What this is for. This diagnostic shows which Unit 6 (Energy and Momentum of Rotating Systems) topics to revisit. It has at least one question per topic and two short written questions. These are original Marlbridge practice questions, not past exam questions. It is not calibrated, so it does not give a predicted score.
How to sit it. Allow about 30 minutes without notes. A calculator is fine for arithmetic. Use g = 9.8 m/s², radians and rad/s. You may use I = ½MR² for a uniform disc or solid cylinder.
Question 1 (multiple choice · 6.1)
A flying disc, modelled as a uniform disc of mass 0.18 kg and radius 0.13 m, moves at 10 m/s and spins at 40 rad/s about its centre. What fraction of its total kinetic energy is rotational?
- (A) 0.12
- (B) 0.14
- (C) 0.21
- (D) 0.27
Answer and explanation
Answer: (A). K_trans = ½(0.18)(10)² = 9.0 J. I = ½MR² = 1.52 × 10⁻³ kg·m², so K_rot = ½Iω² = 1.22 J. The total is 10.2 J, and 1.22 ÷ 10.2 = 0.12.
- (B) is K_rot ÷ K_trans, not a fraction of the total.
- (C) uses I = MR², as if all the mass were on the rim.
- (D) uses I = MR² and divides by K_trans.
If you missed this: read “Moving and spinning: translation plus rotation” in the Topic 6.1 study guide.
Question 2 (multiple choice · 6.2)
A wheel with I = 0.70 kg·m² starts from rest on a frictionless axle. The only torque on it, in its direction of turning, is 8.0 N·m from θ = 0 to θ = 2.0 rad, then falls in a straight line to zero at θ = 5.0 rad. What is the wheel’s angular speed at θ = 5.0 rad?
- (A) 6.3 rad/s
- (B) 6.8 rad/s
- (C) 8.9 rad/s
- (D) 11 rad/s
Answer and explanation
Answer: (C). The work is the area under the τ–θ graph: a rectangle (16 J) plus a triangle (½ × 3.0 × 8.0 = 12 J), so W = 28 J. Then ½(0.70)ω² = 28 gives ω = √80 = 8.9 rad/s.
- (A) drops the ½ in ½Iω².
- (B) counts only the rectangle; the falling torque still does work.
- (D) treats the torque as 8.0 N·m all the way to 5.0 rad (40 J).
If you missed this: read “Area under a torque–angle graph” in the Topic 6.2 study guide.
Question 3 (multiple choice · 6.3)
Take +x to the right, +y up and +z out of the page. A 0.40 kg puck is at r = (3.0 î + 4.0 ĵ) m and moves with v = (2.0 î) m/s. What is its angular momentum about the origin?
- (A) 3.2 kg·m²/s in the +z direction
- (B) 3.2 kg·m²/s in the −z direction
- (C) 4.0 kg·m²/s in the −z direction
- (D) 2.4 kg·m²/s in the −z direction
Answer and explanation
Answer: (B). L = r × p = m(x v_y − y v_x) k̂ = 0.40 × (3.0 × 0 − 4.0 × 2.0) k̂ = −3.2 k̂ kg·m²/s. The lever arm is 4.0 m, the perpendicular distance from O to the line of motion. Seen from +z, the puck goes clockwise round O: the −z sense.
- (A) has the wrong direction; use the right-hand rule.
- (C) uses |r| = 5.0 m instead of the perpendicular distance.
- (D) uses x = 3.0 m, which lies along the velocity and contributes nothing.
If you missed this: read “Angular momentum of a point object” in the Topic 6.3 study guide.
Question 4 (multiple choice · 6.3)
Take counterclockwise as positive. A wheel with I = 0.15 kg·m² is turning at −2.0 rad/s. A net torque then acts: it rises in a straight line from 0 to +1.2 N·m between t = 0 and t = 0.50 s, then falls in a straight line to 0 at t = 1.5 s. What is the angular velocity at t = 1.5 s?
- (A) +4.0 rad/s
- (B) +6.0 rad/s
- (C) +8.0 rad/s
- (D) +10 rad/s
Answer and explanation
Answer: (A). The angular impulse is the area of the triangle under the τ–t graph: ½ × 1.5 × 1.2 = 0.90 N·m·s = ΔL. So Δω = 0.90 ÷ 0.15 = +6.0 rad/s, and ω = −2.0 + 6.0 = +4.0 rad/s.
- (B) is Δω alone. The wheel did not start from rest.
- (C) ignores the negative sign of the starting angular velocity.
- (D) uses the rectangle 1.2 × 1.5 instead of the triangle.
If you missed this: read “The rotational impulse–momentum theorem” in the Topic 6.3 study guide.
Question 5 (multiple choice · 6.4)
A merry-go-round turns freely on a frictionless axle. A child sitting at its rim holds a heavy bag, then simply lets go, so the bag leaves with the rim’s velocity. What happens to the angular velocity of the merry-go-round and child?
- (A) It increases, because the rotational inertia falls and angular momentum is conserved.
- (B) It decreases, because the bag takes angular momentum away.
- (C) It increases, because the bag pushes back on the rim as it leaves.
- (D) It stays the same, because the bag carries away exactly the angular momentum it had.
Answer and explanation
Answer: (D). Before release, the bag’s angular momentum is m_bag R²ω. Just after, it moves at Rω, so it still has m_bag R²ω about the axle. No force acts between bag and rim at release, so the merry-go-round and child keep (I_mgr + I_child)ω: their ω does not change.
- (A) removes the bag’s rotational inertia but not its angular momentum.
- (B) forgets that the bag also takes its rotational inertia, in proportion.
- (C) would need the child to throw the bag backward.
If you missed this: read “Choosing the system” in the Topic 6.4 study guide.
Question 6 (multiple choice · 6.5)
A uniform solid cylinder rolls without slipping at 3.0 m/s onto a rough ramp, and rolls up it without slipping. How high does its centre rise before it stops?
- (A) 0.23 m
- (B) 0.46 m
- (C) 0.69 m
- (D) 0.92 m
Answer and explanation
Answer: (C). Rolling, K = ½Mv² + ½(½Mr²)(v/r)² = ¾Mv². Static friction does no work, so ¾Mv² = Mgh and h = 3v²/(4g) = 0.69 m.
- (A) uses only the rotational energy, ¼Mv².
- (B) uses only the translational energy, v²/(2g).
- (D) uses I = Mr², the value for a hoop.
If you missed this: read “Kinetic energy of a body that moves and spins” in the Topic 6.5 study guide.
Question 7 (multiple choice · 6.5)
Take +x to the right. A ball spinning clockwise (the sense of rolling to the right) is set down on a level floor with its centre at rest. Which statement is correct while it slips?
- (A) v_cm = rω from the moment the ball touches the floor, because the floor is rough.
- (B) Kinetic friction points to the left; it slows the spin, and the centre stays at rest until rolling begins.
- (C) Kinetic friction points to the right; it speeds up the centre and slows the spin, but the total kinetic energy stays constant.
- (D) Kinetic friction points to the right; it speeds up the centre, slows the spin, and the total kinetic energy falls.
Answer and explanation
Answer: (D). The bottom of a ball spinning clockwise moves to the left relative to the floor, so kinetic friction acts to the right. It accelerates the centre to the right, and its torque about the centre opposes the spin. The contact point slides, so friction dissipates energy until v_cm = rω.
- (A) is the end state. While slipping, v_cm and ω are not linked.
- (B) gets the direction wrong: friction opposes the sliding of the contact point, and any net force moves the centre.
- (C) forgets that sliding friction dissipates energy.
If you missed this: read “Rolling while slipping” in the Topic 6.5 study guide.
Question 8 (multiple choice · 6.6)
A probe is launched straight up from the surface of an airless fictional planet of mass M and radius R. Its launch speed is √(GM/R), the circular-orbit speed at the surface. How high above the surface does it rise?
- (A) R/2
- (B) R
- (C) 2R
- (D) It never comes back.
Answer and explanation
Answer: (B). With U = −GMm/r: ½m(GM/R) − GMm/R = −GMm/r, so r = 2R from the centre, a height of R.
- (A) uses h = v²/(2g) with constant g = GM/R². Gravity weakens with height.
- (C) is the distance from the centre, not the height above the surface.
- (D) needs the escape speed √(2GM/R).
If you missed this: read “Gravitational potential energy with the zero at infinity” in the Topic 6.6 study guide.
Question 9 (short answer · 6.4)
A uniform rod (mass 1.2 kg, length 0.80 m, I = ML²/3 about one end) hangs at rest from a frictionless pivot at its top end. A 0.050 kg ball moving horizontally at 8.0 m/s hits the bottom end at right angles to the rod and bounces straight back at 4.0 m/s.
(a) Explain why the angular momentum of rod + ball about the pivot is conserved in the impact, but their linear momentum is not. (b) Find the rod’s angular speed just after the impact. (c) Find how high the rod’s centre of mass rises, and the largest angle the rod swings from vertical.
Answer and explanation
(a) The pivot pushes on the rod during the impact, so linear momentum changes. But that force acts at the axis, so its torque about the pivot is zero. Gravity on the hanging rod acts through the pivot line, and the brief impact makes the ball’s weight negligible.
(b) I = 1.2 × 0.80² ÷ 3 = 0.256 kg·m². Before: L = mvL = 0.050 × 8.0 × 0.80 = 0.32 kg·m²/s. After, the ball has −0.050 × 4.0 × 0.80 = −0.16 kg·m²/s. So the rod has 0.48 kg·m²/s and ω = 0.48 ÷ 0.256 = 1.9 rad/s (1.875 rad/s).
(c) K_rod = ½(0.256)(1.875)² = 0.45 J = MgΔh, so Δh = 0.038 m. The centre is L/2 below the pivot: 0.40(1 − cos θ) = 0.038, so θ = 25°.
If you missed this: read “Choosing the system” in the Topic 6.4 study guide.
Question 10 (short answer · 6.2)
A grinding wheel (I = 0.050 kg·m²) spins at 120 rad/s with its motor off. A tool pressed on the rim, 0.10 m from the axle, exerts a 4.0 N friction force opposing the rotation. Ignore bearing friction.
(a) Find the torque of the friction force, taking the direction of spin as positive. (b) Find the work done by friction in stopping the wheel, and the angle the wheel turns through. (c) Use angular impulse to find the time it takes to stop.
Answer and explanation
(a) τ = −rF = −0.10 × 4.0 = −0.40 N·m.
(b) W = ΔK = 0 − ½(0.050)(120)² = −360 J. For a constant torque W = τΔθ, so Δθ = 360 ÷ 0.40 = 900 rad (about 143 turns).
(c) τΔt = ΔL = 0 − 0.050 × 120 = −6.0 kg·m²/s, so Δt = 15 s (check: 60 rad/s average × 15 s = 900 rad).
If you missed this: read “The rotational work–energy theorem” in the Topic 6.2 study guide.
Your next step
Mark each question. Count a short answer as missed if any part went wrong.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 6.1 Rotational Kinetic Energy | 1, 9(c) | Topic 6.1 study guide |
| 6.2 Torque and Work | 2, 10 | Topic 6.2 study guide |
| 6.3 Angular Momentum and Angular Impulse | 3, 4, 10(c) | Topic 6.3 study guide |
| 6.4 Conservation of Angular Momentum | 5, 9 | Topic 6.4 study guide |
| 6.5 Rolling | 6, 7 | Topic 6.5 study guide |
| 6.6 Motion of Orbiting Satellites | 8 | Topic 6.6 study guide |
How to use your result
- Read every explanation, even for questions you got right.
- Missed one question in a topic? Read the named section, then try that topic’s practice set. Missed two or more? Work through the whole study guide, then its practice set and checklist.
- Missed questions across several topics? Start with Topics 6.1 and 6.3: every later topic uses their two tools.
- Got everything right? Go to the mixed unit review, where each question combines topics.
Your result is a guide to what to study next, not a predicted exam score.