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On this page

  • 1. Centre of Mass (the "balance point")
  • 2. Moment of Inertia III (rotational "heaviness")
  • 3. Rotational Kinetic Energy
  • 4. Rolling Without Slipping
  • 5. Torque and Angular Acceleration
  • 6. Angular Momentum L\mathbf{L}L
  • 7. Collisions in Rotation
  • 8. Key take-aways

Mechanics

Chapter 12: Rotation of a Rigid Body

Evan Luo · May 12, 2025

Mechanics

Chapter 12: Rotation of a Rigid Body

Evan LuoMay 12, 2025

4 min read

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

rcm=1M∑imi ri\mathbf{r}_{\text{cm}} = \frac{1}{M}\sum_i m_i\,\mathbf{r}_ircm​=M1​i∑​mi​ri​

where M=∑imiM = \sum_i m_iM=∑i​mi​. You measure each ri\mathbf{r}_iri​ 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.

Because it is uniform, you can cut it into rectangles, find the COM of each rectangle, and use the formula above. The COM lands at point B on the diagram.

Click to reveal

2. Moment of Inertia III (rotational "heaviness")

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

I=∑imiri2I = \sum_i m_i r_i^2I=i∑​mi​ri2​

with rir_iri​ measured to the axis of rotation.

Why it matters.

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

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

The hoop - its mass lives farther from the centre, so III is MR2MR^2MR2 versus 12MR2\tfrac{1}{2}MR^221​MR2 for the disk.

Click to reveal

3. Rotational Kinetic Energy

When every bit of a rigid body spins with angular speed ω\omegaω,

Krot=12Iω2(always positive)K_{\text{rot}} = \tfrac{1}{2}I\omega^2 \quad\text{(always positive)}Krot​=21​Iω2(always positive)

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

Ktotal=12Mvcm2+12Iω2K_{\text{total}} = \tfrac{1}{2}Mv_{\text{cm}}^2 + \tfrac{1}{2}I\omega^2Ktotal​=21​Mvcm2​+21​Iω2

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


4. Rolling Without Slipping

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

vcm=Rωv_{\text{cm}} = R\omegavcm​=Rω

because in one full turn the centre travels one circumference 2πR2\pi R2πR.

  • Top edge briefly moves twice as fast (vtop=2Rωv_{\text{top}} = 2R\omegavtop​=2Rω)
  • Bottom point is instantaneously at rest (vbottom=0v_{\text{bottom}} = 0vbottom​=0)

Practice: A bowling ball (solid sphere) starts 0.72 m0.72\text{ m}0.72 m high and rolls down. Find its speed at the bottom.

Use energy: Mgh=12Mv2+12Iω2Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2Mgh=21​Mv2+21​Iω2 with I=25MR2I = \tfrac{2}{5}MR^2I=52​MR2 and v=Rωv = R\omegav=Rω. Solving gives v≈3.17 m/sv \approx 3.17\text{ m/s}v≈3.17 m/s.

Click to reveal

5. Torque and Angular Acceleration

The rotational twin of F=maF = maF=ma is

α=∑τI\alpha = \frac{\sum \tau}{I}α=I∑τ​

where τ=rF⊥\tau = rF_\perpτ=rF⊥​ is torque and α\alphaα 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 mR = 0.06\text{ m}R=0.06 m).

Write Newton's 2nd law for each block plus rotational ∑τ=Iα\sum\tau = I\alpha∑τ=Iα for the pulley, relate a=Rαa = R\alphaa=Rα, solve → a≈2.6 m/s2a \approx 2.6\text{ m/s}^2a≈2.6 m/s2.

Click to reveal

6. Angular Momentum L\mathbf{L}L

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

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

Linitial=Lfinal\mathbf{L}_{\text{initial}} = \mathbf{L}_{\text{final}}Linitial​=Lfinal​

Dramatic example: A star shrinks from radius 7.0×105 km7.0 \times 10^5\text{ km}7.0×105 km (1 rev per 30 days) to 1.6×103 km1.6 \times 10^3\text{ km}1.6×103 km. Its huge drop in III makes ω\omegaω skyrocket to ≈ 4.4 rpm!


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 III is I1+I2I_1 + I_2I1​+I2​
  • You can find the shared final ω\omegaω from I1ω1=(I1+I2)ωfI_1\omega_1 = (I_1 + I_2)\omega_{\text{f}}I1​ω1​=(I1​+I2​)ωf​

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 ω\omegaω?

Compute I=12MR2I = \tfrac{1}{2}MR^2I=21​MR2 for each disk, apply conservation → ωf≈7.3 rad/s\omega_{\text{f}} \approx 7.3\text{ rad/s}ωf​≈7.3 rad/s. About 27% of the mechanical energy turns into heat/sound etc.

Click to reveal

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 (KrotK_{\text{rot}}Krot​) but follows familiar conservation ideas
  3. Rolling links translation and rotation with vcm=Rωv_{\text{cm}} = R\omegavcm​=Rω
  4. Net torque changes angular motion; no torque → constant ω\omegaω
  5. Angular momentum is the king of rotational conservation laws, just like linear ppp in straight-line motion

Source: https://notes.ohevan.com/notes/mechanics/12-rotation-of-a-rigid-body

© 2026 Evan Luo. All rights reserved.

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·Last edited May 17, 2025
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© 2026 Evan Luo. All rights reserved.