Rotational Kinetic Energy — Worked Solutions

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

Question 1

A fixed-axis wheel reverses from +10 rad/s to −10 rad/s with the same inertia. How do the endpoint kinetic energies compare?

1. Use K_rot = ½Iω². Why: Rotational kinetic energy is a scalar with squared angular velocity.

2. Squaring +10 and −10 gives the same value, so endpoint energies match. Why: Changing turning sense alone does not change energy at equal angular speed.

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Question 2

A point mass moves in a circle about a fixed axis. You calculate both ½mv² and ½Iω². How should they be used?

1. Substitute v = rω and I = mr² into ½mv². Why: Both expressions describe the same point object’s motion about a fixed axis.

2. The result is ½Iω²; do not add the expressions. Why: Adding equal descriptions would count the same kinetic energy twice.

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Question 3

Which inertia belongs in K_total = ½Mv_cm² + ½Iω²?

1. Separate motion into center translation and motion relative to the center. Why: The two contributions must refer to distinct parts of the motion description.

2. Use I_cm in the rotation term. Why: An external-axis inertia already includes some center motion and can cause double counting if used with the translation term.

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Question 4

Wheel A has I = 0.020 kg·m² and wheel B has I = 0.080 kg·m². Both turn about fixed axles and each has 4.0 J of kinetic energy. Find their angular speeds.

1. Rearrange K = ½Iω² to |ω| = √(2K/I). Why: Both wheels have supplied fixed-axis energy and inertia.

2. Wheel A: √[2(4.0)/0.020] = 20 rad/s. Why: Its smaller inertia permits greater angular speed at the same energy.

3. Wheel B: √[2(4.0)/0.080] = 10 rad/s. Why: Four times the inertia halves angular speed because energy depends on ω².

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Question 5

A 2.0 kg wheel has center-of-mass rotational inertia 0.040 kg·m². In trial A its center speed is 3.0 m/s and angular speed is 15 rad/s. In trial B the center speed is 6.0 m/s and angular speed is 7.5 rad/s. Draw scaled translation and rotation energy bars for each trial. A classmate predicts equal total kinetic energies because one speed doubled while the other halved. Evaluate the prediction without assuming a rolling constraint.

1. Trial A has K_trans = ½(2.0)(3.0²) = 9.0 J and K_rot = ½(0.040)(15²) = 4.5 J, totaling 13.5 J. Why: Center translation and rotation about the center are distinct contributions, so they add.

2. Trial B has K_trans = ½(2.0)(6.0²) = 36 J and K_rot = ½(0.040)(7.5²) = 1.125 J, totaling 37.125 J. Why: Doubling center speed quadruples its energy term, while halving angular speed quarters the other term. These changes do not generally cancel.

3. Use one bar scale, such as 4.5 J per division: A has bars 2 and 1 divisions; B has bars 8 and 0.25 divisions. Reject the equal-total prediction. Why: A common scale makes the changed partition visible. The decomposition is valid without a no-slip relationship between the two speeds.

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