Rotation II: Torque, Rotational Dynamics, and Problem Solving

Rotation II: Torque, Rotational Dynamics, and Problem Solving

Introduction

This study guide covers the fundamental concepts of rotational motion in physics, focusing on torque, rotational dynamics, and practical problem-solving strategies. These topics are essential for understanding how objects rotate, the forces involved, and how to analyze rotational systems using Newton's laws.

Torque

Definition and Calculation

Torque is a measure of the tendency of a force to rotate an object about an axis. It is a vector quantity and plays a central role in rotational dynamics.

• Definition: Torque ($\tau$) is defined as the product of the force and the lever arm (distance from the axis of rotation), and depends on the angle at which the force is applied.

• Formula:

  • $\tau = F_{\perp} r = F r \sin \theta$

• Units: Newton-meter (N·m)

• Direction: Determined by the right-hand rule; counterclockwise (CCW) is positive, clockwise (CW) is negative.

Example: Pushing a door at different points and angles changes the torque produced, affecting how easily the door rotates.

Rotational Dynamics

Newton's Second Law for Rotation

Rotational dynamics extends Newton's laws to rotating systems, relating torque to angular acceleration and moment of inertia.

• Newton's Second Law for Rotation:

  • $\sum \tau = I \alpha$

  where $I$ is the moment of inertia and $\alpha$ is the angular acceleration.

• Moment of Inertia ($I$): Quantifies an object's resistance to changes in rotational motion. For discrete masses:

  • $I = \sum m r^2$

• Comparison: Analogous to mass in linear motion; higher $I$ means harder to accelerate rotationally.

Example: A solid cylinder has a different moment of inertia than a hollow cylinder of the same mass and radius.

Moments of Inertia for Common Objects

The moment of inertia depends on the mass distribution relative to the axis of rotation. Below is a summary table for various shapes:

Object	Moment of Inertia ($I$)	Axis Location
Thin hoop, radius $R$	$I = M R^2$	Through center
Solid cylinder, radius $R$	$I = \frac{1}{2} M R^2$	Through center
Hollow cylinder, inner radius $R_1$, outer radius $R_2$	$I = M \frac{R_1^2 + R_2^2}{2}$	Through center
Uniform sphere, radius $R$	$I = \frac{2}{5} M R^2$	Through center
Long uniform rod, length $L$	$I = \frac{1}{12} M L^2$	Through center
Rectangular plate, length $l$, width $w$	$I = \frac{1}{12} M (l^2 + w^2)$	Through center

Additional info: Table entries inferred from standard physics references.

Problem Solving in Rotational Dynamics

General Strategy

Solving rotational dynamics problems involves identifying forces, drawing diagrams, and applying Newton's laws for both translation and rotation.

• Draw a free-body diagram (FBD) and label all forces at their correct locations.

• Determine if the system is translating, rotating, or both.

• Apply Newton's Second Law for translation ($\sum F = ma$) and/or rotation ($\sum \tau = I \alpha$) as needed.

• Identify the axis of rotation and calculate the moment of inertia.

• Check if torques are balanced or if there is net angular acceleration.

Example: For a see-saw, analyze torques about the pivot to determine the mass needed for rotation.

Worked Examples

Example 1: Door and Zombie

Problem: A force of 30 N is applied at a 30° angle from normal at a radius of 0.4 m. What minimum force must you apply at the edge (0.8 m) at normal incidence to hold the door?

• Calculate torque from the zombie: $\tau_Z = F_Z r_Z \cos 30°$

• Set torques equal for equilibrium: $F_{ME} (0.8) = (30) \cos 30°$

• Solution: $F_{ME} = 13.0$ N

Example 2: See-Saw Balance

Problem: A 1.0 m see-saw of mass 10 kg is at a 30° angle. The pivot is 0.6 m from one end. What mass should be added to the raised end to start rotation?

• Set up torque balance: $m g \cos 30° (0.4) > (10 \text{ kg}) g \cos 30° (0.1)$

• Solve for $m$: $m > 2.5$ kg

Example 3: Merry-Go-Round Acceleration

Problem: A 200 N force is applied to the edge of a 200 kg, 1 m radius merry-go-round (solid cylinder). Find (a) angular acceleration, (b) time to reach 10 rad/s.

• Moment of inertia: $I = \frac{1}{2} M R^2$

• Torque: $\tau = F r$

• Angular acceleration: $\alpha = \frac{F r}{I} = \frac{200 \times 1}{100} = 2$ rad/s2

• Time to reach $\omega = 10$ rad/s: $t = \frac{\omega - \omega_0}{\alpha} = \frac{10}{2} = 5$ s

Example 4: Falling Bucket and Cylinder

Problem: A 2 kg bucket is attached to a massless string wrapped around a 4 cm diameter cylinder. How long does it take to fall 2 meters?

• Apply energy conservation or rotational equations to relate linear and angular acceleration.

• Moment of inertia for cylinder: $I = \frac{1}{2} M R^2$

• Relate torque from tension to angular acceleration: $\tau = T r = I \alpha$

• Use kinematic equations to solve for time.

• Additional info: Full solution requires combining rotational and translational motion equations.

Summary Table: Rotational Problem Solving Steps

Step	Description
1. Identify objects	List all objects in the system
2. Draw diagram/FBD	Label all forces at correct locations
3. Determine motion type	Is the system translating, rotating, or both?
4. Apply Newton's Laws	Use $\sum F = ma$ and/or $\sum \tau = I \alpha$
5. Solve for unknowns	Calculate forces, torques, accelerations, etc.

Additional info: This guide expands on brief lecture notes to provide a self-contained reference for rotational dynamics in introductory physics.