AP Physics 1: Rotational Motion - The Night Before 🚀

Hey there, future physics master! Let's get you prepped for the AP Physics 1 exam with a high-impact review of rotational motion. We'll break down the key concepts, highlight connections, and get you feeling confident. Let's do this!

Torque: The Force Behind Rotation

What is Torque?

Torque (𝜏) is the rotational equivalent of force. It's what causes objects to rotate. Think of it as the "twisting force." 💡

• Key Factors Affecting Torque:
  • Force (F): The magnitude of the force applied.
  • Radius (r): The distance from the pivot point (axis of rotation) to where the force is applied.
  • Angle (θ): The angle between the force vector and the radius vector.

Calculating Torque

The formula for torque is:

$$\tau = rF\sin(\theta)$$

Where:

• 𝜏 = torque (measured in Newton-meters, Nm)
• F = force (in Newtons, N)
• r = distance from the pivot (in meters, m)
• θ = angle between the force and radius vectors

Image courtesy of MIT Scripts.

Torque and Rotation

• No Rotation: If the net torque on an object is zero (∑𝜏 = 0), the object is in rotational equilibrium. It either won't rotate, or it will rotate at a constant angular velocity.

• Rotation: If there's a net torque (∑𝜏 ≠ 0), the object will experience an angular acceleration (it will start rotating or change its rate of rotation).

Example Problem

Let's look at a classic example:

• A 120 N weight is placed 3m from the pivot. What force (F) must be applied 4m from the pivot to keep the beam balanced?
• Solution:
  • Torque due to weight = (120 N) * (3 m) = 360 Nm (counter-clockwise, positive)
  • Torque due to applied force = F * (4 m) (clockwise, negative)
  • For equilibrium, torques must be equal: 360 Nm = F * 4 m
  • Solving for F: F = 90 N

Moment of Inertia (I): Resistance to Rotation

The moment of inertia (I) is the rotational equivalent of mass. It tells you how hard it is to change an object's rotational motion. It depends not only on the mass but also on how that mass is distributed around the axis of rotation.

• Point Mass: For a single point mass, I = mr², where m is the mass and r is the distance from the axis of rotation.

• General Rule: The closer the mass is to the axis of rotation, the smaller the moment of inertia, and the easier it is to rotate the object.

Image courtesy of ScienceABC.

Angular Acceleration (α)

Angular acceleration (α) is the rate of change of angular velocity. It's the rotational equivalent of linear acceleration. If there's a net torque on an object, it will experience angular acceleration.

Newton's Second Law for Rotation

The relationship between torque, moment of inertia, and angular acceleration is given by:

$$\sum\tau = I\alpha$$

This is the rotational equivalent of F = ma!

• ∑𝜏 = net torque
• I = moment of inertia
• α = angular acceleration

We can rearrange this to solve for angular acceleration:

$$\alpha = \frac{\sum\tau}{I}$$

Example Problem

Let's work through a quick example:

• A rod of length L and mass M is pivoted at one end and released from a horizontal position. What is its initial angular acceleration? (I = ⅓ML²)
• Solution:
  • Torque due to weight = (Mg)(L/2) (force acts at the center of the rod)
  • ∑𝜏 = Iα => (Mg)(L/2) = (⅓ML²)α
  • Solving for α: α = (3g)/(2L)

Final Exam Focus 🎯

Okay, let's talk strategy for the exam. Here's what to focus on:

• High-Priority Topics:

  • Torque: Understanding how force, distance, and angle affect torque. Be ready to calculate it in various scenarios.
  • Rotational Equilibrium: Knowing when torques are balanced and an object isn't rotating or is rotating at a constant rate.
  • Moment of Inertia: Understanding how mass distribution affects rotational inertia.
  • Newton's Second Law for Rotation: Applying ∑𝜏 = Iα to solve problems involving angular acceleration.

• Common Question Types:

  • Multiple Choice: Conceptual questions about torque direction, moment of inertia, and factors affecting rotational motion.
  • Free Response: Problems involving calculating torques, analyzing rotational equilibrium, and applying Newton's Second Law for Rotation.

Practice Question

Practice Questions

Okay, let's test your knowledge! Here are some practice questions to get you ready:

Multiple Choice Questions

1. A force F is applied at a distance r from the axis of rotation. If the force is applied at a distance of 2r from the axis of rotation, what is the new torque? (A) 0.25𝜏 (B) 0.5𝜏 (C) 2𝜏 (D) 4𝜏

2. Two objects have the same mass, but object A has a larger moment of inertia than object B. If the same torque is applied to both objects, which one will have a larger angular acceleration? (A) Object A (B) Object B (C) They will have the same angular acceleration (D) Cannot be determined without more information

3. A uniform beam is balanced on a pivot. A weight is placed on one side of the beam. What can be done to balance the beam? (A) Place a weight further from the pivot on the same side (B) Place a weight closer to the pivot on the same side (C) Place a weight on the other side, closer to the pivot (D) Place a weight on the other side, further from the pivot

Free Response Question

A uniform rod of mass M and length L is initially at rest in a vertical position, pivoted at its top. It is then allowed to fall. The moment of inertia of the rod about its end is (1/3)ML². Assume no friction at the pivot.

(a) Calculate the torque on the rod about the pivot when the rod is at an angle θ with respect to the vertical.

(b) Calculate the angular acceleration of the rod at this angle θ.

(c) Determine the linear acceleration of the center of mass of the rod at this angle θ.

(d) What is the angular velocity of the rod when it is horizontal?

Scoring Rubric:

(a) 2 points * 1 point for using the correct force (weight of the rod, Mg) * 1 point for using the correct lever arm (L/2)sinθ * Correct answer: τ = (MgL/2)sinθ

(b) 2 points * 1 point for using the correct expression for torque from part (a) * 1 point for using the correct moment of inertia (1/3)ML² * Correct answer: α = (3g/2L)sinθ

(c) 2 points * 1 point for relating linear acceleration to angular acceleration (a = rα) * 1 point for using the correct radius (L/2) * Correct answer: a = (3g/4)sinθ

(d) 3 points * 1 point for using conservation of energy * 1 point for correct potential energy (MgL/2) * 1 point for correct rotational kinetic energy (1/2)Iω² * Correct answer: ω = √(3g/L)