The word rotational and angular are used interchangeably.
Rotational Kinetic Energy
Kinetic energy can be translational, rotational, or both.
If a marbles slides down a ramp it is said to have translational (linear) kinetic energy equal to (1/2)mv2.
If a marble rolls down a ramp is is said to have both translational and rotational kinetic energy. See this detailed chart to learn more about the variable used in the formula below.
KErotational = (1/2)Iω2
Note all other types of energy (Potential, elastic, etc) don’t have a rotational equation counterpart. It’s just for kinetic energy.
- Use the law of conservation of energy → Ei = Ef
- If something is rolling and moving linear, be sure to include both rotational and translational energy in the equation.
Rotational Energy Example
A marble starts from rest and rolls down an incline. What was the speed at the bottom of the incline.
Solution: Energy is conserved. Specifically, potential energy turns into linear and rotational kinetic energy. Mathematically this should look like:
(1) Ei = Ef
(2) mgh = (1/2)mv2 + (1/2)Iω2
(3) mgh = (1/2)mv2 + (1/2)[(2/5)mr2](v2/r2)
(4) mgh = (1/2)mv2 + (1/5)mv2
(5) mgh = (7/10)mv2
(6) v = [(10/7)gh]1/2
This is arguably the most commonly missed topic on the AP exam.
Linear momentum is p = mv, thus angular momentum is L = Iω.
Change in angular momentum = Angular Impulse = ∆L = I∆ω = τext(t)
Just like how a linear impulse is caused by a net external force, an angular impulse is caused by a net external torque.
Linear and angular momentum are conserved separately. This means you should NOT combine the conservation of linear momentum (Pi = Pf) with the conservation of angular momentum (Lf = Li).
You can, however use linear variables in angular momentum. Makes senses?
Let’s make a formula for the angular momentum of a particle traveling at velocity v → L = Iω → L = (mr2)·(v/r). This is still angular momentum, despite substituting in linear variables. Note that the mr2 is the moment of inertia of a point mass.
A disk of mass 2 kg and radius 1 meter, is rotating at 6 rad/s. Another disk of mass 5 kg is dropped on top of the rotating disk. What is the new angular velocity of the two disk system? What is the ratio of energy before and after the collision?
To solve this problem, we can apply conservation of angular momentum:
Lf = Li –> .5m1r2(ωi) = (.5m1r2 + .5m2r2)(ωf)
Substitute in values and solve for ωf –> ωf = 1.09 rad/s
Does this makes sense? Yes! Since angular momentum is conserved velocity of the system should decrease as the moment of inertia increases.
Part 2: The energy is purely rotational:
KEi/KEf –> [.5Iiωi2]/[.5Ifωf2] –> miωi2/mfωf2
Substitute numbers and solve the ratio: [2*62]/[7*1.092] = 8.7
For the AP Exam…
Angular momentum is arguably the most misunderstood concept on the AP Physics 1 Exam. Here are a couple of tips:
- Use conservation of angular momentum separately from conservation of linear momentum
- Angular momentum is always conserved in a collision
- Total angular momentum of a system should not change if there is no external force.
- Angular momentum is not conserved if there is an external force on the system. This is an impulse.
- It is okay to substitute linear variables into the formulas for angular momentum. For example you can substitute ω with v/r.
- Practicing the questions below will help you increase your undetstnaing.
1. A child of mass 3 kg rotates on a platform of 10kg. They start walking towards the center while the platform is rotating. Which of the following could possibly decrease the total angular momentum of the child-platform system?
(a) The child walks towards the center of the platform
(b) The child walks towards the edge of the platform
(c) The child walks in a circle, opposite to the rotational direction of the platform
(d) The child walks in a circle, along with the rotational direction of the platform
(e) none of the above will change the total angular momentum of the system
2. A ice skater that is spinning in circles has an initial rotational inertia Ii. You can approximate her shape to be a cylinder. She is spinning with velocity ωi. As she extends her arms she her rotational inertia changes by a factor of x and her angular velocity changes by a factor of y. Which one of the following options best describe x and y.
(a) x = 1, y < 1
(b) x = 2; y = 1/2
(c) x > 1; y < 1
(d) x < 1; y > 1
(e) x >1; y > 1
3. FRQ Question (3 parts)
A system consists of two small disks, of masses m and 2m, attached to ends of a rod of negligible mass of length 3x. The rod is free to turn about a vertical axis through point P. The first mass, m, is located x away from point P, and therefore the other mass, of 2m, is 2x from point P. The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is . At time t = 0, the rod has an initial counterclockwise angular velocity ωi about P. The system is gradually brought to rest by friction.
Develop expressions for the following quantities in terms of µ, m, x, g, and ωi.
(a) The initial angular momentum of the system about the axis through P
(b) The frictional torque acting on the system about the axis through P
(c) The time T at which the system will come to rest
4. A sphere of mass M and radius r, and rotational inertia I is released from the top of a inclinded plane of height h. The surface has considerable friction. Using only the variable mentioned, derive an expression for the center of mass of the sphere.
- (e) There are no external torques that would cause a change in angular momentum of the system.
- (d) Angular momentum is conserved thus I and ω and inversely proportional
- 3 Parts:
- (a) L = 6mx2ωi
- (b) net = -4µmgx
- (c) t = 3ωix / [2µg]
- [ (2Mghr2) / ( I + Mr2) ]