PHSX 220 Physics I (w/ Calculus) · Unit 3: Momentum, rotation, gravitation · on Exam 3
Ch 10.4–10.8 Rotational dynamics
Study notes for Rotational dynamics in PHSX 220 Physics I (w/ Calculus) at Montana State University: the big ideas, key formulas, a worked example, common mistakes and an exam tip. The same topic has endless practice questions and flashcards in Mathub.
Big ideas
- Moment of inertia $I = \sum m_ir_i^2$ (or $\int r^2\,dm$) is rotational mass: it depends on how far the mass sits from the axis. Point mass $mr^2$; hoop $MR^2$; solid disk/cylinder $\tfrac12 MR^2$; solid sphere $\tfrac25 MR^2$; thin rod about center $\tfrac1{12}ML^2$, about end $\tfrac13 ML^2$.
- Newton's second law for rotation: $\sum\tau = I\alpha$ about a fixed axis. Torque from a force at radius $r$: $\tau = rF\sin\theta$.
- Rotational kinetic energy $K = \tfrac12 I\omega^2$. Work by a torque $W = \tau\theta$; power $P = \tau\omega$. The work–energy theorem holds with these.
- Parallel-axis theorem: $I = I_{cm} + Md^2$.
- Pulley with mass: the two rope tensions differ, and $(T_1 - T_2)R = I\alpha$ with $a = R\alpha$.
Key formulas
Moment of inertia
$$I = \sum m_i r_i^2;\quad I_{\text{disk}} = \tfrac12 MR^2,\ I_{\text{sphere}} = \tfrac25 MR^2,\ I_{\text{hoop}} = MR^2,\ I_{\text{rod,center}} = \tfrac1{12}ML^2$$
Dynamics and energy
$$\sum\tau = I\alpha,\qquad K_{\text{rot}} = \tfrac12 I\omega^2,\qquad W = \tau\theta,\qquad P = \tau\omega$$
Worked example
A 4.0 kg solid disk of radius 0.20 m is spun up by a 10 N tangential force at its rim. Find $\alpha$.
Show the solution
$I = \tfrac12(4)(0.04) = 0.08\ \text{kg·m}^2$; $\tau = 10(0.20) = 2.0$ N·m; $\alpha = 2.0/0.08 = 25\ \text{rad/s}^2$.
Common mistakes
- Using the wrong moment of inertia (disk vs hoop) or the wrong axis.
- Using the diameter instead of the radius in I.
- Assuming the rope tension equals the hanging weight when a massive pulley accelerates.
Exam tipThe rotational world is a dictionary: m→I, F→τ, a→α, v→ω, x→θ, p→L. Every linear formula you know has a twin.
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