# Rotational Motion Speed Review

###### Jason Kuma

Writer | Coach | Builder | Fremont, CA

###### Article Content
If you’re studying AP Physics 1, you have most likely heard or rotation, angular, and even centripetal motion. Are they all the same things? This is a pretty confusing topic for most. So let’s break it down.

### 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.

KE_{\text{rotational}} = \frac{1}{2}I \omega ^2

Note all other types of energy (Potential, elastic, etc) don’t have a rotational equation counterpart. It’s just for kinetic energy.

#### Solving Problems

1. Use the law of conservation of energyE_i = E_f
2. 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. Initial and final energy equality: E_i = E_f
2. Energy components: mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
3. Substituting moment of inertia for a sphere: mgh = \frac{1}{2}mv^2 + \frac{1}{2} \left[ \frac{2}{5}mr^2 \right] \left( \frac{v^2}{r^2} \right)
4. Simplifying: mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2
5. Combining like terms: mgh = \frac{7}{10}mv^2
6. Solving for v: v = \sqrt{\frac{10}{7}gh}

### 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.

\text{Change in angular momentum} = \text{Angular Impulse} = \Delta L = I\Delta \omega = \tau \cdot t

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 v \rightarrow L = I\omega \rightarrow L = (mr^2) \cdot \left(\frac{v}{r}\right). This is still angular momentum, despite substituting in linear variables. Note that the mr^2 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.

1. L_f = L_i \rightarrow 0.5m_1r^2\omega_i = (0.5m_1r^2 + 0.5m_2r^2)\omega_f.
2. Substitute in values and solve for \omega_f \rightarrow \omega_f = 1.09, \text{rad/s}.
3. 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:

1. \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}
2. Substitute numbers and solve the ratio: \frac{ 2 \cdot 6^2}{7 \cdot 1.09^2} = 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:

1. Use conservation of angular momentum separately from conservation of linear momentum
2. Angular momentum is always conserved in a collision
3. Total angular momentum of a system should not change if there is no external force.
4. Angular momentum is not conserved if there is an external force on the system. This is an impulse.
5. It is okay to substitute linear variables into the formulas for angular momentum. For example you can substitute ω with v/r.
6. Practicing the questions below will help you increase your understanding.

### Practice Questions for Mastery

Question 1
Solve Type - Conceptual
A child of mass 3 kg rotates on a platform of 10 kg. 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?
1. The child walks towards the center of the platform.

2. The child walks towards the edge of the platform.

3. The child walks in a circle, opposite to the rotational direction of the platform.

4. The child walks in a circle, along with the rotational direction of the platform.

5. none of the above will change the total angular momentum of the system.

Question 2
Solve Type - Proportional Analysis
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.
1. x = 1, y < 1

2. x = 2; y = 1/2

3. x > 1; y < 1

4. x < 1; y > 1

5. (e) x >1; y > 1

Question 3
Solve Type - Mathematical
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. Derive an expressions for the following quantities in terms of µ, m, x, g, and ωi.
Question 4
Solve Type - Mathematical
A sphere of mass M and radius r, and rotational inertia I is released from the top of a inclined plane of height h. The surface has considerable friction. Using only the variable mentioned, derive an expression for the sphere's center of mass velocity.
###### Jason Kuma

Writer | Coach | Builder | Fremont, CA

## Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy

Unit 5 – Momentum

Unit 6 – Torque

Unit 7 – Oscillations

Unit 8 – Fluids

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##### Made By Nerd-Notes.com
KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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