Rotational Dynamics and Moment of Inertia: Study Notes

Section 7.5 Rotational Dynamics and Moment of Inertia

Rotational Dynamics and Moment of Inertia

Rotational dynamics describes how objects rotate under the influence of forces and torques. The moment of inertia quantifies how mass is distributed relative to the axis of rotation, affecting how easily an object can be spun.

• Torque (τ) causes an angular acceleration (α).

• Tangential acceleration is given by: $a_t = \frac{F}{m}$

• Tangential and angular accelerations are related by: $a_t = \alpha r$ so $\alpha = \frac{F}{mr}$

• Force perpendicular to the radial line exerts torque: $\tau = rF$

• Relationship with angular acceleration: $\alpha = \frac{\tau}{mr^2}$

Newton's Second Law for Rotational Motion

Newton's Second Law for Rotational Motion

For a rigid body rotating about a fixed axis, the object can be considered as a collection of particles, each experiencing torque and sharing the same angular acceleration.

• Torque for each particle: $\tau_i = m_i r_i^2 \alpha$

• Net torque for all particles: $\tau_{net} = \sum m_i r_i^2 \alpha = (\sum m_i r_i^2) \alpha$

• The proportionality constant $\sum m_i r_i^2$ is the moment of inertia (I): $I = \sum m_i r_i^2$

• Units: kg·m2

• Moment of inertia depends on the axis of rotation.

• Newton's second law for rotation: $\alpha = \frac{\tau_{net}}{I}$

• A net torque causes angular acceleration.

Interpreting the Moment of Inertia

Physical Meaning and Examples

The moment of inertia is the rotational equivalent of mass. It depends not only on the object's mass but also on how the mass is distributed around the rotation axis.

• Objects with mass concentrated far from the axis have a larger moment of inertia and are harder to spin.

• Example: A merry-go-round is harder to spin when people sit far from the center (higher inertia) than when they sit close to the center (lower inertia).

Synthesis 7.2 Linear and Rotational Dynamics

Analogies Between Linear and Rotational Dynamics

Variables in linear dynamics have direct analogs in rotational dynamics. Newton's second law for rotational dynamics is expressed using these variables.

• Linear dynamics: Net force ($F_{net}$), mass ($m$), acceleration ($a$)

• Rotational dynamics: Net torque ($\tau_{net}$), moment of inertia ($I$), angular acceleration ($\alpha$)

• Newton's second law: Linear: $F_{net} = ma$ Rotational: $\tau_{net} = I\alpha$

The Moments of Inertia of Common Shapes

Table: Moments of Inertia for Uniform Objects

The moment of inertia depends on both the shape of the object and the axis about which it rotates. Below are common formulas for objects of uniform density and mass $M$.

Object and Axis	Picture	I
Thin rod (any cross section), about center	[Image: rod about center]	$\frac{1}{12}ML^2$
Thin rod (any cross section), about end	[Image: rod about end]	$\frac{1}{3}ML^2$
Plane or slab, about center	[Image: slab about center]	$\frac{1}{12}ML^2$
Plane or slab, about edge	[Image: slab about edge]	$\frac{1}{3}ML^2$
Cylinder or disk, about center	[Image: cylinder]	$\frac{1}{2}MR^2$
Cylindrical hoop, about center	[Image: hoop]	$MR^2$
Solid sphere, about diameter	[Image: sphere]	$\frac{2}{5}MR^2$
Spherical shell, about diameter	[Image: shell]	$\frac{2}{3}MR^2$

Section 7.6 Using Newton's Second Law for Rotation

Problem-Solving Approach 7.1: Rotational Dynamics Problems

Solving rotational dynamics problems involves a systematic approach similar to linear dynamics.

• Strategize: Model the object as a simple shape. Identify what is rotating, the axis, and the forces causing rotation.

• Prepare: Define coordinates and symbols, list known information, identify axes, forces, and calculate torques and their signs.

• Solve: Use Newton's second law for rotation: $\tau_{net} = I\alpha$ or $\alpha = \frac{\tau_{net}}{I}$

• Find the moment of inertia using formulas or tables.

• Use rotational kinematics to find angular positions and velocities.

• Assess: Check units, reasonableness, and answer the question.

Worked Examples

Example 7.16: Angular Acceleration of a Falling Pole

This example models a caber toss, where a uniform pole falls under gravity, rotating about its end.

• Strategy: Model the pole as a uniform thin rod rotating about one end.

• Preparation: Two forces act: gravity at the center of mass and ground force at the end (no torque from ground force).

• Torque due to gravity: $\tau_{net} = \left(\frac{L}{2}\right) w \sin\theta = \frac{mgL}{2} \sin\theta$

• Moment of inertia (rod about end): $I = \frac{1}{3}mL^2$

• Angular acceleration: $\alpha = \frac{\tau_{net}}{I} = \frac{mgL \sin\theta}{2 \cdot \frac{1}{3}mL^2} = \frac{3g \sin\theta}{2L}$

• Time to rotate by 5°: $\Delta \theta = \frac{1}{2} \alpha (\Delta t)^2$ $\Delta t = 0.40$ s

• Assessment: The result does not depend on mass, only on length and angle. The angular acceleration is modest, giving time for safety.

Example 7.18: Spinning Up a Flywheel

Calculates the time required for a motor to spin up a heavy flywheel to a high angular speed.

• Strategy: Find moment of inertia, use torque to find angular acceleration, then time to reach final speed.

• Preparation: Convert rpm to rad/s: $f = 13,000 \text{ rev/min} = 217 \text{ rev/s}$ $\omega = 2\pi f = 1360 \text{ rad/s}$

• Moment of inertia (cylinder): $I = \frac{1}{2}MR^2 = \frac{1}{2}(540)(0.30)^2 = 24.3 \text{ kg·m}^2$

• Angular acceleration: $\alpha = \frac{\tau}{I} = \frac{2.8}{24.3} = 0.115 \text{ rad/s}^2$

• Time to reach speed: $\Delta t = \frac{\Delta \omega}{\alpha} = \frac{1360}{0.115} = 11,800 \text{ s} = 3.3 \text{ hours}$

• Assessment: The long time is expected due to large inertia and modest torque.

Example Problem: Angular Acceleration of a Falling Baseball Bat

Finds the angular acceleration of a bat falling from vertical, modeled as a uniform cylinder.

• Center of gravity: $L/2 = 0.43$ m

• Torque: $\tau = mg \sin 20^\circ \cdot \frac{L}{2} = 1.18$ Nm

• Moment of inertia (about end): $I = \frac{1}{3}mL^2 = 0.20$ kg·m2

• Angular acceleration: $\alpha = \frac{\tau}{I} = \frac{1.18}{0.20} = 5.8$ rad/s2

• Mass cancels out; only length matters for acceleration.

Constraints Due to Ropes and Pulleys

When a pulley turns without the rope slipping, the rope's speed matches the rim speed of the pulley. The attached object must have the same speed and acceleration as the rope.

• Speed constraint: $v_{rim} = \omega R$

• Acceleration constraint: $a_{rim} = \alpha R$

Example 7.19: Time for a Bucket to Fall

Analyzes the time for a bucket to fall when attached to a rope wound around a rotating cylinder.

• Strategy: Treat the fall of the bucket and spin of the cylinder as connected problems.

• Preparation: Rope tension acts upward on bucket and downward on cylinder; forces are equal in magnitude.

• Newton's second law for bucket: $ma_y = T - mg$

• Torque on cylinder: $\tau = TR$

• Moment of inertia (cylinder): $I = \frac{1}{2}MR^2$

• Angular acceleration: $\alpha = \frac{TR}{I} = \frac{2T}{MR}$

• Constraint (no slip): $a_y = -\alpha R$

• Solving for acceleration: $a_y = -\frac{g}{1 + M/2m}$

• Time to fall: $\Delta t = \sqrt{\frac{2\Delta y}{a_y}} = 1.1$ s

• Assessment: If cylinder mass $M = 0$, $a_y = -g$ (free fall). With cylinder, acceleration is reduced and fall takes longer.