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5.3Angular momentum

The angular momentum the product of moment of inertia and angular velocity, L = Iω of a rigid body of moment of inertia I that rotates about a fixed axis with angular speed ω is defined to be

L=I ω

(Angular momentum is a vector.)

If the body is a point particle the angular momentum is L = I ω = m r 2 ω = m(ω r) r = mvr, an expression we will use in Unit E in relation to the Bohr model. The unit for angular momentum is kg m2 s–1. (This is equivalent to J s.)

Just as there is a relation between linear momentum and net force in particle mechanics, there is a relation between angular momentum and net torque:

τ net =Δ L Δ t

The net torque on a body is equal to the rate of change of the angular momentum of the body.

And similarly, we have the law of conservation of angular momentum when the net torque on a system is zero, the angular momentum is conserved; that is, it stays constant:

KEY POINT

When the net torque on a system is zero, the angular momentum is conserved; that is, it stays constant.

In Figure 5.34 the skater brings her arms closer to her body thus reducing her moment of inertia. Because there are no external torques on her body her angular momentum is conserved and so her angular speed increases.

Figure 5.34: a A skater rotates about her axis. b When she brings her arms in, the moment of inertia is reduced and the angular velocity increases in order to conserve angular momentum.

Rewriting the last equation as

Δ L=τ net Δ t

and calling the quantity τ net Δt the angular impulse the change in angular momentum J, we see that the change in angular momentum is the angular impulse applied. This is in analogy with linear motion where F net Δt is the linear impulse applied, which in that case equals the change in linear momentum.

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Linear motion | Rotational motion
p=m v | L=l ω
F net =Δ p Δ t | τ net =Δ L Δ t
Δ p=F net Δ t | Δ L=τ net Δ t
E K=p 2 2 m | E K=L 2 2 l

You should be able to prove the last formula in the table.

CHECK YOURSELF 5

Figure 5.35 shows the elliptical path of a planet around the sun. SA = 3 SP.

Figure 5.35: For check yourself question 5.

What is the ratio of the speed of the planet at A to that at P?

Worked example 5.12

A spherical star of mass M and radius R rotates about its axis. The star explodes, ejecting mass into space. The star that is left behind has mass M 10 and radius R 20. Calculate the new angular speed of the star in terms of the original angular speed ω.

Answer

Angular momentum is conserved and so

2 5 M R 2 ω=2 5 M 10(R 20)2 ω′⇒ω′=10×20 2 ω=4000 ω

Worked example 5.13

A horizontal disc of mass M = 0.95 kg and radius R = 0.35 m rotates about its vertical axis with angular speed 3.6 rad s–1. A piece of modelling clay of mass m = 0.30 kg lands vertically on the disc, attaching itself at a point a distance r = 0.25 m from the centre. Find the new angular speed of the disc.

Answer

The moment of inertia of the disc is

I=1 2 M R 2=1 2×0.95×0.35 2=0.0582 kg m 2

The original angular momentum of the disc is

L in=I ω=0.0582×3.6=0.210 J s

The final angular momentum is

L in=I ω′+m r 2 ω′=(0.0582⏟disc +0.30×0.25 2⏟clay )ω′

Angular momentum will be conserved because there are no external torques on the disc–clay system. So

(0.0582+0.30×0.25 2)ω′=0.210 ω′=2.7 rad s−1

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Worked example 5.14

A torque of 12 N m acts on a body of moment of inertia 4.0 kg m2 for 3.0 s. The body is initially at rest.

aWhat is the final angular velocity of the body?

bHow many revolutions did the body make during the 3.0 s?

Answer

aThe applied angular impulse is τΔt = 12 × 3.0 = 36 J s and equals the change in angular momentum. Thus

I ω = 36 J s and so ω=36 4.0=9.0 rad s−1.

bThe angular acceleration of the body is

α=τ I=12 4.0=3.0 rad s−2

From ω 2 = 2αΔθ we find

Δ θ=ω 2 2 α=81 6.0=13.5 rad⇒13.5 2 π=2.15 revolutions

Worked example 5.15

A solid object at rest explodes into two fragments X and Y of moment of inertia 8.0 kg m2 and 16 kg m2 respectively. Fragment X rotates with angular velocity 5.0 rad s−1 about a vertical axis. The fragments just rotate with no translational motion.

aExplain why fragment Y must also rotate about a vertical axis but in the opposite sense to that of X.

bCalculate the angular velocity of Y.

cCalculate the ratio rotational rotational kinetic energy of X rotational kinetic energy of Y.

Answer

aThe original object has zero angular momentum, and so the fragments must have equal and opposite angular momenta.

bFrom 8.0 × 5.0 = 16 × ω we find ω = 2.5 rad s−1.

cUse E K=L 2 2 I so rotational kinetic energy of X rotational kinetic energy of Y=L 2 2×8,0 L 2 2×16=2.

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TEST YOUR UNDERSTANDING

19A torque of 2.8 N m acts on a body for 3.2 s. What is the change in the angular momentum of the body?

20The angular momentum of a body changed by 24 J s in 4.0 s as a result of the action of a torque. What was the torque?

21Two identical rods, X and Y, each of length 1.20 m and mass 2.40 kg, are made to rotate with angular velocity 4.50 rad s–1, as shown in Figure 5.36.

Figure 5.36: For question 21.

Calculate the kinetic energy and angular momentum of X and Y.

22A star of mass M and radius R explodes radially symmetrically. The star is left with a mass equal to M 10 and a radius equal to R 50. Calculate the ratio of the final to the initial angular velocity of the star. (I=2 5 M R 2)

23The angular momentum of a body is 240 J s, and its rotational kinetic energy is 920 J. What is its moment of inertia?

24A disc of moment of inertia 2.4 kg m2 is rotating about a vertical axis through its centre with angular velocity 1.2 rad s–1. A second nonrotating disc of moment of inertia 1.8 kg m2 falls on the first disc and sticks to it as shown in Figure 5.37.

aWhat is the angular velocity of the two discs?

bHow much kinetic energy was lost?

Figure 5.37: For question 24.

25A battery-driven toy car of mass 0.18 kg is placed on a circular track that is part of a horizontal ring of moment of inertia 0.20 kg m2 and radius 0.50 m (Figure 5.38). The ring can rotate about a vertical axis without friction.

Figure 5.38: For question 25.

The car is started, and its speed relative to the ground is measured to be 0.80 m s–1. Calculate the angular speed of the ring.

26A cylinder of mass M and radius R can rotate about a horizontal axle through its centre as shown in Figure 5.39. A rope is wrapped around the cylinder, and a block of mass m hangs from the rope. The block is released from rest. The moment of inertia of the cylinder is 1 2 M R 2.

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Figure 5.39: For question 26.

Use the following data:

M=4.0 kg R=0.20 m m=2.0 kg g=10 m s−2

aFind the linear acceleration of the falling block and the angular acceleration of the cylinder.

bCalculate the tension in the string.

cCalculate the linear velocity of the block and the angular velocity of the cylinder 2.0 s after releasing the block.

dFind the change in the angular momentum of the cylinder and in the angular momentum of the block 2.0 s after releasing the block.

eWhat is the net torque on the system?

fWhat is the total angular impulse delivered to the system in 2.0 s?

NATURE OF SCIENCE

Engineers apply the laws of physics to design structures and machines that make our lives safer, more comfortable and more enjoyable. The great technological advances in energy production, food production, transportation, telecommunications, space exploration, weather prediction, medicine, the entertainment industry and myriad other cases are all the result of the application of the basic laws and principles of the sciences to practical problems.

Links

The concept of angular momentum has played a crucial role in the development of quantum theory. Bohr asserted that the angular momentum of an electron orbiting the nucleus of a hydrogen atom is an integral multiple of a basic quantity known as the Planck constant h:m v r=n h 2 π. Using this assertion Bohr was able to show that the energy of the electron in the hydrogen atom is quantised and that only certain orbits are allowed. See Unit E.