The rotational equivalent of linear momentum is angular momentum. It is defined as the product of moment of inertia and angular velocity.
Key Concepts
Angular momentum is the product of moment of intertia and angular velocity:
\(L=I\omega\)
- \(L\) is angular momentum (kgm2s-1)
- \(I\) is moment of inertia (kgm2)
- \(\omega\) is angular velocity (rads-1)
Angular momentum is a conserved quantity: total angular momentum remains constant.
Ice skaters make good use of this effect when spinning.
With limbs spread out, the moment of inertia is high as more mass is at a distance from the pivot. This results in a low angular velocity.
With the limbs closer to the body, the moment of inertia decreases. Since angular momentum is conserved, this increases the angular velocity:
\(I_1\omega_1=I_2\omega_2\)
Conservation of angular momentum also explains why the addition of a child to a spinning roundabout results in a reduction in angular velocity:
\(L_\textrm{initial}=L_\textrm{final}\)
\(I_\textrm{roundabout}\omega_\textrm{initial}=(I_\textrm{child} +I_\textrm{roundabout})\omega_\textrm{final}\)
Now consider the angular momentum equivalent of an explosion, where two objects begin to rotate simultaneously in opposite directions.
Rotational kinetic energy is calculated as follows:
\({E_\textrm{K}}_\textrm{rot}={1\over 2}I\omega ^2\)
- \({E_\textrm{K}}_\textrm{rot}\) is rotational kinetic energy (J)
- \(I\) is moment of inertia (kgm2)
- \(\omega\) is angular velocity (rads-1)
Energy is another conserved quantity, with energy changed equal to work done:
\(\Delta {E_\textrm{K}}_\textrm{rot}=\Delta W\)
Reconsidering our roundabout example, a child would need to do work to move from position A to position B:
- The moment of inertia decreases as the child moves from A to B due to the reduction in radius.
- As a result of conservation of angular momentum, the angular velocity increases by the same proportion.
- Since rotational kinetic energy is proportional to the square of the angular velocity, this effect dominates, resulting in an increase in the rotational kinetic energy.
- The child must do work equal to this change in energy.
Although not articulated, in the Mechanics topic it was assumed that all moving bodies were sliding. In the Engineering topic, we assume that bodies roll instead.
Bodies are made to roll by the action of a torque:
\(\Gamma ={\Delta L \over t}\)
- \(\Gamma\) is the torque applied (Nm)
- \(\Delta L\) is change in angular momentum (kgm2s-1)
- \(t\) is the time taken (s)
This equation is another statement of Newton's second law: \(\Gamma=I\alpha = I{(\omega_f-\omega_i)\over t}={I\omega_f-I\omega_i\over t}\)
If given the same push, a cylinder of equal mass to a hoop would roll more quickly. This is because:
- the push gives each body an equal torque over an equal time, and so each receives the same angular momentum
- the hoop has the larger moment of inertia (more mass is at a larger distance from the centre)
- angular momentum = Iω so large I means small ω
Use flashcards to practise your recall.
Use quizzes to practise application of theory.
START QUIZ!
In this Algodoo simulation two discs are connected by a friction-free bearing. Suddenly the friction is increased and the discs turn together. The moment of inertia of the big disc is 4 times that of the small one.
If the initial angular velocity of the big one is 1 rads-1 what is the final angular velocity?
Angular momentum is conserved, so \(I\omega_\text{initial}=I\omega_\text{final}\):
\(1\times 4 = (1+4)\omega_f\)
\(ω_f = {4\over 5}\) rads-1
In this Algodoo simulation two discs are connected by a friction-free bearing. Suddenly the friction is increased and the discs turn together. The moment of inertia of the big disc is 4 times that of the small one.
If the initial angular velocity of the small one is 1 rads-1, what is the final angular velocity?
Angular momentum is conserved, so \(I\omega_\text{initial}=I\omega_\text{final}\):
\(1\times 1 = (1+4)\omega_f\)
\(ω_f = {1\over 5}\) rads-1
The Algodoo simulation shows two discs connected by a deactivated motor. The moment of inertia of the big disc is 4 times that of the small one. Suddenly the motor is switched on causing the small disc to rotate at -1.2 rads-1.
What is the angular velocity of the big disc?
Angular momentum is conserved, so remains constant at 0.
\(0 = 4ω +1\times (- 1.2)\)
\(ω = {1.2\over 4}\) rads-1
The image is a still from an Algodoo simulation.
The body is stationary but suddenly the motor on the right disc starts to turn the disc clockwise.
What happens?
Angular momentum is conserved so some anticlockwise motion is required... but about where?
Because there is no external resultant force acting, the centre of mass of the whole system must remain stationary.
In this Algodoo simulation, two rods of equal density are connected by a motor which is suddenly switched on.
If the angular velocity of the short rod is -1.6 rads-1, the angular velocity of the long one is
Moment of inertia of a rod is proportional to \(ML^2\) (NB: Not required to memorise \(I = {1 \over 12}{ML^2}\))
The long rod has 2x mass and 2x length so 8x moment of inertia.
Angular momentum is conserved so \(0 = 1 \times (-1.6) + 8ω_f\)
\(ω_f = {1.6\over 8}\) rads-1
Two identical balls are connected by a spring in this Algodoo simulation.
If the length of the spring is doubled, the ratio \(w_\text{final}\over \omega_\text{initial}\) is
Angular momentum is conserved.
Moment of inertia is proportional to \(r^2\) so is \(\times 4\)
\(\omega\) is \(\div4\)
Two identical balls are connected by a spring in this Algodoo simulation.
If the length of the spring is doubled, the ratio of speed of the balls before / speed after is:
Angular momentum is conserved.
Moment of inertia is proportional to \(r^2\)so is \(\times 4\)
\(ω\) is \(\div4\) BUT \(v = ωr\) so if:
- \(r\) is \(\times2\)
- \(\omega\) is \(\div 4\)
- \(\Rightarrow v\) is \(\div 2\)
A hollow cylinder of radius 10 cm and mass 2 kg rolls down a slope.
At the bottom of the slope the angular velocity of the cylinder is 2 rads-1. What is the rotational kinetic energy of the cylinder?
\(I = mr^2 = 2 \times 0.1^2 = 0.02 \) kgm2
\(E_{k_\text{rot}} = {1\over2}Iω^2 = {1\over2} \times 0.02 \times 2^2 = 0.04\)J
A constant force of 6 N acts on a wheel of radius 50 cm and mass 1 kg.
How much rotational kinetic energy is gained per revolution (in Joules)?
Gain in kinetic energy = work done = \(Tθ =(6 \times 0.5) \times 2π\)
By moving her arms and legs an ice skater halves her moment of inertia.
By what factor does her rotational kinetic energy increase?
Angular momentum is conserved so \(ω \) is x2
Initial kinetic energy \(={1\over2}Iω^2 \)
Final kinetic energy \(= {1\over 2} \times {I\over2} \times (2ω)^2 =2\)x initial
How much of Angular momentum and rotational kinetic energy have you understood?
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