These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. This set is for the calculus-based Physics C: Mechanics course. Use g = 9.8 m/s². Take counterclockwise torques as positive unless a question says otherwise. All rotation is in one plane. Round final answers to 2 significant figures unless told otherwise. All data are invented for practice.
Question 1 (multiple choice · foundation)
A record-player turntable rotates at a constant 3.5 rad/s counterclockwise. Its motor exerts a torque of 0.42 N·m counterclockwise. Friction in the bearing is the only other torque. What is the friction torque?
- (A) Zero
- (B) 0.42 N·m clockwise
- (C) 1.47 N·m clockwise
- (D) It cannot be found without the rotational inertia of the turntable.
Answer and explanation
Answer: (B). The angular velocity is constant, so the turntable is in rotational equilibrium: Στ = 0.42 + τ_f = 0, giving τ_f = −0.42 N·m, that is 0.42 N·m clockwise.
- (A) assumes nothing opposes a rotation that is already happening. If friction were zero, the motor’s torque would be unbalanced and ω would increase.
- (C) multiplies the torque by ω (0.42 × 3.5). Angular velocity does not enter the equilibrium condition.
- (D) The rotational inertia matters only when ω changes (Topic 5.6). With α = 0, Iα = 0 for any I.
Question 2 (multiple choice · core)
Several forces act on a flat rigid plate. Their vector sum is zero. The net torque about point P is 4.0 N·m counterclockwise. Point Q is 0.50 m from P. What is the net torque about Q?
- (A) Zero
- (B) 2.0 N·m counterclockwise
- (C) 4.0 N·m counterclockwise
- (D) 8.0 N·m counterclockwise
Answer and explanation
Answer: (C). Moving the torque point by d changes the net torque by −d × ΣF: τ_Q = τ_P − d × ΣF. Here ΣF = 0, so τ_Q = τ_P = 4.0 N·m counterclockwise. The forces form a couple, whose torque is the same about every point.
- (A) assumes ΣF = 0 forces Στ = 0. A couple is the counter-example; the two conditions are independent.
- (B) multiplies the torque by the distance PQ (4.0 × 0.50). The distance only matters through d × ΣF, which is zero.
- (D) divides by the distance (4.0 ÷ 0.50). There is no reason for the torque to grow as Q moves away.
Question 3 (multiple choice · core)
A thin rod of length L lies along the x-axis from x = 0 to x = L. Its linear mass density is λ(x) = bx, where b is a positive constant. At what x must a single support be placed for the rod to balance horizontally?
- (A) L/2
- (B) 2L/3
- (C) L/3
- (D) 3L/4
Answer and explanation
Answer: (B). The rod balances on a support under its center of mass, where the weight’s torque about the support is zero. M = ∫₀ᴸ bx dx = bL²/2 and ∫₀ᴸ x(bx) dx = bL³/3, so x_cm = (bL³/3) ÷ (bL²/2) = 2L/3.
- (A) treats the rod as uniform. More mass sits near x = L, so the balance point moves to the right.
- (C) is the right distance measured from the wrong end (L − 2L/3).
- (D) is the result for λ ∝ x², which concentrates even more mass near x = L.
Question 4 (calculation · core)
Take +x to the right and +y upward, origin at the hinge. A uniform drawbridge of mass 900 kg and length 8.0 m is hinged at its lower end. It is held at rest at 30° above the horizontal, rising to the right, by a horizontal cable attached to its upper end and running left to a tower. Find (a) the tension in the cable and (b) the size and direction of the force from the hinge.
Worked solution
- Extended force diagram: weight Mg = 8820 N acting at the midpoint; tension T to the left at the upper end; hinge force (H, V) at the origin.
- Torques about the hinge (the hinge force drops out). The weight’s lever arm is the horizontal distance to the midpoint: (L/2) cos 30° = 4.0 × 0.866 = 3.46 m, clockwise: −8820 × 3.46 = −3.06 × 10⁴ N·m. The horizontal cable’s lever arm is the height of the upper end: L sin 30° = 4.0 m; pulling left above the hinge, it gives +4.0T.
- (a) 4.0T − 3.06 × 10⁴ = 0, so T = 7.6 × 10³ N (7640 N). In symbols, T = Mg/(2 tan 30°).
- (b) ΣF_x = 0: H − T = 0, so H = 7640 N to the right. ΣF_y = 0: V − 8820 = 0, so V = 8820 N upward.
- Size: √(7640² + 8820²) = 1.2 × 10⁴ N, at tan⁻¹(8820/7640) = 49° above the horizontal, pointing up and to the right. Note that this is steeper than the bridge itself (30°): the hinge force does not act along the bridge.
Suggested mark points (4): 1 for both lever arms; 1 for the torque equation and T; 1 for the two force components from ΣF = 0; 1 for the size and direction of the hinge force.
Common error: using the full length 8.0 m as the cable’s lever arm gives T = 3.8 × 10³ N. The cable is horizontal, so its lever arm is the vertical height of its attachment point.
Question 5 (constructed response · core)
Take +x to the right, origin at the left end, counterclockwise positive. A rod of length L hangs horizontally from two vertical strings, A at x = 0 and B at x = L. Its linear mass density is λ(x) = λ₀(1 + x/L).
(a) Derive the mass M and the position of the center of mass in terms of λ₀ and L. (b) Derive the tensions T_A and T_B in terms of M and g. (c) Evaluate M, T_A and T_B for λ₀ = 2.0 kg/m and L = 1.5 m. (d) A small block of mass equal to M is hung from the rod so that the two tensions become equal. Find where it must hang.
Worked solution and suggested Marlbridge rubric
(a) M = ∫₀ᴸ λ₀(1 + x/L) dx = λ₀(L + L/2) = 3λ₀L/2. ∫₀ᴸ xλ₀(1 + x/L) dx = λ₀(L²/2 + L²/3) = 5λ₀L²/6. So x_cm = (5λ₀L²/6) ÷ (3λ₀L/2) = 5L/9.
(b) Torques about A: T_B L − Mg(5L/9) = 0, so T_B = 5Mg/9. Forces: T_A + T_B = Mg, so T_A = 4Mg/9.
(c) M = 1.5 × 2.0 × 1.5 = 4.5 kg; Mg = 44.1 N. T_B = 25 N (24.5 N); T_A = 20 N (19.6 N). The denser end hangs on B, and B carries more, as expected.
(d) The tensions are equal when the combined center of mass is at L/2 (then torques about the midpoint balance). M(5L/9) + Md = 2M(L/2), so d = L − 5L/9 = 4L/9 = 0.67 m from A. Each string then carries (2M)g/2 = 44 N.
| Point | What earns it |
|---|---|
| 1 | (a) Integrates λ for M and xλ for the first moment |
| 1 | (a) x_cm = 5L/9 |
| 1 | (b) Torque equation about a string end with the weight at x_cm, giving T_B = 5Mg/9 |
| 1 | (b)–(c) T_A from ΣF = 0, and correct numbers with units |
| 1 | (d) Uses “equal tensions ⇔ combined center of mass at the midpoint” (or an equivalent torque equation) to get 4L/9 |
Alternative method. In (b), taking torques about B first gives T_A = Mg(L − 5L/9)/L = 4Mg/9 directly; either order earns the points.
Question 6 (experimental design · stretch)
Take +x to the right, origin at the hinge, counterclockwise positive. A uniform board of length 1.20 m is hinged at its left end. A student holds it horizontal with a vertical force sensor placed a distance x from the hinge and records the sensor force F. The fictional data are:
| x (m) | 0.40 | 0.60 | 0.80 | 1.00 | 1.20 |
|---|---|---|---|---|---|
| F (N) | 36.9 | 24.3 | 18.5 | 14.6 | 12.3 |
(a) Derive an expression for F in terms of the board’s mass M, g, its length L and x. (b) State what to plot to get a straight line through the origin. (c) Use the data to find M. (d) For x = 0.40 m and x = 1.20 m, find the force from the hinge, with direction. Use your model, not the data. (e) At what x does the hinge exert no force? Explain why.
Worked solution and suggested Marlbridge rubric
(a) Torques about the hinge: Fx − Mg(L/2) = 0, so F = MgL/(2x).
(b) Plot F (N) on the vertical axis against 1/x (m⁻¹) on the horizontal axis. The slope is MgL/2.
(c) Values of 1/x: 2.50, 1.67, 1.25, 1.00, 0.833 m⁻¹. The best-fit line through the origin has slope ≈ 14.7 N·m. So M = 2 × 14.7 ÷ (9.8 × 1.20) = 2.5 kg.
(d) With M = 2.5 kg, Mg = 24.5 N and MgL/2 = 14.7 N·m. Let the hinge force be V (positive upward); ΣF_y = 0 gives V = Mg − F.
- x = 0.40 m: F = 36.75 N, so V = −12 N: 12 N downward. The sensor is close to the hinge and pushes up more than the weight, so the hinge must hold the board down.
- x = 1.20 m: F = 12.25 N, so V = 12 N upward.
(e) V = 0 when F = Mg, which gives x = L/2 = 0.60 m. The sensor is then directly under the center of mass, so it balances both the force and the torque of the weight on its own. (The data agree: 24.3 N ≈ 24.5 N.)
| Point | What earns it |
|---|---|
| 1 | (a) Torques about the hinge with the weight at L/2 |
| 1 | (b) F against 1/x, slope identified as MgL/2 |
| 1 | (c) M = 2.5 kg from the slope (2.4–2.6 kg accepted) |
| 1 | (d) Both hinge forces, with correct directions from ΣF_y = 0 |
| 1 | (e) x = L/2 with the reason that the support is under the center of mass |
Alternative method for (c). Plotting Fx against x should give a horizontal line at MgL/2; its mean value (≈ 14.7 N·m) gives the same M. Using a single data point earns the mark only if the result is within range and the student notes a graph is more reliable.
Question 7 (explanation · stretch)
A rigid spanner is tossed across a workshop. It tumbles end over end in a vertical plane, spinning at 12 rad/s about its center of mass. Ignore air resistance. A student says:
“The spanner’s center of mass is accelerating downward, so it is not in equilibrium. So its angular velocity must be changing as well.”
Evaluate the claim. Then describe a different situation in which the reverse is true: translational equilibrium without rotational equilibrium.
Model answer and suggested Marlbridge rubric
Model answer. The first sentence is half right: the only force is gravity, so ΣF = Mg ≠ 0 and the center of mass accelerates downward at g. The spanner is not in translational equilibrium.
The second sentence does not follow. Rotation about the center of mass depends on the net torque about the center of mass, not on the net force. Gravity acts on every piece of the spanner, and its total torque about the center of mass is −Mg × (horizontal distance from the center of mass to itself) = 0. So Στ = 0, and by Newton’s first law in rotational form the spanner keeps turning at 12 rad/s. After 0.50 s its vertical velocity has changed by 4.9 m/s, but its spin has not changed.
The two equilibrium conditions are independent: one can hold while the other fails.
Reverse case. A couple: for example, two people push a board lying on ice with equal and opposite forces along parallel lines that do not coincide. ΣF = 0, so the center of mass stays at rest, but Στ ≠ 0, so the board starts to rotate.
| Point | What earns it |
|---|---|
| 1 | Agrees ΣF ≠ 0, so no translational equilibrium |
| 1 | Gravity acts at the center of mass, so its torque about the center of mass is zero |
| 1 | Concludes ω stays constant (Newton’s first law in rotational form) and the claim is false |
| 1 | Gives a valid example of ΣF = 0 with Στ ≠ 0 (a couple), with a reason |
How did you do?
- Q1 or Q7 wrong: re-read “Newton’s first law in rotational form” and “Equilibrium of one kind only” in the study guide.
- Q2 wrong: go through the derivation in “Choosing the point for torques”.
- Q3 or Q5 wrong: revisit “Extended force diagrams” and Worked example 1. Find x_cm by integration before taking torques.
- Q4 or Q6 incomplete: compare with Worked example 2. Use lever arms, then ΣF = 0 for the pivot force.
Then tick off the topic checklist.