Rotational Motion · quiz worked solutions

Tried the quiz first? Good. Here’s how to work through each answer. Follow the link beside a question to jump back to its original spot. Questions marked Extra exploration go beyond this AP Physics 1 course; Book controls questions explain the interface.

Angular Forces · question 1

Back to the question · AP Topic 5.6

Question: A uniform solid disk wheel of mass 35.5 kg and diameter 0.75 m starts from rest and rolls without slipping 10 m in 10 s on level ground. A constant driving torque acts at its axle; rolling resistance is negligible. What driving torque is required?

Answer: 3.99 N·m.

1. Find a = 2s/t² = 2(10)/10² = 0.20 m/s² and α = a/R with R = 0.375 m. Why: The wheel starts from rest and rolls without slipping.

2. Static friction supplies ma, while the axle torque supplies both translation and spin: τ = (mR + I/R)a. Why: Combining ΣF = ma and τ − fR = Iα avoids counting only the spin.

3. For a solid disk I = ½mR², so τ = 1.5mRa = 1.5(35.5)(0.375)(0.20) = 3.99375 N·m. Why: Ideal static friction does not imply zero acceleration or zero axle work.

Angular Forces · question 2

Back to the question · AP Topic 5.2

Question: If a 1 m long fan blade has a rotational velocity of 3 rad/s , how much faster is the tip traveling versus a point in the middle of the fan blade?

Answer: 1.5 m/s faster.

1. Use tangential speed v = rω. Why: All points share angular speed, not tangential speed.

2. The tip is at 1 m and midpoint at 0.5 m, giving Δv = (1 − 0.5)(3) = 1.5 m/s. Why: The farther point travels a longer arc each second.

Angular Forces · question 3

Back to the question · AP Topic 5.3

Question: What torque is generated in the following set-up? Positive torque results in a Counter-Clockwise acceleration.

Answer: +6 N·m, counterclockwise.

1. Use the pictured 2 m radius and signed tangential moments. Why: The top 15 N force turns clockwise; the 10 N and 8 N forces turn counterclockwise.

2. Add τ = 2(−15 + 10 + 8) = +6 N·m. Why: The plus sign follows the question’s counterclockwise convention.

Angular Forces · question 4

Back to the question · AP Topic 5.3

Question: To balance the moment of a 100 N force perpendicular to a bat at a point 0.80 m from the batter’s hands, what torque magnitude must the batter exert about the hands?

Answer: 80 N·m.

1. Use the lever arm 0.80 m and perpendicular force 100 N. Why: For a perpendicular force, the full distance is the moment arm.

2. Find τ = rF = (0.80)(100) = 80 N·m, opposite the applied moment. Why: Balancing requires equal magnitude and opposite rotational tendency.

Angular Forces · question 5

Back to the question · AP Topic 5.6

Question: A uniform solid disk wheel has mass 200 kg and radius 2.0 m. Its current angular speed is 8.0 rad/s, and a constant net torque of 800 N·m acts in its rotation direction for 10 s. What is the tangential speed of a point 0.50 m from the axis then?

Answer: 14 m/s.

1. Find disk inertia I = ½(200)(2²) = 400 kg·m² and α = 800/400 = 2 rad/s². Why: The given torque is the net torque.

2. After 10 s, ω = 8 + 2(10) = 28 rad/s; at r = 0.50 m, v = rω = 14 m/s. Why: Use the point’s 0.50 m radius rather than the disk’s outer 2 m radius for its speed.

Angular Momentum · question 1

Back to the question · AP Topic 6.3

Question: Two particles have momenta of magnitudes 4 and 5 kg·m/s. Their straight paths have perpendicular offsets of 3 and 4 m from the same origin. Find the ratio of their angular-momentum magnitudes, particle 1 to particle 2.

Answer: 3/5.

1. Use L = p r_perp for each straight path. Why: The offset is the perpendicular distance from the origin to the motion line.

2. Find L₁ = 4(3) = 12 and L₂ = 5(4) = 20 kg·m²/s; L₁/L₂ = 12/20 = 3/5. Why: Angular momentum depends on the selected origin as well as linear momentum.

Angular Momentum · question 2

Back to the question · AP Topic 6.3

Question: Two particles have momenta of magnitudes 4 and 5 kg·m/s and perpendicular path offsets of 3 and 4 m from O. Their angular-momentum vectors point respectively into and out of the page. With out of the page positive, find the net angular momentum about O.

Answer: +8 kg·m²/s.

1. Compute magnitudes L₁ = 12 and L₂ = 20 kg·m²/s. Why: Each magnitude uses momentum times perpendicular offset.

2. With out of page positive, add −12 + 20 = +8. Why: Opposite rotational senses subtract rather than add magnitudes.

Angular Momentum · question 3

Back to the question · AP Topic 6.4

Question: A man sits on a spinning stool with his arms folded in. When he extends his arms, what happens?

Answer: Moment of inertia increases and angular speed decreases.

1. Choose man plus stool, with negligible external torque. Why: That makes total angular momentum Iω constant.

2. Extending the arms moves mass farther from the axis and increases I, so ω must decrease. Why: Constant angular momentum does not require constant rotational kinetic energy.

Angular Momentum · question 4

Back to the question · AP Topic 6.3

Question: Which of the following has the largest angular momentum given a constant R → and p → .

Answer: Diagram C.

1. Compare L = rp sinθ at fixed r and p. Why: Only the angle changes among the diagrams.

2. C has perpendicular r and p, so sinθ = 1 and L is largest. Why: Antiparallel vectors in D give zero, despite their large drawn lengths.

Angular Momentum · question 5

Back to the question · Extra exploration

Question: A spinning wheel has its angular momentum vector pointing in the + z direction.If you press gently on the axle in the - x direction, as shown in the diagram, in what direction does the axle move?

Answer: The axle moves toward −y in the pictured setup.

1. The force acts at the top of the axle, so r points +z and F points −x; τ = r × F points −y. Why: A force direction alone is not enough; its point of application fixes the torque.

2. Use ΔL = τΔt to tip the original +z angular momentum toward −y. Why: The axle follows that tip under the gyroscope’s stated gentle-precession model; this three-dimensional motion is enrichment.

Circular Motion · question 1

Back to the question · AP Topic 2.9

Question: On a level, unbanked road, which real force supplies the inward horizontal resultant that keeps a car on a circular path?

Answer: Static friction from the road.

1. Choose the car and identify real horizontal forces. Why: On a level curve, weight and normal are vertical.

2. Road friction supplies the inward resultant mv²/R. Why: Centripetal names the required resultant direction, not a separate interaction to add to the FBD.

Circular Motion · question 2

Back to the question · AP Topic 2.9

Question: Jake rides his bicycle at the same centripetal-acceleration magnitude, 3 m/s², on circles of radii r_i = 0.10 m and r_f = 0.30 m. What is the speed ratio v_i/v_f?

Answer: v_i/v_f = 0.577.

1. Use v = √(a_c r) at the same acceleration magnitude. Why: The bicycle’s speed must change when radius changes.

2. Find v_i/v_f = √(0.10/0.30) = 1/√3 ≈ 0.577. Why: This compares speeds, not their squares.

Circular Motion · question 3

Back to the question · AP Topic 2.9

Question: A proposed circular-orbit model uses orbital radius 1.00 × 10⁶ m, speed 20 m/s and planet mass 3000 kg. What value of G would those data imply from v² = GM/r, and is it compatible with the measured universal G = 6.67430 × 10⁻¹¹ N·m²/kg²?

Answer: 1.33×10⁵ N·m²/kg², incompatible with measured G.

1. Rearrange v² = GM/r to G = v²r/M. Why: The circular model sets gravity equal to the needed centripetal resultant.

2. Insert (20²)(1.00×10⁶)/3000 = 1.333×10⁵ N·m²/kg². Why: The huge disagreement with 6.67430×10⁻¹¹ shows that these proposed orbit data do not describe physical universal gravity.

Circular Motion · question 4

Back to the question · AP Topic 2.9

Question: On a level unbanked curve of radius R, a bicycle has maximum nonskid speed v. If the wet-road static friction coefficient becomes one third of its former value, what is the new maximum speed at the same radius?

Answer: v/√3.

1. At the nonskid limit, μₛmg = mv²/R, so v_max = √(μₛgR). Why: Static friction supplies the inward resultant.

2. Replacing μₛ by μₛ/3 multiplies the maximum speed by √(1/3). Why: A one-third force limit produces a square-root speed change.

Circular Motion · question 5

Back to the question · AP Topic 2.9

Question: A 2.0 kg box moves on the inside of a vertical circular loop of radius 10 m. Its speed at the top is 20 m/s. With g the gravitational-acceleration magnitude, what is the inward normal force there?

Answer: 60.38 N inward at g = 9.81 m/s².

1. At the loop top, both weight and normal point inward. Why: Choose toward the center as the positive radial direction.

2. Use N + mg = mv²/R = (2)(20²)/10 = 80 N, so N = 80 − 19.62 = 60.38 N. Why: The centripetal requirement is the total, not the normal force alone.

Moment of Inertia · question 1

Back to the question · AP Topic 5.4

Question: The object rotates about the axis shown. Its parameters are given in the image. What is the moment of inertia?

Answer: 10.8 kg·cm², or 0.00108 kg·m².

1. Read the solid cone’s m = 4 kg, radius 3 cm and central symmetry axis. Why: Its height does not enter inertia about this axial direction.

2. Use I = (3/10)mr² = 0.3(4)(3²) = 10.8 kg·cm². Why: Converting cm² to m² requires multiplying by 10⁻⁴; this particular shape formula is supplied in the retained reference table.

Moment of Inertia · question 2

Back to the question · AP Topic 5.4

Question: The object rotates about the axis shown. Its parameters are given in the image. What is the moment of inertia?

Answer: 3.9 kg·m².

1. Read the thick cylindrical shell’s mass 25 kg, inner radius 0.25 m and outer radius 0.50 m. Why: The stated central axis runs along the cylinder’s length.

2. Use I = ½m(r₁² + r₂²) = ½(25)(0.25² + 0.50²) = 3.90625 kg·m². Why: Both radii contribute for a thick shell.

Moment of Inertia · question 3

Back to the question · AP Topic 5.4

Question: The object rotates about the axis shown. Its parameters are given in the image. What is the moment of inertia?

Answer: About 1.67 kg·m².

1. Read the uniform cylindrical rod’s m = 5 kg, length 1 m, radius 0.05 m and transverse end axis. Why: The axis is at an end, not at the center.

2. Use I = mL²/3 + mR²/4 = 5/3 + 5(0.05²)/4 = 1.66979 kg·m². Why: The slender-rod limit mL²/3 alone already rounds to the keyed 1.67.

Moment of Inertia · question 4

Back to the question · AP Topic 5.4

Question: A juggler is spinning a thin disk-like plate on a stick. If the plate’s moment of inertia is 25 000 g ⋅ mm 2 , and if the mass of the plate is 5 grams , what is the radius of the plate?

Answer: 10 cm.

1. Use the disk relation I = ½mr² with I = 25,000 g·mm² and m = 5 g. Why: Keeping both in the same mass unit is valid.

2. Find r² = 2I/m = 10,000 mm², so r = 100 mm = 10 cm. Why: Taking a square root gives length rather than area.

Moment of Inertia · question 5

Back to the question · AP Topic 5.4

Question: A soccer ball, modeled as a thin spherical shell of mass 2.0 kg and radius 0.15 m, spins about an axis through its center. What is its moment of inertia?

Answer: 0.030 kg·m².

1. Use the specified thin spherical-shell model I = (2/3)mr². Why: A shell and a solid sphere have different mass distributions.

2. Calculate (2/3)(2.0)(0.15²) = 0.030 kg·m². Why: The 0.018 distractor uses the solid-sphere coefficient instead.

FFR - Rotational Energy · question 1

Back to the question · AP Topic 6.2

Question: A uniform solid disk wheel rolls without slipping up a 15° ramp at an initial center-of-mass speed of 5.0 m/s. Its mass is 3.0 g and radius 0.50 m. With no energy losses, how far along the ramp does it travel before stopping?

Answer: 7.4 m along the ramp.

1. For a rolling solid disk, K = ½mv² + ½(½mR²)(v/R)² = ¾mv². Why: No slipping lets angular speed be replaced by v/R.

2. Set mgd sin15° = ¾m(5²), giving d = 3(25)/(4·9.81·sin15°) ≈ 7.384 m. Why: Both translation and spin supply the climb; the very small mass cancels.

FFR - Rotational Energy · question 2

Back to the question · AP Topic 6.2

Question: A uniform thin-rod propeller of length 2.0 m and mass 15 kg starts from rest and reaches 120 rpm after turning 5.0 revolutions. What constant net torque acts on it?

Answer: 12.6 N·m.

1. Find I = mL²/12 = 15(2²)/12 = 5 kg·m² and ω_f = 120·2π/60 = 4π rad/s. Why: The thin rod rotates about its center and rpm must be converted.

2. Use τΔθ = ΔK_rot with Δθ = 5(2π), so τ = [½(5)(4π)²]/(10π) = 4π ≈ 12.57 N·m. Why: Rotational work gives the constant net torque without needing elapsed time.

FFR - Rotational Energy · question 3

Back to the question · AP Topic 6.2

Question: A uniform solid sphere of radius 0.50 m and mass 1.0 kg starts from rest and rolls without slipping down a hill with a vertical drop of 6.0 m. With no losses, what is its center-of-mass speed at the bottom?

Answer: 9.17 m/s.

1. For a solid sphere, I = (2/5)mR² and no slipping gives K_total = (7/10)mv². Why: The gravitational drop becomes both translational and rotational kinetic energy.

2. Set mg(6) = (7/10)mv², giving v = √(10·9.81·6/7) ≈ 9.17 m/s. Why: Ignoring spin would overestimate the bottom speed.

FFR - Rotational Energy · question 4

Back to the question · Extra exploration

Question: A barrel modeled as a thin cylindrical shell of mass 10 kg and radius 1.0 m reaches a rough level floor with center-of-mass speed 50 m/s and zero spin. It slips until rolling without slipping begins. Neglect its drop height and rolling resistance. What is its angular speed when rolling begins?

Answer: 25.0 rad/s.

1. Let friction impulse be J: mv_f = mv₀ − J and Iω_f = JR. Why: Friction slows the center of mass while giving the initially nonspinning barrel torque.

2. For a thin shell I = mR² and v_f = Rω_f, solve v_f = v₀/2 = 25 m/s. Why: The final rolling condition couples the two impulse equations.

3. Divide by R = 1 m to get ω_f = 25 rad/s. Why: Mechanical energy decreases during slipping; a lossless rolling-energy equation cannot solve this transition.

FFR - Rotational Energy · question 5

Back to the question · Extra exploration

Question: A uniform solid sphere of radius 0.25 m and mass 0.50 kg slides at 5.0 m/s with zero spin. It then enters a rough level floor and slips until rolling without slipping begins. What is its final center-of-mass speed?

Answer: 3.57 m/s.

1. Write mv_f = mv₀ − J and Iω_f = JR with I = (2/5)mR². Why: Contact friction transfers momentum and spins up the sphere.

2. Apply v_f = Rω_f: v_f = v₀/[1 + I/(mR²)] = 5/(1 + 2/5) = 25/7 ≈ 3.571 m/s. Why: This is quantitative slipping-to-rolling enrichment, with kinetic energy lost as heat.

Model references

These original worked steps use the stated question models. Physics references: Torque, Rolling motion. AP topic labels follow the supplied-edition College Board CED, Effective Fall 2024, ©2026, pinned in the course audit.