Energy and Momentum of Rotating Systems: Unit Diagnostic (Physics 1 Unit 6)

This diagnostic is for the algebra-based Physics 1 course, Unit 6 (Energy and Momentum of Rotating Systems). It shows you which topics to revisit. Allow about 30 minutes, then use the table at the end.

These are original Marlbridge practice questions, not past exam questions. They have not been calibrated against real exam results, so your total does not predict an exam score. Use a scientific calculator, g = 9.8 m/s² and G = 6.67 × 10⁻¹¹ N·m²/kg², with angles in radians. Axles are frictionless unless stated, and all planets are fictional.

Question 1 (multiple choice · 6.1)

A light rod carries two small beads, each r from the axle, and spins at ω. In a second set-up the same beads sit r/2 from the axle and the rod spins at 2ω. How does the second set-up’s rotational kinetic energy compare with the first?

(A) It is 4 times as large
(B) It is 2 times as large
(C) It is the same
(D) It is ¼ as large

Answer and explanation

Answer: (C). K = ½Iω² with I = Σmr². Halving r divides I by 2² = 4. Doubling ω multiplies ω² by 4. The two factors cancel.

(A) only counts the change in ω.
(B) treats I as proportional to r instead of r².
(D) only counts the change in I.

If you missed this: read “Comparing scenarios: factors of change” in the Topic 6.1 study guide.

Question 2 (multiple choice · 6.2)

A wheel (I = 0.20 kg·m²) starts at rest; its direction of turning is positive. A motor exerts +5.0 N·m while it turns from 0 to 3.0 rad, then a brake exerts −3.0 N·m from 3.0 rad to 5.0 rad. What is its angular speed at 5.0 rad?

(A) 12 rad/s
(B) 9.5 rad/s
(C) 6.7 rad/s
(D) 14 rad/s

Answer and explanation

Answer: (B). The work is the signed area under the torque–angle graph: 5.0 × 3.0 = +15 J, then −3.0 × 2.0 = −6.0 J. Net work 9.0 J = ½Iω², so ω = √(2 × 9.0 ÷ 0.20) = 9.5 rad/s.

(A) ignores the brake (15 J).
(C) forgets the ½ in ½Iω².
(D) adds the two areas (21 J). The brake’s work is negative.

If you missed this: read “When the torque changes: area under the graph” in the Topic 6.2 study guide.

Question 3 (multiple choice · 6.3)

Take counterclockwise as positive. A wheel with I = 0.40 kg·m² turns at +6.0 rad/s. A constant torque of −1.2 N·m then acts on it for 4.0 s. What is its angular velocity at the end?

(A) 0
(B) +18 rad/s
(C) −12 rad/s
(D) −6.0 rad/s

Answer and explanation

Answer: (D). L₀ = 0.40 × 6.0 = +2.4 kg·m²/s. The angular impulse is −1.2 × 4.0 = −4.8 N·m·s. So L = 2.4 − 4.8 = −2.4 kg·m²/s and ω = −6.0 rad/s. The torque stopped the wheel after 2.0 s, then reversed it.

(A) assumes the torque only stops the wheel. It keeps acting afterwards.
(B) treats the angular impulse as positive.
(C) forgets L₀: the theorem gives the change in L.

If you missed this: read “The rotational impulse–momentum theorem” and Worked example 1 in the Topic 6.3 study guide.

Question 4 (multiple choice · 6.4)

Take counterclockwise (seen from above) as positive. A platform (I = 400 kg·m²) is at rest on a frictionless vertical axle. A 60 kg student, treated as a point, stands on its edge, 2.0 m from the axle, then walks counterclockwise around the edge at 1.5 m/s relative to the ground. What is the platform’s angular velocity?

(A) −0.45 rad/s
(B) +0.45 rad/s
(C) 0, because nothing outside the platform pushes it
(D) −0.75 rad/s

Answer and explanation

Answer: (A). System: student + platform. Foot–platform friction is internal, so L stays zero. The student has L = mvr = +180 kg·m²/s, so the platform needs −180 kg·m²/s: ω = −180 ÷ 400 = −0.45 rad/s.

(B) has the wrong direction: the total must stay zero.
(C) forgets that the student’s feet push the platform backwards.
(D) gives the platform the student’s angular speed, v / r. Angular momenta cancel, not angular velocities.

If you missed this: read “Choosing the system” in the Topic 6.4 study guide.

Question 5 (multiple choice · 6.4)

A turntable with I = 0.030 kg·m² spins at 10 rad/s. A 0.050 kg lump of putty is thrown horizontally at 4.0 m/s, aimed straight at the axle. It hits the rim, 0.20 m from the axle, and sticks. What is the new angular speed?

(A) 10 rad/s
(B) 5.0 rad/s
(C) 9.4 rad/s
(D) 11 rad/s

Answer and explanation

Answer: (C). The putty moves straight at the axle, so its angular momentum about it is zero. L = 0.30 kg·m²/s before and after. Afterwards I = 0.030 + 0.050 × 0.20² = 0.032 kg·m², so ω = 0.30 ÷ 0.032 = 9.4 rad/s.

(A) forgets that the putty adds rotational inertia.
(B) averages 10 rad/s with the putty’s zero.
(D) adds mvr = 0.040 kg·m²/s as if the putty moved tangentially.

If you missed this: read “Total angular momentum of a system” and Worked example 2 in the Topic 6.4 study guide.

Question 6 (multiple choice · 6.5)

A uniform solid cylinder (I = ½MR²) rolls without slipping, and its total kinetic energy is 24 J. A thin hoop (I = MR²) of the same mass rolls without slipping at the same speed. What is the hoop’s total kinetic energy?

(A) 24 J
(B) 32 J
(C) 48 J
(D) 16 J

Answer and explanation

Answer: (B). For rolling, K = ½(1 + β)Mv². For the cylinder, 1.5 × ½Mv² = 24 J, so ½Mv² = 16 J. The hoop has β = 1, so K = 2 × 16 = 32 J.

(A) assumes the same mass and speed give the same energy.
(C) applies the hoop’s factor of 2 to the cylinder’s total instead of to ½Mv².
(D) is only the translational part.

If you missed this: read “Kinetic energy of a rolling object” in the Topic 6.5 study guide.

Question 7 (multiple choice · 6.6)

A satellite moves to a new circular orbit with 4 times the radius of its old one around the same planet. By what factors do its speed and its angular momentum about the planet’s centre change?

(A) Speed × ¼; angular momentum unchanged
(B) Speed × ½; angular momentum × ½
(C) Speed × 2; angular momentum × 8
(D) Speed × ½; angular momentum × 2

Answer and explanation

Answer: (D). v² = GM/r, so v ∝ 1/√r and the speed halves. L = mvr changes by ½ × 4 = 2.

(A) treats v as proportional to 1/r and assumes L is conserved between orbits. It is constant only within one orbit.
(B) forgets that r also changes in mvr.
(C) assumes a bigger orbit is faster. Weaker gravity farther out needs less speed.

If you missed this: read “Comparing orbits of different radius” in the Topic 6.6 study guide.

Question 8 (short answer · 6.3)

A wheel (I = 0.50 kg·m²) starts at rest; its direction of spin is positive. A motor exerts +2.0 N·m for 3.0 s. A brake then exerts a constant −1.5 N·m until the wheel stops.

(a) Find the wheel’s angular momentum and angular speed at t = 3.0 s. (b) Find how long the brake takes to stop the wheel. (c) Sketch the angular momentum–time graph from t = 0 until the wheel stops, giving the slope of each part.

Worked answer

(a) Starting from rest, L = τΔt = 2.0 × 3.0 = 6.0 kg·m²/s, so ω = 6.0 ÷ 0.50 = 12 rad/s.

(b) The brake must remove 6.0 kg·m²/s: Δt = ΔL ÷ τ = −6.0 ÷ (−1.5) = 4.0 s. The wheel stops at t = 7.0 s.

(c) A straight line rising from 0 to 6.0 kg·m²/s between 0 and 3.0 s (slope +2.0 N·m), then a straight line falling to zero at 7.0 s (slope −1.5 N·m). Each slope is the net torque.

Check yourself: 1 point each for 6.0 kg·m²/s with 12 rad/s, 4.0 s, the shape, and slopes equal to the torques.

If you missed this: read “Reading an angular momentum–time graph” and Worked example 2 in the Topic 6.3 study guide.

Question 9 (short answer · 6.5)

A uniform solid cylinder of mass 3.0 kg and radius 0.10 m (I = ½MR² = 0.015 kg·m²) rolls without slipping across a level floor at 2.0 m/s.

(a) Find its angular speed and its total kinetic energy. (b) It rolls onto a level patch of frictionless ice. State what happens to v_cm and ω on the ice, and explain. (c) It then meets a frictionless icy ramp. Find the greatest height its centre rises, and its angular speed there.

Worked answer

(a) ω = v / R = 2.0 ÷ 0.10 = 20 rad/s. K = ½ × 3.0 × 2.0² + ½ × 0.015 × 20² = 6.0 + 3.0 = 9.0 J.

(b) Both stay constant. No horizontal force acts, so v_cm is fixed, and no force has a torque about the centre, so ω is fixed. v_cm = Rω still holds: rolling on level ground needs no friction.

(c) Gravity and the normal force act through the centre, so there is no torque and ω stays 20 rad/s. Only the translational 6.0 J becomes potential energy: h = v² ÷ 2g = 0.20 m. (On a rough ramp the spin energy would convert too: 0.31 m.)

Check yourself: 1 point each for 20 rad/s with 9.0 J, both constant on the ice with reasons, ω unchanged on the ramp, and 0.20 m from translational energy only.

If you missed this: read “Friction in ideal rolling does no work” and Worked example 2 in the Topic 6.5 study guide.

Question 10 (short answer · 6.6)

A 600 kg satellite moves in a circular orbit of radius 1.2 × 10⁷ m around a fictional planet of mass 4.5 × 10²⁴ kg.

(a) Find its orbital speed and the escape speed at that radius. (b) Find the total mechanical energy of the satellite–planet system. (c) Find the least energy that must be supplied for the satellite to escape from this orbit, and show that it equals the satellite’s kinetic energy in orbit.

Worked answer

(a) GM = 6.67 × 10⁻¹¹ × 4.5 × 10²⁴ = 3.0 × 10¹⁴ N·m²/kg. v = √(GM / r) = 5.0 × 10³ m/s. v_esc = √(2GM / r) = 7.1 × 10³ m/s, which is √2 times the orbital speed.

(b) K = ½mv² = 7.5 × 10⁹ J and U_g = −GMm / r = −1.5 × 10¹⁰ J, so E = −7.5 × 10⁹ J.

(c) To escape, E must rise to zero, so at least 7.5 × 10⁹ J is needed. That equals K, because E = −K in a circular orbit. (Do not use ½m(v_esc − v)²: subtract energies, not speeds.)

Check yourself: 1 point each for both speeds, negative E, the condition E = 0, and the link to E = −K.

If you missed this: read “Circular orbits: everything is constant” and “Escape speed” in the Topic 6.6 study guide.

Your next step

Mark each question right or wrong. Count Questions 8 to 10 as right only if you earned every point.

Topic	Question(s)	If you missed it, read
6.1 Rotational kinetic energy	1	Topic 6.1 study guide
6.2 Torque and work	2	Topic 6.2 study guide
6.3 Angular momentum and angular impulse	3, 8	Topic 6.3 study guide
6.4 Conservation of angular momentum	4, 5	Topic 6.4 study guide
6.5 Rolling	6, 9	Topic 6.5 study guide
6.6 Motion of orbiting satellites	7, 10	Topic 6.6 study guide

How to use your result

Missed one question in a topic? Re-read the section named in the answer, then try that topic’s practice set.
Missed two or more questions in a topic? Start again with that study guide and its worked examples before mixed practice.
Guessed? Treat the question as missed.
Look for patterns. Errors in Questions 2 to 4 often come from signs; errors in Questions 1, 6 and 7 from a missing square or square root.
All correct? Move straight to the mixed unit review.