Newton’s Second Law in Rotational Form — Worked Solutions

These solutions belong to Newton’s Second Law in Rotational Form. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

A rigid wheel’s inertia doubles while its net torque stays the same. How does α change?

1. Use α = τ_net/I for the fixed rigid configuration. Why: Inertia and torque refer to the same rotation axis.

2. Replacing I with 2I gives α_new = α/2. Why: At fixed net torque the relationship is inverse.

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Question 2

A hanging mass is released from rest and accelerates downward while driving a pulley with rotational inertia. How does string tension compare with mg?

1. For the mass, write mg − T = ma with downward positive. Why: The falling object needs a positive net downward force.

2. Since acceleration is downward, T is less than mg. Why: Tension both opposes the fall and drives the pulley’s angular acceleration.

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Question 3

A graph plots net torque vertically against angular acceleration horizontally. What is the slope for a rigid system?

1. Write τ_net = Iα in the graph form y = slope × x. Why: Torque is on the vertical axis and angular acceleration on the horizontal axis.

2. The slope is I, with units kg·m². Why: This graph estimates inertia from several torque/acceleration trials.

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Question 4

Compare two separate trials of a rotating frame. In trial A its masses remain fixed in a configuration with I = 0.050 kg·m². In trial B they remain fixed farther outward, with I = 0.080 kg·m². Apply constant net torque +0.40 N·m for 2.0 s in each trial. Compare α and the change in ω.

1. Use α = τ_net/I with each fixed configuration. Why: Both trials have the same torque but different inertias.

2. Initially α = 0.40/0.050 = 8.0 rad/s², so Δω = 8.0(2.0) = 16 rad/s. Why: Constant torque and inertia give constant angular acceleration.

3. Outward masses give α = 0.40/0.080 = 5.0 rad/s² and Δω = 10 rad/s. Why: Greater inertia reduces the response without changing the net torque.

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Question 5

A fixed-axis frame has the same inertia in three trials. Measured applied driving torques are 0.20, 0.40, and 0.60 N·m, and the corresponding angular accelerations are 1.0, 3.0, and 5.0 rad/s². A constant unknown bearing torque opposes the drive. Choose axes for a linear graph that would let you infer both inertia and bearing torque, and find them from the data. A classmate labels the vertical axis “net torque” without changing the plotted values. Explain the error and where a graph of true net torque would meet the axis.

1. Plot applied driving torque vertically against angular acceleration horizontally. The data have slope (0.60 − 0.20)/(5.0 − 1.0) = 0.10 kg·m². Why: The equation is τ_applied − τ_bearing = Iα, so τ_applied = Iα + τ_bearing. Its slope measures inertia even when the constant resistance is unknown.

2. Use any point: τ_bearing = 0.20 − 0.10(1.0) = 0.10 N·m. Why: The vertical intercept is the applied torque needed to balance the bearing torque when α = 0, not an additional inertia.

3. Subtract 0.10 N·m from each applied value to obtain true net torques 0.10, 0.30, and 0.50 N·m. Their graph is τ_net = 0.10α and passes through the origin. Why: Simply relabeling applied torque leaves the resistance unaccounted for. An ideal constant-inertia true-net graph has zero intercept.

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