AP C → USAPhO intro · 40 min
A concentrated test of torque arms, rolling constraints, angular momentum, and rotational energy.
Mark the axis before writing torque.
State the rolling constraint with signs.
For angular momentum, name the axis and external torque check.
Question 1
A uniform beam of weight 80 N and length L is pivoted at one end and held horizontal by a vertical rope at the other. Find rope tension.
Take torque about the pivot.
T L = 80(L/2), so T = 40 N.
Trap tested: Using force balance alone.
Question 2
A 30 N force is applied at 60° to a 0.40 m wrench. Find torque magnitude.
Use sin of the angle between r and F.
τ = rF sin 60° = (0.40)(30)(0.866) = 10.4 N·m.
Trap tested: Forgetting the perpendicular component.
Question 3
A solid disk rolls without slipping from rest through a vertical drop h. Find v at the bottom.
Use I = ½MR² and v = ωR.
Mgh = ½Mv² + ½(1/2MR²)(v/R)² = 3/4 Mv², so v = √(4gh/3).
Trap tested: Ignoring rotational kinetic energy.
Question 4
In ideal rolling without slipping down a fixed incline, does static friction dissipate mechanical energy?
Look at the contact point.
No. The contact point is instantaneously at rest, so static friction does no work in the ideal model, though it provides torque.
Trap tested: Equating any friction with energy loss.
Question 5
A puck of mass m moving tangentially at speed v sticks to the rim of a disk of radius R and moment of inertia I about its center. (a) Find the final angular speed. (b) State whether kinetic energy is conserved and justify.
Conserve angular momentum about the disk center.
External torque about the disk center is negligible during the short collision, so angular momentum is conserved. Initial L = mvR. Final inertia is I + mR², so ω = mvR/(I + mR²). Kinetic energy is not conserved because the puck sticks; mechanical energy is dissipated internally.
Trap tested: Conserving kinetic energy in a sticking rotational collision.
Rubric
Question 6
A uniform rod of length L pivots about one end and is released from horizontal. (a) Find its angular speed when vertical. (b) Explain why using constant angular acceleration would be a modeling error.
The center of mass drops L/2; I_end = 1/3 ML².
The center of mass drops L/2, so Mg(L/2) = (1/2)(1/3ML²)ω² and ω = √(3g/L). Constant angular acceleration is invalid because the torque Mg(L/2)sin(θ) changes as the rod rotates.
Trap tested: Trying constant-torque kinematics even though torque changes with angle.
Rubric
score What it means
Move to mixed USAPhO-intro mechanics and timed solution writing.
Identify whether torque, rolling, or angular momentum caused the misses.
Work torque-arm and rolling-constraint lessons untimed before full mocks.
after Turn the mock into data
If you missed a question, do not just reread the solution. Open the linked lesson, redo one drill, then come back to the mock.