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. “Small” objects may be treated as point masses. Rods are thin. Round final answers to 2 significant figures unless told otherwise.
Question 1 (multiple choice · foundation)
Two small objects are fixed to a light frame in the xy-plane: 2.0 kg at (0.30 m, 0.40 m) and 1.0 kg at (−0.20 m, 0.10 m). The frame turns about the y-axis. What is its rotational inertia about that axis?
- (A) 0.33 kg·m²
- (B) 0.22 kg·m²
- (C) 0.55 kg·m²
- (D) 0.80 kg·m²
Answer and explanation
Answer: (B). The perpendicular distance to the y-axis is |x|. I = 2.0 × 0.30² + 1.0 × 0.20² = 0.18 + 0.04 = 0.22 kg·m².
- (A) uses the y-coordinates, which are the distances to the x-axis: 2.0 × 0.40² + 1.0 × 0.10².
- (C) uses the distance to the origin, which is the right distance only for the z-axis: 2.0 × 0.50² + 1.0 × (√0.05)².
- (D) uses mr instead of mr²: 2.0 × 0.30 + 1.0 × 0.20, which does not even have the unit kg·m².
Question 2 (multiple choice · core)
A uniform disk of mass M and radius R turns about an axis perpendicular to the disk, through a point on its rim. About its central axis, I = ½MR². What is its rotational inertia about the rim axis?
- (A) ½MR²
- (B) MR²
- (C) (3/2)MR²
- (D) 2MR²
Answer and explanation
Answer: (C). The rim axis is parallel to the central axis and a distance d = R from it, and the central axis passes through the centre of mass. I = ½MR² + MR² = (3/2)MR².
- (A) ignores the shift of axis. I about a non-cm axis is always larger.
- (B) is Md² alone; it leaves out I_cm.
- (D) adds MR² to the hoop value MR², using the wrong I_cm.
Question 3 (multiple choice · core)
Disk A is uniform. Disk B has the same mass M and radius R, but its density increases steadily from the centre to the rim. Both turn about their central axes. Which statement about I_B is correct?
- (A) I_B = ½MR², because the two disks have the same mass and radius.
- (B) ½MR² < I_B < MR²
- (C) I_B < ½MR², because B’s centre has less mass.
- (D) I_B > MR², because B’s mass is concentrated near the rim.
Answer and explanation
Answer: (B). B has more of its mass at large r than A, so I_B > I_A = ½MR². But no part of B is farther than R from the axis, so I_B < MR², the value if all the mass were at the rim (a hoop).
- (A) assumes I depends only on M and R. The distribution matters.
- (C) has the effect backwards: moving mass outwards raises I.
- (D) is impossible: MR² is the largest I any object of mass M fitting inside radius R can have about that axis.
Question 4 (calculation · core)
A uniform rod has mass 2.4 kg and length 1.5 m. It turns about an axis perpendicular to the rod through a point L/4 from one end.
(a) By integration, derive I about this axis in terms of M and L. (b) Check your answer with the parallel axis theorem, using I_cm = ML²/12. (c) Evaluate I, and compare it with the values about the centre and about an end.
Worked solution and suggested Marlbridge rubric
(a) Put the axis at the origin, so the rod runs from x = −L/4 to x = 3L/4, with λ = M/L. I = ∫ x² (M/L) dx from −L/4 to 3L/4 = (M/3L)[(3L/4)³ − (−L/4)³] = (M/3L)(27L³/64 + L³/64) = (M/3L)(28L³/64) = 7ML²/48.
(b) d = L/2 − L/4 = L/4. I = ML²/12 + M(L/4)² = 4ML²/48 + 3ML²/48 = 7ML²/48. ✓
(c) I = 7 × 2.4 × 1.5² ÷ 48 = 0.79 kg·m² (0.7875). About the centre: ML²/12 = 0.45 kg·m². About an end: ML²/3 = 1.8 kg·m². The value lies between the two, as expected, since the axis is between the centre and the end.
| Point | What earns it |
|---|---|
| 1 | (a) Sets up ∫x² λ dx with λ = M/L and limits that cover the rod once |
| 1 | (a) Evaluates to 7ML²/48 |
| 1 | (b) Parallel axis with d = L/4 gives the same result |
| 1 | (c) 0.79 kg·m² and a correct comparison with both 0.45 and 1.8 kg·m² |
Question 5 (derivation · core)
(a) A thin cylindrical shell (a tube with very thin walls) has mass M, radius R and length h. Show that its rotational inertia about its central axis is MR², and explain why h does not appear. (b) A uniform flat ring has mass M, inner radius R₁ and outer radius R₂. By treating it as a set of thin coaxial rings, derive I = ½M(R₁² + R₂²) about its central axis. (c) Show that your result in (b) gives the correct answers for a solid disk and for a thin hoop. (d) For a uniform solid disk of radius R, find what fraction of its rotational inertia comes from the outer part, R/2 < r < R. Comment on your answer.
Worked solution and suggested Marlbridge rubric
(a) Every element dm of the thin shell is the same distance R from the axis, whatever its position along the length. So I = ∫R² dm = R²∫dm = MR². Spreading the mass along the axis (changing h) does not change any distance from the axis, so h does not appear.
(b) Area density σ = M/[π(R₂² − R₁²)]. A ring of radius r and width dr has dm = σ(2πr dr), all at distance r. I = ∫ r² σ 2πr dr from R₁ to R₂ = (πσ/2)(R₂⁴ − R₁⁴) = (πσ/2)(R₂² − R₁²)(R₂² + R₁²). Substitute σ: I = ½M(R₁² + R₂²).
(c) R₁ = 0 gives ½MR₂², the solid disk. R₁ = R₂ = R gives ½M(2R²) = MR², the hoop. ✓
(d) Outer part: I_out = (πσ/2)(R⁴ − R⁴/16) = (15/16)(πσR⁴/2) = (15/16) × ½MR². So 15/16 ≈ 94% of I comes from the outer half of the radius, even though that part holds only 3/4 of the mass. Mass far from the axis dominates because r is squared (and there is more area at large r).
| Point | What earns it |
|---|---|
| 1 | (a) All of dm at distance R, so I = MR², with h irrelevant because no distance from the axis depends on it |
| 1 | (b) Correct dm = σ 2πr dr with σ in terms of M, R₁, R₂ |
| 1 | (b) Integration and factorising to reach ½M(R₁² + R₂²) |
| 1 | (c) Both limits checked |
| 1 | (d) 15/16 with a comment linking it to r² weighting |
Alternative method. For (b), subtracting a disk of radius R₁ from a disk of radius R₂ (same σ) earns both (b) points if the masses of the two disks are written in terms of σ correctly.
Question 6 (constructed response · stretch)
A light horizontal bar can spin about a vertical axle through its centre. Two identical small masses m are clamped on the bar, one each side, both at distance r from the axle. A rotation sensor reports the rotational inertia I of the whole spinning system. The results are:
| r (m) | 0.05 | 0.10 | 0.15 | 0.20 | 0.25 |
|---|---|---|---|---|---|
| I (kg·m²) | 0.0068 | 0.0091 | 0.0127 | 0.0181 | 0.0246 |
(a) Write the predicted relationship between I and r, and explain why a graph of I against r² should be a straight line. (b) Plot the graph (or tabulate the values) and find the slope and intercept of the best-fit line. (c) Use your line to find m, and state what the intercept represents. (d) Predict I for r = 0.30 m. (e) The “small masses” are actually blocks 4 cm wide. Explain, using the parallel axis theorem, whether this would make the graph curve.
Worked solution and suggested Marlbridge rubric
(a) I = I₀ + 2mr², where I₀ is the rotational inertia of the bar and axle. Plotting I (vertical) against r² (horizontal) gives slope 2m and intercept I₀, a straight line.
(b) r² values: 0.0025, 0.0100, 0.0225, 0.0400, 0.0625 m². A least-squares line gives slope ≈ 0.30 kg (0.297) and intercept ≈ 0.0061 kg·m².
(c) 2m = 0.297 kg, so m ≈ 0.15 kg. The intercept, 0.0061 kg·m², is I₀, the rotational inertia of the bar, axle and clamps with the masses at r = 0.
(d) I = 0.0061 + 0.297 × 0.09 = 0.033 kg·m².
(e) Each block has its own I_block,cm about a vertical axis through its centre. By the parallel axis theorem each contributes I_block,cm + mr². The extra terms are constant, so the graph stays straight; only the intercept rises by 2I_block,cm. The slope still gives 2m.
| Point | What earns it |
|---|---|
| 1 | (a) I = I₀ + 2mr² and the slope/intercept meaning |
| 1 | (b) Slope 0.28–0.32 kg and intercept 0.005–0.007 kg·m² from a fitted line |
| 1 | (c) m ≈ 0.15 kg (carry forward) and the intercept identified as I₀ |
| 1 | (d) About 0.033 kg·m² (carry forward) |
| 1 | (e) Uses I_cm + mr² for each block to show the line stays straight, with only the intercept changing |
Question 7 (explanation · stretch)
A rod of length L = 0.90 m lies along x from 0 to L. Its mass per unit length is λ = kx, where k is a constant, and its mass is M = 0.60 kg.
A student finds I₀ about the light end (x = 0) correctly. She then writes: “About the other end, I_L = I₀ + ML², by the parallel axis theorem.”
(a) Find k in terms of M and L, and show that I₀ = ML²/2. (b) Find the centre of mass and I_cm. (c) Find I_L correctly, and evaluate it. Compare with the student’s value. (d) Explain the student’s error.
Model answer and suggested Marlbridge rubric
(a) M = ∫₀ᴸ kx dx = kL²/2, so k = 2M/L². I₀ = ∫₀ᴸ x² kx dx = kL⁴/4 = (2M/L²)(L⁴/4) = ML²/2.
(b) x_cm = (1/M)∫₀ᴸ x · kx dx = kL³/(3M) = 2L/3 = 0.60 m. I_cm = I₀ − M(2L/3)² = ML²/2 − 4ML²/9 = ML²/18.
(c) The dense end is L/3 from the centre of mass: I_L = ML²/18 + M(L/3)² = 3ML²/18 = ML²/6. (Direct check: ∫₀ᴸ (L − x)² kx dx = kL⁴/12 = ML²/6.) Numbers: I₀ = 0.243 kg·m², I_cm = 0.027 kg·m², I_L = 0.081 kg·m². The student’s value is I₀ + ML² = 3ML²/2 = 0.729 kg·m², nine times too big.
(d) The parallel axis theorem relates I about any axis to I about the centre-of-mass axis. The student used it between two off-centre axes. The correct route goes through I_cm. Physically, I_L must be smaller than I₀ because most of the mass is near x = L, close to that axis.
| Point | What earns it |
|---|---|
| 1 | (a) k = 2M/L² and I₀ = ML²/2 by integration |
| 1 | (b) x_cm = 2L/3 and I_cm = ML²/18 |
| 1 | (c) I_L = ML²/6 = 0.081 kg·m², by the parallel axis theorem from I_cm or by direct integration |
| 1 | (d) Identifies that the theorem must start from the cm axis, and gives a physical reason I_L < I₀ |
How did you do?
- Q1 or Q3 wrong: re-read “Point masses: I = mr²” and “Why a hoop beats a disk” in the study guide.
- Q2 or Q4 wrong: revisit “The parallel axis theorem” and the uniform-rod derivation.
- Q5 incomplete: work through “Deriving the standard results” and Worked example 3.
- Q6 or Q7 incomplete: compare with “Linking to experiments” and Worked example 2.
Then tick off the topic checklist.