In Mechanics Tutorial 11 we met linear momentum as the product of mass and velocity, the simple equation being p = mv. Any object moving in a straight line has a momentum, so it is very reasonable that any spinning object has angular momentum in its spin. As we saw in Tutorial 1, we found that the quantity in rotational dynamics that was equivalent to mass was the moment of inertia. So we can write an expression that tell us about angular momentum:
Angular momentum = moment of inertia × angular velocity
L = Iw
[L - angular momentum (kg m2 s-1); I - moment of inertia (kg m2); w - angular velocity (rad s-1)]
Why are the units for L kg m2 s-1?
Cricketers add spin to the ball when they bowl by giving the ball a quick flick just as the ball leaves their hand. The idea is to make it change the direction of flight as it bounces. (I was hopeless at cricket and never got the knack... For this and many other reasons I developed a heart-felt loathing for cricket, which I have to this day.)
A disc of moment of inertia 10 kg m2 rotates at an angular velocity of 20 rad s-1. What is its angular momentum?
Like momentum in a straight line, angular momentum is conserved, which means that momentum before = momentum after, provided that there has been no angular impulse. You can try this for yourself on a swivel chair. Spin yourself around. Put your legs out and you will find yourself spinning more slowly. Then tuck your legs in, and you will spin faster. You are changing the moment of inertia. If the moment of inertia increases, the angular velocity must go down to keep the angular momentum the same.
High divers and ice skaters use the ideas of angular momentum as part of their acts, although not that many know about the physics.
A ballet dancer spins about her vertical axis at 1 revolution per second with her arms outstretched. With her arms folded, her moment of inertia decreases to 40 % of what it was. What is her new rate of turning?
Let her original moment of inertia be I, so her new moment of inertia = 0.4 I
Angular momentum is conserved.
I ×1 s-1 = L = 0.4 I ×w
w = 1 ÷ 0.4 = 2.5 revolutions per second.
A potter in an African village makes large clay pots on a stone wheel. The wheel rotates freely on a central bearing and is driven by the potter, who applies a tangential force repeatedly to its rim using his foot until the wheel reaches its normal working angular speed. He then stops driving and throws a lump of clay onto the centre of the wheel.
The normal working angular speed of the wheel is 5.0 rad s-1. The moments of inertia of the wheel and the clay about the axis of rotation are 1.6 kg m2 and 0.25 kg m2, respectively. When the clay is added, the angular speed of the wheel changes suddenly. The net angular impulse is zero. Calculate the angular speed of the wheel immediately after the clay has been added.
(AQA past question)
In Mechanics we saw that impulse was change in momentum. We could rewrite Newton II as:
Therefore impulse Dp = FDt
We can do exactly the same with angular momentum. We call the change in angular momentum the angular impulse. Instead of Force, we have torque. Therefore:
Angular impulse (kg m2 rad s-1) = torque applied (N m) × the time interval (s)
In physics code we write it as:
A small torque applied for a long time interval will have the same effect as a large torque applied for a short time interval.
Here is a rotating sign that rotates in windy conditions. You often see them outside garages.
On a still day, a gust of wind from a passing vehicle imparts an angular impulse of 1.2 kg m2 rad s-1 to the sign, which accelerates from rest during a time of 2.8 s. The moment of inertia of the sign about its axis of rotation is 4.8 × 10-2 kg m2.
Assuming that the frictional couple acting on the sign is negligible, calculate:
(a) the angular momentum acquired by the sign as a result of the angular impulse, showing your reasoning clearly,
(b) the angular speed of the sign immediately after the impulse has been imparted.
(c) What was the torque that acted on the sign?
(AQA Past Question)
Work and Power in Rotating Objects
In linear motion we have seen that work done can be related by the simple equation:
Work done (J) = Force (N) × distance moved in the direction of a force (m)
W = Fs
We can use the same idea to work out the work done in a rotating system:
Work done (J) = torque (N m) × angle rotated (rad)
W = tq
A torque of 135 N m is required to turn a nut through half a turn. What is the work done by the mechanic?
In linear motion we also found a useful relationship linking force and power:
Power (W) = force (N) × speed (m s-1).
We can derive a similar expression for rotational motion:
Work done = energy used.
Power = energy used ÷ time interval
Power = (torque x angle rotated) ÷ time interval
So it does not take a genius to see:
Power (W) = torque (N m) × angular velocity (rad s-1)
P = tw
A motor gives out a torque of 150 Nm at a speed of 3000 rpm. What is its power?
The diagram shows a human centrifuge used in pilot training to simulate the large "g" forces experienced by pilots during aerial manoeuvres. The trainee sits in the capsule at the end of the rotating centrifuge arm, which is driven by an electric motor.
When working at maximum power, the motor is capable of increasing the angular speed of the arm from its minimum working speed of 1.6 rad s-1 to its maximum speed of 7.4 rad s-1 in 4.4 s.
The net power needed to achieve this acceleration is 150kW.
(a) Assuming that this power remains constant during the acceleration, calculate the energy supplied to the centrifuge by the motor.
(b) Hence estimate the moment of inertia of the rotating system. (AQA past question)
All rotating machines have a certain amount of friction. We treat friction as an opposing rotational couple which provides a torque in the opposite direction. We use the equation:
t = Ia
to work out the rate of change of angular velocity.
A roller is running freely on its bearings and is turning at an angular velocity of 8.5 rad s-1. The moment of inertia of the roller is 2.5 kg m2. If the frictional couple is 0.67 N m,
(a) what is the angular acceleration?
(b) How many revolutions does the roller need to come to rest?
(a) Use the relationship
t = Ia
The frictional couple should be negative, as it is opposing the turning.
-0.67 N m = 2.5 kg m2 a
a = -0.268 rad s-2 (minus because the roller is slowing down)
(b) To work out the angle covered, we use:
w22 = w12 + 2aq
0 = 8.5 rad s-1 + 2 × -0.268 × q
q = 15.9 rad
No of revolutions = 15.9 rad ÷ 2p rad = 2.5 revolutions
A rotating flower bed forms a novelty feature in the annual display of a horticultural society. The circular platform supporting the plants floats in a water tank and is caused to rotate by means of four water jets directed at the rim of the platform.
Each of the four jets exerts a tangential force of 0.60 N on the platform at a distance of 1.8 m from the axis of the rotation. The platform rotates at a steady angular speed, making one complete revolution in 110 s.
(i) the total torque exerted on the platform by the four jets,
(ii) the power dissipated by the frictional couple acting on the rotating platform, showing your reasoning.
(b) When the water jets are switched off, all the kinetic energy of the loaded platform is dissipated as heat by the frictional couple and the platform comes to rest from its normal steady speed in 12 s.
(i) The kinetic energy of the loaded platform when rotating at its normal steady speed is 1.5 J.
Show that this value is consistent with your answer to part (a) (ii).
(ii) Calculate the moment of inertia of the loaded platform.
(AQA Past Question)