AP Physics 1 Unit 5 FRQ Practice: Torque and Rotational Dynamics Free Response

Free-Response Question · Unit 5

Scenario: A uniform horizontal beam of length L = 4.0 m and mass m_beam = 10 kg is pivoted at its left end (point P). A block of mass m_block = 5.0 kg hangs from the right end of the beam. A vertical cable, attached to the beam at its midpoint (d = 2.0 m from the pivot), holds the beam in horizontal equilibrium.

— AP Physics 1 style problem · Topics 5.3, 5.5

Draw a free-body diagram of the beam, showing all forces acting on it (the beam's weight, the weight of the hanging block, the tension in the cable, and the reaction force at the pivot). Indicate where each force acts on the beam.

✓ Model answer (earns the points)

The beam has four forces acting on it:

1. Beam's weight (W_beam = m_beam·g = 100 N), acting DOWNWARD at the beam's center of mass — for a uniform beam, that's the geometric center, 2.0 m from the pivot.

2. Weight of the hanging block (W_block = m_block·g = 50 N), acting DOWNWARD at the right end of the beam — 4.0 m from the pivot. The block hangs by a string, but it pulls down on the beam with its full weight.

3. Tension T in the cable, acting UPWARD at the midpoint of the beam — 2.0 m from the pivot. The cable is vertical, so the tension is straight up.

4. Reaction force R at the pivot, acting at the left end of the beam — the pivot can exert forces in any direction; in this problem, R will be vertical (since all other forces are vertical).

Why it scores: (1) Identifies all four forces. (2) Each force has correct direction (down for weights, up for tension). (3) Each force is placed at the correct location (beam weight at center, block weight at right end, tension at midpoint, pivot force at left end).

Using rotational equilibrium (Στ = 0) about the pivot, calculate the tension in the cable.

✓ Model answer (earns the points)

Take counterclockwise as positive. About the pivot (left end of beam), the pivot's reaction force R produces ZERO torque (zero lever arm — that's why we chose this pivot).

Torque from beam's weight (acts at center, 2.0 m from pivot, downward → CW → negative):

τ_beam = −W_beam · 2.0 = −(100)(2.0) = −200 N·m

Torque from block's weight (acts at right end, 4.0 m from pivot, downward → CW → negative):

τ_block = −W_block · 4.0 = −(50)(4.0) = −200 N·m

Torque from tension (acts at midpoint, 2.0 m from pivot, upward → CCW → positive):

τ_T = +T · 2.0 = +2.0T

Set Στ = 0:

+2.0T − 200 − 200 = 0

2.0T = 400

T = 200 N

Answer: The tension in the cable is 200 N.

Why it scores: (1) Chooses the pivot strategically to eliminate the unknown pivot reaction force. (2) Identifies each torque with the correct lever arm AND sign (CCW vs CW). (3) Places the beam's weight at the CENTER (uniform beam). (4) Sets up and solves Στ = 0 correctly. (5) Reports answer with units.

A student claims: "If we move the cable's attachment point CLOSER to the pivot, the tension in the cable will DECREASE because the cable is doing less rotational work." Is the student correct? Justify your answer with a calculation.

✓ Model answer (earns the points)

The student is incorrect. Moving the cable closer to the pivot actually INCREASES the tension — the opposite of what the student claims.

Why? The downward torques from the weights are unchanged (the weights are still at their same positions). The cable still has to balance those torques. But with a smaller lever arm, the cable needs MORE force to produce the same torque.

Calculation: Suppose we move the cable to d = 1.0 m from the pivot (half its original distance). Using Στ = 0:

T · d = (100)(2.0) + (50)(4.0)

T(1.0) = 200 + 200 = 400

T = 400 N

The tension DOUBLED (from 200 N to 400 N) when we moved the cable from d = 2.0 m to d = 1.0 m.

In general: T = 400/d. Smaller d means LARGER T. This is the same principle as why a wrench has a long handle — a longer lever arm lets you produce a given torque with less force. Conversely, a shorter lever arm requires more force.

Why it scores: (1) Clearly states the student is incorrect. (2) Re-applies Στ = 0 with a smaller d. (3) Calculates the new T and shows it's LARGER, not smaller. (4) Explains the general relationship T = 400/d (or equivalent reasoning). (5) Bonus: connects to the wrench/lever-arm intuition.

How to score points on AP Physics 1 torque FRQs

Always draw a free-body diagram first. Show all forces, including the pivot reaction. Place each force at the correct location.
For uniform beams, weight acts at the center of mass — usually the geometric center.
Choose your pivot strategically. Pick one through an unknown force (like the pivot reaction) to eliminate it from the torque equation.
Track signs carefully. Counterclockwise = +, clockwise = − (or pick your own convention and stick with it).
For each torque, multiply the FORCE by the LEVER ARM (perpendicular distance from pivot to line of force).
For "is the student correct?" questions, do the math. Verbal arguments alone usually don't earn full credit — show a calculation.
Units matter. "200" is incomplete; "200 N" is correct.