Rotational Equilibrium and Newton’s First Law in Rotational Form — Worked Solutions

These solutions belong to Rotational Equilibrium and Newton’s First 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 wheel is in rotational equilibrium. Which statement must be true?

1. Rotational equilibrium means Στ = 0. Why: The torques balance about the stated axis.

2. With constant I, α = 0 and ω stays constant, including a possible nonzero value. Why: Equilibrium does not require rest or translational equilibrium.

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Question 2

Two equal opposite forces act along separated parallel lines on a bar. What can happen?

1. The equal opposite forces cancel in ΣF. Why: Their vector sum controls the center-of-mass acceleration.

2. Separated lines of action can give turning effects that add. Why: Their locations determine torques, so zero net force does not force zero net torque.

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Question 3

Why choose a support point as the axis when solving a beam in equilibrium?

1. Choose the axis through the support force’s application point. Why: That makes the perpendicular distance to its line of action zero.

2. The support contributes no torque about that point but remains in ΣF. Why: Choosing a convenient axis removes an unknown torque term without removing the physical force.

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Question 4

A uniform 2.0 m board weighs 60 N and is pivoted 0.80 m from its left end. A 100 N load hangs from the left end. What downward load at the right end balances it?

1. Locate the board’s center 1.0 m from the left, or 0.20 m right of the pivot. Why: Uniform-board weight acts at the center of mass, which is not at the pivot here.

2. Balance torques: 100(0.80) = 60(0.20) + W(1.20). Why: The left load turns counterclockwise; board weight and the right load turn clockwise.

3. Solve W = (80 − 12)/1.20 = 56.7 N. Why: The unknown must supply the remaining clockwise torque.

4. Force balance gives F_pivot = 100 + 60 + 56.7 = 216.7 N upward. Why: Static equilibrium also needs zero net vertical force.

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Question 5

A uniform meterstick is horizontal and balanced on a pivot at its 0.40 m mark. A 0.20 kg mass hangs at 0.10 m and a 0.10 kg mass hangs at 0.90 m. The meterstick’s mass is unknown; neglect attachment masses. Infer it without using a scale. Then predict which way the stick would initially turn if the pivot moved to 0.50 m while both hanging masses stayed in place. Explain why the meterstick’s weight matters in one configuration but has zero torque in the other.

1. At the 0.40 m pivot, the known left load has lever arm 0.30 m, the right load 0.50 m, and the meterstick’s center 0.10 m to the right. Why: A uniform stick’s weight acts at its 0.50 m center of mass, not automatically at its support.

2. Balance counterclockwise and clockwise torques: 0.20g(0.30) = 0.10g(0.50) + mg(0.10). Cancel g and solve m = 0.10 kg. Why: Gravity supplies all three downward forces, so its common factor cancels while the lever arms remain.

3. With the pivot at 0.50 m, the stick’s own lever arm is zero. The left and right load torques have magnitudes 0.20g(0.40) and 0.10g(0.40); the left wins, so the left end initially moves downward, counterclockwise as viewed with the left end to the left. Why: Equal new lever arms make the unequal hanging masses decisive, and the stick’s center force acts through the new pivot.

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