Rotational MechanicsPart IV – Angular Momentum

Pre AP Physics

Angular Momentum and Its Conservation

In an analogy with linear momentum, we can define angular momentum L:

Since = I and since , We can then

write the total torque as being the rate of change of angular momentum.

If the net torque on an object is zero, the total angular momentum is constant.

f i

t

L = I

f iI

t

f iI L

t t

Angular Momentum and Its Conservation

Another way of expressing the angular momentum L of a particle relative to a point O is the cross product of the particle's position r relative to O with the linear momentum p of the particle.

vmrprL

Angular momentum of a particle

The value of the angular momentum depends on the choice of the origin O, since it involves the position vector relative to the origin

The units of angular momentum: kgm2/s

NOTE:

If = v/r and I = mr2, then L = mr2 OR L = mvr

Change in angular momentum

τ (t) = Iωf – Iωi τ = torque

t = time

I = moment of inertia

ωf = final angular velocity

ωi = initial angular velocity

The angular momentum of a system remains unchanged unless an external torque acts on it

Spinning ice skater

Arms extended, inertia is larger, velocity is smaller

Pull arms in tight, inertia decreases, velocity increases

Therefore, systems that can change their rotational inertia through internal forces will also change their rate of rotation:

If the ice cap at the South Pole melted and the water wereuniformly distributed over the earth's oceans, what wouldhappen to the earth’s angular velocity? Would it increase,decrease or stay the same?

Example #1

REASONING AND SOLUTION

Consider the earth to be an isolated system. Note that the earth rotates about an axis that passes through the North and South poles and is perpendicular to the plane of the equator. If the ice cap at the South Pole melted and the water were uniformly distributed over the earth's oceans, the mass at the South Pole would be uniformly distributed and, on average, be farther from the earth's rotational axis. The moment of inertia of the earth would, therefore, increase.

Example #1 con’t

Since the earth is an isolated system, any torques involved in the redistribution of the water would be internal torques; therefore, the angular momentum of the earth must remain the same. If the moment of inertia increases, and the angular momentum is to remain constant, the angular velocity of the earth must decrease.

Rotational Inertia, I, INCREASES

BUT angular momentum, L, stays the same

Since L = I

Angular velocity, , DECREASES

Example #2 – Pulling Through a HoleA particle of mass m moves with speed vi in a circle of

radius ri on a frictionless tabletop. The particle is attached to a massless string that passes through a hole in the table as shown. The string is pulled slowlydownward until the particle is a distance rf from the hole and continues to rotate in a circle of that radius. (A) Find the final velocity vf. (B) Find the tension T in the string when the particle moves in a circle of radius r in terms of the angular momentum L.

Lf = Li(A) mvfrf = mviri

i if

f

v rv

r

22mv m LT

r r mr

(B)

2

3

L

mrL = mvr v = L/mr

T = centripetal Force

A skater has a moment of inertia of 3.0 kg m2 when her arms are outstretched and 1.0 kg m2 when her arms are brought in close to her sides. She starts to spin at the rate if 1 rev/s when her arms are outstretched, and then pulls her arms to her sides. (A) What is her final angular speed? (B) How much work did she do?

Example #3

(A)

(B)

Lf = Li Iff = Iii

if i

f

I

I

2

2

31 / sec

1f

kg mrev

kg m

3 rev/sec

Work (rotational) = KE (rotational) Work-Energy Theorem

2 21 1

2 2f f i iWork I W = (.5)(1.0)(3 x 2π)2 – (.5)(3)(1 x 2π)2

118 J

Angular MomentumAngular momentum depends on linear momentum and

the distance from a particular point. It is a vector quantity with

symbol L. If r and v are then the magnitude of angular

momentum with respect to point Q is given by L = p∙r = m v r.

In the case shown below L points out of the page. If the mass

were moving in the opposite direction, L would point into the

page. The SI unit for angular

momentum is the kg m2 / s. (It has

no special name.) Angular momentum

is a conserved quantity. A torque is

needed to change L, just a force is

needed to change p. Anything

spinning has angular has angular

momentum. The more it has, the

harder it is to stop it from spinning.

r

v

mQ

If r and v are not then the angle between these two

vectors must be taken into account. The general definition of

angular momentum is given by a vector cross product:

This formula works regardless of the angle. The magnitude

of the angular momentum of m relative to point Q is: L = r p sin= m v r.

r

v

m

Q

vmrprL

Now if a net force acts on an object, it must accelerate,

which means its momentum must change. Similarly, if a net

torque acts on a body, it undergoes angular acceleration, which

means its angular momentum changes. Recall, angular

momentum’s magnitude is given by

So, if a net torque is applied, angular velocity must

change, which changes angular momentum.

So net torque is the rate of change of angular

momentum, just as net force is the rate of change of linear

momentum.

L = mvrif v and r are mutually perpendicular

Here is yet another pair of similar equations, one linear, one rotational. From the formula v = r, we get

L = mvr = mr (r) = m r2 = I

This is very much like p = m v, and this is one reason I is defined the way it is.

In terms of magnitudes, linear momentum is inertia times speed, and angular momentum is rotational inertia times angular speed.

NOTE:

If = v/r and I = mr2, then L = mr2 OR L = mvr

Suppose Mr. Stickman is sitting on a stool that swivels holding a

pair of dumbbells. His axis of rotation is vertical. With the weights

far from that axis, his moment of inertia is large. When he pulls his

arms in as he’s spinning, the weights are closer to the

axis, so his moment of inertia gets much

smaller. Since L = I and L is

conserved, the product of I and is a

constant. So, when he pulls his arms in,

I goes down, goes up, and he starts

spinning much faster.

I = L = I

Since an external net force is required to change

the linear momentum of an object.

An external net torque is required to change the

angular momentum of an object.

It is easier to balance on a moving bicycle than on

one at rest.

• The spinning wheels have angular momentum.

• When our center of gravity is not above a point of

support, a slight torque is produced.

• When the wheels are at rest, we fall over.

• When the bicycle is moving, the wheels have

angular momentum, and a greater torque is required

to change the direction of the angular momentum.

Angular momentum is conserved when no external

torque acts on an object.

Angular momentum is conserved for systems in

rotation.

The law of conservation of angular momentum

states that if no unbalanced external torque acts on a

rotating system, the angular momentum of that system is

constant.

With no external torque, the product of rotational

inertia and rotational velocity at one time will be the same

as at any other time.

Rotational speed is controlled by variations in the

body’s rotational inertia as angular momentum is conserved

during a forward somersault. This is done by moving some

part of the body toward or away from the axis of rotation.

Example #4

What is the angular momentum of a 0.210-kg ball rotating on the end of a thin string in a circle of radius 1.10 m at an tangential speed of 5.25 m/s?

m = 0.210 kg; r = 1.10 m; v = 5.25 m/s

L = mvr

L = (0.210 kg)(5.25 m/s)(1.10 m)

L = 1.21 kg∙m2/s

Example #5

What is the angular momentum of a 0.210-kg ball rotating on the end of a thin string in a circle of radius 1.10 m at an angular speed of 10.4 rads/sec?

L = I = mr2

L = (0.210 kg)(1.10 m)2(10.4 rads/s)

L = 2.64 kg-m2/s

m = 0.210 kg; r = 1.10 m; = 10.4 rads/sec

Example #6

What is the angular momentum of a 2.8-kg uniform cylindrical grinding wheel of radius 18 cm when rotating at 1500 rpm?

L = I

L = 7.1 kg-m2/s

m = 2.8 kg; r = 0.18 m; = 1500 rpm

212cylinderI mr

21500

60rpm

= 50 rads/sec

21

22.8 0.18 50 / secL kg m rads

An artificial satellite is placed into an elliptical orbit about the earth. Telemetry data indicate that its point of closest approach (called the perigee) is rP = 8.37 × 106 m from the center of the earth, and its point of greatest distance (called the apogee) is rA = 25.1 × 106 m from the center of the earth. The speed of the satellite at the perigee is vP = 8450 m/s. Find its speed vA at the apogee.

Example #7

LApogee = LPerigee IAA = IPP

2 2A PA P

A P

v vmr mr

r r

P PA

A

r vv

r

6

6

m / s8.37 10 m 8450

25.1 10 mAv

2820 m/s

A horizontal disk of rotational inertia 4.25kg ⋅ m2 with respect to its axis of symmetry is spinning counterclockwise about its axis of symmetry, as viewed from above, at 15.5 revolutions per second on a frictionless massless bearing. A second disk, of rotational inertia 1.80 kg ⋅ m2 with respect to its axis of symmetry, spinning clockwise as viewed from above about the same axis (which is also its axis of symmetry) at 14.2 revolutions per second, is dropped on top of the first disk. If the two disks stick togetherand rotate as one abouttheir common axis ofsymmetry, at what newangular velocity would thecombined disks move inrads/sec?

Example #8

15.5 2

seci

revs rads

rev

= +31 rads/s (counterclockwise)

14.2 2

seci

revs rads

rev

= −28.4 rads/s (clockwise)

L Before = L After

I11 + (−I22) = I1Final + I2Final

Example #8 Continued

I11 + (−I22) = I1Final + I2Final

1 1 2

1 2

Final

I I

I I

2 2

2 2

4.25 31 / 1.80 28.2 /

4.25 1.80Final

kg m rads s kg m rads s

kg m kg m

Final = 41.9 rads/s

Example #8 Continued

The Vector Nature of Rotational Motion

The vector directions of the angular velocity vector and the angular momentum vector Lare along the axis of rotation. Applying the right-hand rule gives the direction.

Right-Hand Rule Grasp the axis of rotation with your right hand, so that your fingers circle the axis in the same sense as the rotation. Your extended thumb points along the axis in the direction of the angular velocity vector.

Angular acceleration arises when the angular velocity changes, and the acceleration vector also points along the axis of rotation. The acceleration vector has the same direction as the change in the angular velocity.

Angular acceleration and angular

momentum vectors also point along

the axis of rotation.

Example #9 Tipping the WheelA student sitting on a stool on a frictionless turntable is

holding a rapidly spinning bicycle wheel. Initially, the axis of the wheel is horizontal, with the angular momentum vector Lpointing to the right.

What happens when the student tips the wheel so that the spin axis is vertical, with the wheel spinning counterclockwise?

Answer

The system is free to rotate about the vertical axis (no vertical torques) and initially the angular momentum is zero along that axis. Therefore, vertical angular momentum is conserved and the final angular momentum must also be zero.

In the final state, the wheel has a large angular momentum pointing vertically upward, so the stool and student must rotate clockwise to have an equal and opposite downward angular momentum.