Chapter 12: Rotation of a Rigid Body

1. Centre of Mass (the "balance point")

What it is. Imagine replacing a whole object with a single point that behaves as if the entire mass were concentrated there. That special point is the centre of mass (COM).

Formula for several parts

$$\mathbf{r}_{\text{cm}} = \frac{1}{M}\sum_i m_i\,\mathbf{r}_i$$

where $M = \sum_i m_i$. You measure each $\mathbf{r}_i$ from a common origin.

Key ideas (plain-speak)

• Heavier pieces pull the COM closer to them
• For uniform (same-density) objects, the COM sits at the geometric centre
• If you track the COM, you know how the whole object "moves through space"

Try it yourself - locate the COM of the L-shaped plate shown in the slides.

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2. Moment of Inertia $I$ (rotational "heaviness")

What it is. In rotation, mass that sits far from the pivot is harder to spin. Moment of inertia captures that:

$$I = \sum_i m_i r_i^2$$

with $r_i$ measured to the axis of rotation.

Why it matters.

• Big $I$ ⟹ object resists angular acceleration (it feels "sluggish")
• Small $I$ ⟹ easier to start or stop spinning
• Tables of $I$ for common shapes (solid disk, ring, sphere...) save you the integration

Quick check: Which has the larger $I$ about the same axle: a solid disk or a thin hoop of the same mass and radius?

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3. Rotational Kinetic Energy

When every bit of a rigid body spins with angular speed $\omega$,

$$K_{\text{rot}} = \tfrac{1}{2}I\omega^2 \quad\text{(always positive)}$$

If the object is also sliding, total kinetic energy is the sum of straight-line and rotational parts:

$$K_{\text{total}} = \tfrac{1}{2}Mv_{\text{cm}}^2 + \tfrac{1}{2}I\omega^2$$

Example: A 400 g disk rolls; you'll add both terms to find its speed at various points on a slope (see slide example).

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4. Rolling Without Slipping

For wheels, balls, or disks that roll, not slide:

$$v_{\text{cm}} = R\omega$$

because in one full turn the centre travels one circumference $2\pi R$.

• Top edge briefly moves twice as fast ($v_{\text{top}} = 2R\omega$)
• Bottom point is instantaneously at rest ($v_{\text{bottom}} = 0$)

Practice: A bowling ball (solid sphere) starts $0.72\text{ m}$ high and rolls down. Find its speed at the bottom.

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5. Torque and Angular Acceleration

The rotational twin of $F = ma$ is

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

where $\tau = rF_\perp$ is torque and $\alpha$ is angular acceleration.

Pulley insight. A massive pulley needs different string tensions on each side to create a net torque and spin.

Try it: Work out the acceleration of two blocks connected over a 3 kg pulley (solid disk, $R = 0.06\text{ m}$).

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6. Angular Momentum $\mathbf{L}$

For one particle: $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. For a rigid body: $\mathbf{L} = I\boldsymbol{\omega}$.

Conservation law. In an isolated system (no external torque)

$$\mathbf{L}_{\text{initial}} = \mathbf{L}_{\text{final}}$$

Dramatic example: A star shrinks from radius $7.0 \times 10^5\text{ km}$ (1 rev per 30 days) to $1.6 \times 10^3\text{ km}$. Its huge drop in $I$ makes $\omega$ skyrocket to ≈ 4.4 rpm!

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7. Collisions in Rotation

During a fast collision, external torques are negligible, so total angular momentum stays the same even if energy does not.

• If two disks stick together, their combined $I$ is $I_1 + I_2$
• You can find the shared final $\omega$ from $I_1\omega_1 = (I_1 + I_2)\omega_{\text{f}}$

Check-your-understanding: The 10 cm, 3.0 kg disk (10 rad/s) meets a 7.5 cm, 2.0 kg disk at rest. What is the final $\omega$?

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8. Key take-aways

1. Centre of mass tells you where the "mass acts"; moment of inertia tells you how that mass is distributed
2. Rotation brings new energy ($K_{\text{rot}}$) but follows familiar conservation ideas
3. Rolling links translation and rotation with $v_{\text{cm}} = R\omega$
4. Net torque changes angular motion; no torque → constant $\omega$
5. Angular momentum is the king of rotational conservation laws, just like linear $p$ in straight-line motion