Rotational Inertia — Worked Solutions

These solutions belong to Rotational Inertia. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

A point mass is moved from radius r to 2r about the same axis. How does its rotational inertia change?

1. Write the original contribution as I = mr². Why: A point-object contribution weights the squared perpendicular distance.

2. At 2r, I_new = m(2r)² = 4I. Why: Squaring the radius gives a factor of four while mass is unchanged.

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Question 2

Which conditions justify I = I_cm + Md²?

1. Identify a center-of-mass axis and a parallel displaced axis. Why: The parallel-axis theorem requires this specific geometry.

2. Use I = I_cm + Md² with the total mass and axis separation. Why: An arbitrary known off-center inertia cannot replace I_cm in the formula.

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Question 3

A hoop and a disk have equal mass and outer radius. Which has greater inertia about its center axis?

1. Each mass contribution grows with squared distance from the axis. Why: Compare distributions about the same center axis.

2. A hoop puts its mass at the outer radius, so its inertia exceeds the disk’s. Why: Some disk mass lies closer to the axis and contributes less per kilogram.

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Question 4

Two 0.50 kg point masses are attached at the two endpoints of a massless 0.80 m rod. Find I about a perpendicular axis through its center and about a parallel axis through one end.

1. At the center each mass has r = 0.40 m. Why: The center axis bisects the massless 0.80 m rod.

2. I_cm = 2(0.50)(0.40²) = 0.160 kg·m². Why: Add the two point-object contributions about the same axis.

3. Use M = 1.0 kg and d = 0.40 m: I_end = I_cm + Md² = 0.320 kg·m². Why: The end axis is parallel to the center-of-mass axis.

4. Check directly: 0.50(0²) + 0.50(0.80²) = 0.320 kg·m². Why: Agreement between the theorem and the point sum verifies the axis distances.

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Question 5

A light rotating frame has inertia 0.010 kg·m². Two 0.20 kg point masses are attached at equal adjustable radii r, and a known constant net torque 0.040 N·m drives each fixed configuration. You can determine angular acceleration α for several radii. Choose between plotting α against r² or plotting 1/α against r² if your goal is a straight-line test of the point-mass inertia model. Derive the predicted slope and intercept for your choice, and explain what mistake results from ignoring the frame’s inertia.

1. Write I_total = 0.010 + 2(0.20)r² = 0.010 + 0.40r², then α = 0.040/I_total. Why: The frame and two point masses all resist the same angular acceleration about the stated axis.

2. Rearrange to 1/α = I_total/0.040 = 0.25 + 10r². Choose 1/α vertically against r² horizontally, with slope 10 s²/m² and intercept 0.25 s², treating radians as dimensionless. Why: Reciprocating acceleration converts the inverse inertia response into a linear relation. In contrast, α versus r² is curved.

3. Retain the nonzero intercept. Ignoring the frame predicts a line through zero and overestimates acceleration, especially at small r. Why: The point masses’ contribution shrinks near the axis, while the frame’s own inertia remains.

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