FULL REVIEW

Full Review: Rotational Motion — Foundational

Review the essential ideas, relationships, and problem-solving tools for rotational motion.



TIME

45–60 minutes



BEST FOR

A complete topic review



FINISH WITH

A readiness check

After this full review, you'll be able to...

recall the essential ideas, apply them to representative problems, and determine what to study next.

Choose how you want to review

Start the Review

Watch the Video Review

Download the PDF

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: Concept-first 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 45-60 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. Classify the motion from the signs before calculating.

A disk has ω = −8.0 rad/s and α = +2.0 rad/s². Is it rotating clockwise or counterclockwise, and is its angular speed increasing or decreasing?

Reveal Answers

Clockwise; angular speed is decreasing.

Why it works: Negative ω means clockwise. Because ω and α have opposite signs, the magnitude |ω| decreases. A positive α does not by itself mean “speeding up.”

ACTIVITY 2

Recall Activity 2

Compare points on one rigid disk at the same instant.

Point B is twice as far from the fixed axis as point A. Compare θ, ω, α, v_t, a_t, and a_r.

Reveal Answers

θ, ω, and α are equal for both points. At B, v_t, a_t, and a_r are each twice their values at A.

Why it works: Rigid-body points share angular variables. The linear magnitudes v_t = rω, a_t = rα, and a_r = rω² scale with radius at a given instant.

ACTIVITY 3

Recall Activity 3

Choose the rotation axis before evaluating torque.

A 30 N force acts on a 0.40 m wrench. Its line of action passes through the bolt axis. What torque does it produce about the bolt?

Reveal Answers

0 N·m.

Why it works: The perpendicular lever arm is zero, so τ = r_⊥F = 0 even though the force and position-vector magnitudes are nonzero.

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?

← View Review Map

Continue to Core Concepts →

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

Rotational kinematics and radius-dependent linear motion

For a rigid body rotating about a fixed axis, angular position θ locates its orientation and Δθ = θ_f − θ_i is signed angular displacement, not total angular distance. Radians arise from θ = s/r and must be used in s = rθ, v_t = rω, and a_t = rα. Average rates use finite changes; instantaneous rates use instantaneous-rate reasoning. With counterclockwise positive, the signs of ω and α jointly determine speeding up or slowing down. Constant-angular-acceleration equations are valid only when α is constant. Tangential acceleration changes speed; radial acceleration points inward and changes direction. When both are present, their perpendicular vector sum has magnitude √(a_t² + a_r²).

θ = s/r; ω_avg = Δθ/Δt; ω = dθ/dt; α_avg = Δω/Δt; α = dω/dt. For constant α: ω_f = ω_i + αt; θ_f = θ_i + ω_it + ½αt²; ω_f² = ω_i² + 2α(θ_f − θ_i); θ_f − θ_i = ½(ω_i + ω_f)t. Also 1 rev = 2π rad = 360°; v_t = rω; a_t = rα; a_r = v_t²/r = rω².

A wheel starts with ω_i = −12.0 rad/s and has constant α = +3.00 rad/s² for 2.00 s. Then ω_f = −6.00 rad/s and Δθ = (−12.0)(2.00) + ½(3.00)(2.00)² = −18.0 rad. It remains clockwise but slows. At r = 0.250 m at the final instant, v_t = 1.50 m/s, a_t = 0.750 m/s², a_r = 9.00 m/s², and a = 9.03 m/s².

Never treat rev as rad: multiply rev by 2π. On graphs, slope of θ(t) is ω; slope of ω(t) is α; signed area under ω(t) is Δθ; signed area under α(t) is Δω.

KEY CONCEPT 2

Moment of inertia and rotational kinetic energy

Moment of inertia I measures rotational inertia about a specified axis. It depends on mass, its distribution, and axis location; it is not an axis-independent property. For particles, sum m_ir_i²; for a continuous body, integrate r² dm. Standard results must be paired with their exact geometry and axis. For parallel axes, I = I_CM + Md², where d is their perpendicular separation. A compound body’s total I is the sum of component moments about the same axis. Rotational kinetic energy K_rot = ½Iω² is a nonnegative scalar.

I = Σm_ir_i²; I = ∫r² dm; I = I_CM + Md²; K_rot = ½Iω². Standard axes: hoop/thin cylindrical shell, central symmetry axis: MR²; solid disk/cylinder, central symmetry axis: ½MR²; solid sphere, diameter: ⅖MR²; thin spherical shell, diameter: ⅔MR²; thin rod length L, axis through center perpendicular to rod: 1/12ML²; thin rod, axis through one end perpendicular to rod: ⅓ML²; rectangular thin plate sides a and b, central perpendicular axis: 1/12M(a² + b²). SI unit: kg·m².

A 4.00 kg, 0.300 m solid disk and a 1.00 kg point mass fixed 0.200 m from its center rotate together at 10.0 rad/s. About the disk axis, I_total = ½(4.00)(0.300²) + (1.00)(0.200²) = 0.220 kg·m². Thus K_rot = ½(0.220)(10.0²) = 11.0 J.

Write the object and axis beside every I formula. For the parallel-axis theorem, the known center-of-mass axis and new axis must be parallel; d is not a radius unless the geometry makes it so.

KEY CONCEPT 3

Torque, rotational dynamics, work, and power

Torque describes a force’s turning effect about a selected axis: vector τ = r × F and magnitude τ = rF sin θ = r_⊥F = rF_⊥, where θ is between r and F and r_⊥ is the perpendicular distance to the line of action. With counterclockwise positive, sum signed torques. A force through the axis has zero torque. For a rigid body about a fixed axis with I evaluated about that same axis, Στ = Iα. Rotational work accumulates torque through angular displacement; net rotational work changes K_rot. Instantaneous power is P = τω for torque and angular velocity components about the same axis; its sign indicates energy transfer into or out of rotational motion.

τ = r × F; |τ| = rF sin θ = r_⊥F; τ_net = Στ; Στ = Iα; rotational work changes K_rot; for constant torque, W = τΔθ; W_net = ΔK_rot; P = dW/dt = τω. Torque unit N·m (not called joules); work/energy J; power W.

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.

Use this order: select system → select axis → draw forces and lines of action → set torque sign → compute each torque → sum torques → choose I about the same axis → solve. A massive pulley can require unequal tensions; write separate force equations and the pulley torque equation.

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?

← View Review Map

Start Guided Practice →

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

Worked 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 Answers

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

Guided Problem

Identify each component and calculate every contribution about the same axis.

A thin rod of mass 3.00 kg and length 1.20 m rotates about an axis through one end, perpendicular to the rod. A 2.00 kg point mass is attached at the far end. Find I_total and K_rot at 4.00 rad/s.

Reveal Answers

I_total = 4.32 kg·m²; K_rot = 34.6 J.

Why it works: I_rod,end = ⅓ML² = ⅓(3.00)(1.20²) = 1.44 kg·m². I_point = mr² = (2.00)(1.20²) = 2.88 kg·m². Total I = 4.32 kg·m²; K_rot = ½Iω² = ½(4.32)(4.00²) = 34.56 J.

PRACTICE 3

Independent Problem

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 Answers

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.

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?

← View Review Map

Start the Confidence Check →

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

Kinematics and graphs

Answer both parts. Score 1 point per correct response (2 total).

(a) If ω < 0 and α < 0, is angular speed increasing or decreasing? (b) On an ω-versus-t graph, what does signed area represent?

Reveal Answers

(a) Increasing. (b) Angular displacement Δθ.

Why it works: Same signs mean |ω| increases. The time integral of ω is Δθ.

QUICK CHECK 2

Axis and torque

Answer both parts. Score 1 point per correct response (2 total).

(a) Why must an I formula name its axis? (b) What torque is produced about an axis by a force whose line of action passes through it?

Reveal Answers

(a) I depends on mass distribution relative to the selected axis. (b) Zero.

Why it works: Changing the axis changes the distances r in I = ∫r² dm. A line of action through the axis gives r_⊥ = 0.

QUICK CHECK 3

Dynamics, work, and power

Answer both parts. Score 1 point per correct response (2 total).

(a) State the conditions for Στ = Iα in this chapter. (b) When is P = τω positive?

Reveal Answers

(a) A rigid body rotates about a fixed axis, and both torque and I are evaluated about that axis. (b) When signed τ and ω about that axis have the same sign.

Why it works: The equation is axis-specific. Positive power increases rotational mechanical energy at that instant; negative power removes it.

How did it go?

You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?

← View Review Map

Review the Summary →

Summary

Before moving on, take one final look at the most important ideas from this review.

KEY TAKEAWAY 1

Read signs and radians first

Declare a positive direction, convert rev or rpm to radians, and use ω and α signs together. Constant-α equations require constant angular acceleration.

KEY TAKEAWAY 2

Name the object and axis

Moment of inertia depends on mass distribution and the selected axis. Add compound-body contributions about one common axis before using K_rot or Στ = Iα.

KEY TAKEAWAY 3