Angular Momentum and Angular Impulse — Worked Solutions

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

Question 1

A puck travels along a straight line. Can its angular momentum about a reference point be nonzero?

1. Use L = mvb, where b is distance from the reference point to the straight path. Why: Angular momentum is defined about a chosen point or axis.

2. When the point is off the path, b can be nonzero and L is nonzero. Why: The path need not curve, and nonzero angular momentum does not require a current force.

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Question 2

A net torque–time triangle has duration 0.50 s and peak 4.0 N·m above zero. What angular impulse does it supply?

1. Angular impulse is signed torque–time area. Why: The pulse changes angular momentum, not directly an energy value.

2. J_angular = ½(0.50)(4.0) = +1.0 N·m·s = +1.0 kg·m²/s. Why: The triangular shape halves the rectangular peak-times-duration value.

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Question 3

A graph of angular momentum against time is a horizontal line at a positive value. What is the net external torque?

1. Net external torque is the slope of L against t. Why: Torque measures the rate of change of angular momentum.

2. A horizontal graph has zero slope, so τ_net = 0. Why: Its positive height is angular momentum already present, not torque.

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Question 4

A wheel has I = 0.20 kg·m² and initial angular velocity +5.0 rad/s. A constant braking torque of −0.40 N·m acts for 1.5 s. Find L_f and ω_f.

1. Compute L_i = Iω_i = 0.20(5.0) = +1.0 kg·m²/s. Why: Use the specified axis and positive initial turning sense.

2. J_angular = (−0.40)(1.5) = −0.60 kg·m²/s. Why: The braking torque transfers angular momentum out over the stated time.

3. L_f = 1.0 − 0.60 = +0.40 kg·m²/s; ω_f = L_f/I = +2.0 rad/s. Why: Subtract the impulse from initial L, and check that the remaining positive value has not reversed the spin.

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Question 5

A fixed-axis wheel starts with angular momentum +0.50 kg·m²/s. From t = 0 to 2.0 s its net torque forms a positive triangle, rising from zero to +2.0 N·m at 1.0 s and returning to zero at 2.0 s. From 2.0 to 3.5 s the net torque is −1.0 N·m. Sketch angular momentum against time with labeled values and explain its changing slopes. Does the wheel ever reverse its turning sense during this interval? Justify your answer from the data.

1. The triangular angular impulse is ½(2.0)(2.0) = +2.0 kg·m²/s, so L reaches +2.50 kg·m²/s at 2.0 s. Why: Signed torque–time area changes L, and the positive triangle adds to the initial value.

2. During the triangle, the L graph increases with a slope that rises from zero to +2.0 N·m and falls back to zero. L is +1.50 kg·m²/s at 1.0 s. During the last interval it is a straight line of slope −1.0 N·m, ending at L = 2.50 − 1.0(1.5) = +1.00 kg·m²/s. Why: Torque is the instantaneous slope of L against time. The first two halves have equal triangular areas; the final constant torque gives linear decrease.

3. The wheel does not reverse: L stays positive, with its smallest value +0.50 initially and +1.00 after braking. Why: For positive fixed inertia, L and signed angular velocity share a sign. A negative torque can reduce positive L without making it cross zero.

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