomentumAn object moving with velocity v, also will have angular momentum. The angular momentum depends on the distance to the axis of rotation, even if the object isn’t rotating.

~L = ~R⇥ ~pSo the magnitude of L is given by

L = mvR sin ✓Note that you get the same thing if you consider the moment of inertia of a particle, I=mR2 going in a circle

L = I! = (mR2)(v

R) = mvR

Example 11-12A bullet of mass m and velocity v strikes and becomes imbedded at the edge of a cylinder of mass M and radius R0. The cylinder, initially at rest, begins to rotate about its symmetry axis, which remains fixed in position. Assuming no frictional torque, what is the angular velocity of the cylinder after the collision? Is kinetic energy conserved? conservation of angular momentum

Li = mvR0

Lf = (Icyl + Ib)!

Icyl =1

2MR2

0 Ib = mR20

Lf = Li = mvR0 = (1

2M +m)R2

0!! =

mv

( 12M +m)R0

kinetic energy Ki =1

2mv2

Kf =1

2(1

2M +m)R2

0!2 =

1

2(1

2M +m)R2

0

� mv

( 12M +m)R0

�2=

1

2

m2

( 12M +m)v2

since kinetic energy was not conservedm2

( 12M +m)< m

Chapter 10 – 29,35,41,47,58,71,85

Chapter 11 – 45,58,61

Homework