Overview
Rotational motion includes: non-static torque (pulleys, ball rolling down inclines), angular kinematics, angular momentum, and rotational kinetic energy. This post will cover the latter two.
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.
[katex] KE_{\text{rotational}} = \frac{1}{2}I \omega ^2[/katex]
Note all other types of energy (Potential, elastic, etc) don’t have a rotational equation counterpart. It’s just for kinetic energy.
Solving Problems
- Use the law of conservation of energy → [katex]E_i = E_f[/katex]
- 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:
- Initial and final energy equality: [katex]E_i = E_f[/katex]
- Energy components: [katex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/katex]
- Substituting moment of inertia for a sphere: [katex]mgh = \frac{1}{2}mv^2 + \frac{1}{2} \left[ \frac{2}{5}mr^2 \right] \left( \frac{v^2}{r^2} \right)[/katex]
- Simplifying: [katex]mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2[/katex]
- Combining like terms: [katex]mgh = \frac{7}{10}mv^2[/katex]
- Solving for [katex]v[/katex]: [katex]v = \sqrt{\frac{10}{7}gh}[/katex]
Angular Momentum
This is arguably the most commonly missed topic on the AP exam.
Linear momentum is p = mv, thus angular momentum is L = Iω.
[katex]\text{Change in angular momentum} = \text{Angular Impulse} = \Delta L = I\Delta \omega = \tau \cdot t[/katex]
Just like how a linear impulse is caused by a net external force, an angular impulse is caused by a net external torque.
Important Concepts
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 [katex]v \rightarrow L = I\omega \rightarrow L = (mr^2) \cdot \left(\frac{v}{r}\right)[/katex]. This is still angular momentum, despite substituting in linear variables. Note that the [katex] mr^2 [/katex] is the moment of inertia of a point mass.
Example
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?
Solution:
Part 1: Apply conservation of angular momentum.
- [katex]L_f = L_i \rightarrow 0.5m_1r^2\omega_i = (0.5m_1r^2 + 0.5m_2r^2)\omega_f[/katex].
- Substitute in values and solve for [katex]\omega_f \rightarrow \omega_f = 1.09, \text{rad/s}[/katex].
- 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:
- [katex]\frac{KE_i}{KE_f} \rightarrow \frac{0.5I_i\omega_i^2}{0.5I_f\omega_f^2} \rightarrow \frac{m_i\omega_i^2}{m_f\omega_f^2}[/katex]
- Substitute numbers and solve the ratio: [katex]\frac{ 2 \cdot 6^2}{7 \cdot 1.09^2} = 8.7[/katex]
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 understanding.
Practice Questions for Mastery
The child walks towards the center of the platform.
The child walks towards the edge of the platform.
The child walks in a circle, opposite to the rotational direction of the platform.
The child walks in a circle, along with the rotational direction of the platform.
none of the above will change the total angular momentum of the system.
x = 1, y < 1
x = 2; y = 1/2
x > 1; y < 1
x < 1; y > 1
(e) x >1; y > 1