FOCUSED REVIEW

Focused Review: Rotational Motion — Calculus-Based

Reinforce the highest-leverage ideas and representative problem-solving tools for rotational motion.

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TIME

15–20 minutes

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BEST FOR

A focused chapter review

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FINISH WITH

A four-response readiness check

After this focused review, you’ll be able to…

interpret angular signs, connect angular and tangential quantities, and build an axis-consistent torque–inertia solution.

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Course Alignment

This bundle is designed to complement the unit listed below. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.

RESOURCE: Physics Sensei Mechanics

UNIT: MEC-U08

UNIT: Rotational Motion

TREATMENT: Calculus-based introductory college physics

BEST USED

✓ After studying the unit

✓ Before starting homework

✓ Before a quiz or exam

Your Review Plan

Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.

6 stages • Approximately 15–20 minutes.

①

Warm-Up Check

Activate prior knowledge.

②

Core Concepts

Review the essential ideas.

③

Guided Practice

Apply what you learned.

④

Confidence Check

Confirm your understanding.

Warm-Up Check

Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.

ACTIVITY 1

Recall Activity 1

Use counterclockwise as positive.

For ω = +5.0 rad/s and α = −1.0 rad/s², state direction and whether angular speed increases or decreases.

Reveal Answer

Counterclockwise and slowing.

Why it works: ω is positive, while α is opposite in sign, so |ω| decreases.

ACTIVITY 2

Recall Activity 2

Compare two radii on one rigid disk.

Point B is three times farther from the axis than A. Compare ω and v_t.

Reveal Answer

Both have the same ω; B has three times the tangential speed.

Why it works: Rigid-body points share angular velocity, but v_t = rω.

ACTIVITY 3

Recall Activity 3

Think in terms of line of action.

When does a nonzero force make zero torque about the selected axis?

Reveal Answer

When its line of action passes through the axis.

Why it works: Then r_⊥ = 0 in τ = r_⊥F.

Ready to strengthen your understanding?

You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?

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Core Concepts

Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.

KEY CONCEPT 1

Signed rotational motion in radians

Use radians in angular–linear relationships. With counterclockwise positive, ω gives rotation direction; ω and α with the same sign increase angular speed, while opposite signs decrease it. Constant-α kinematics may be used only after confirming α is constant. Points on a rigid body share θ, ω, and α but have radius-dependent v_t, a_t, and a_r.

Formula: 1 rev = 2π rad; ω_f = ω_i + αt; Δθ = ½(ω_i + ω_f)t; v_t = rω; a_t = rα; a_r = rω².

Example: Convert first: ω_i = −180(2π/60) = −18.85 rad/s and ω_f = +60.0(2π/60) = +6.283 rad/s. Then α = (ω_f − ω_i)/t = +3.927 rad/s². Since α is constant, Δθ = ½(ω_i + ω_f)t = ½(−18.85 + 6.283)(8.00) = −50.27 rad.

Sensei note: A negative α does not automatically mean slowing. Decide from the signs of both ω and α.

KEY CONCEPT 2

Axis-first rotational dynamics

Moment of inertia belongs to an object–axis pair. Torque also depends on the selected axis and line of action. For a rigid body about a fixed axis, sum signed torques and use I about that same axis. Rotational energy and power follow from I, ω, torque, and angular displacement.

Formula: I = Σm_ir_i² or ∫r² dm; K_rot = ½Iω²; τ = rF sin θ = r_⊥F; Στ = Iα; Στ_ext = dL/dt; W_net = ΔK_rot; P = τω. When net external torque is negligible, angular momentum is conserved; for fixed-axis rigid-body rotation, L = Iω.

Example: A solid disk (M = 6.00 kg, R = 0.200 m) has a 15.0 N tangential force producing counterclockwise torque and a 4.00 N tangential resisting force at the rim. I = ½MR² = 0.120 kg·m². τ_net = +(15.0)(0.200) − (4.00)(0.200) = +2.20 N·m, so α = 18.3 rad/s². If this net torque remains constant through +3.00 rad, W_net = +6.60 J. At ω = +8.00 rad/s, P_net = +17.6 W.

Sensei note: Choose the axis before finding lever arms or I. Use signed torques, not a sum of positive magnitudes.

FOCUSED SYNTHESIS

Connect signs, radius, and axis

First establish radians and the rotational sign convention. Then distinguish shared angular quantities from radius-dependent tangential quantities. For dynamics, choose one fixed axis and use it consistently for lever arms, signed torques, and moment of inertia.

Decision rule: Signs describe the rotational state; radius connects angular and tangential quantities; the selected axis controls both torque and moment of inertia.

Ready to apply these ideas?

You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?

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Guided Practice

Now it's time to apply what you've reviewed.

Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.

PRACTICE 1

Guided Example

Use counterclockwise as positive and confirm α is constant.

A rotor changes from 180 rpm clockwise to 60.0 rpm counterclockwise in 8.00 s at constant angular acceleration. Find α and Δθ.

Reveal Answer

α = +3.93 rad/s²; Δθ = −50.3 rad (50.3 rad clockwise).

Why it works: Convert first: ω_i = −180(2π/60) = −18.85 rad/s and ω_f = +60.0(2π/60) = +6.283 rad/s. Then α = (ω_f − ω_i)/t = +3.927 rad/s². Since α is constant, Δθ = ½(ω_i + ω_f)t = ½(−18.85 + 6.283)(8.00) = −50.27 rad.

PRACTICE 2

Independent Check

Draw the disk and use signed torque contributions about its axle.

A uniform solid disk has M = 8.00 kg and R = 0.250 m. A 20.0 N tangential force at the rim acts counterclockwise; a 5.00 N tangential force at the rim and a bearing-friction torque of 0.300 N·m act clockwise. Find α. If ω = +6.00 rad/s at an instant, find net power.

Reveal Answer

α = +13.8 rad/s²; P_net = +20.7 W.

Why it works: I = ½MR² = 0.250 kg·m². τ_net = +(20.0)(0.250) − (5.00)(0.250) − 0.300 = +3.45 N·m. Thus α = τ_net/I = +13.8 rad/s². At ω = +6.00 rad/s, P_net = τ_netω = +20.7 W.

BEFORE YOU CHECK

Audit the setup

Before continuing, verify the sign convention, radian conversion, force line of action, torque sign, and selected moment-of-inertia axis in your two solutions.

You’ve reinforced the essential skills through focused practice. Now confirm your understanding with two short questions.

Ready to check your understanding?

You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?

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Confidence Check

You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.

QUICK CHECK 1

Signs and radius

Answer both parts. Score 1 point each (2 total).

(a) If ω and α have opposite signs, what happens to angular speed? (b) Do two points on one rigid disk have equal tangential speed?

Reveal Answer

(a) It decreases. (b) Not generally; v_t depends on radius.

Why it works: Angular signs control |ω|; linear tangential speed scales as rω.

QUICK CHECK 2

Torque and axis

Answer both parts. Score 1 point each (2 total).

(a) What is the lever arm? (b) Which I belongs in Στ = Iα; Στ_ext = dL/dt?

Reveal Answer

(a) Perpendicular distance from the axis to the force line of action. (b) I evaluated about the same fixed axis used for the torques.

Why it works: Both torque and rotational inertia are axis-specific.

READINESS GUIDE

Interpret your score

Score 1 point per correct response across the two confidence checks (4 total).

4 correct: You’re ready to continue. • 3 correct: Review the missed idea, then continue. • 0–2 correct: Revisit Core Concepts or choose more practice.