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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019-resit/11/b/i/solution

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Past exam of the physics 1 paper of the University of Oxford / 2019 resit / 11 / b / i / Solution by

Codex  0 2026-10-25

Take rightward translation as positive and counterclockwise angular acceleration

α

as positive. Let

f

denote signed ground friction, positive to the right. The source's lower-tangent winding makes the tension exert counterclockwise torque

T r

. For a horizontal pull, translation and rotation obey

M a = T + f, I α = T r + R f, a = - R α .

The final relation is the rolling without slipping constraint. Eliminating

f

and

α

gives

a = \frac{TR ( R - r )}{I + M R ^{2}} > 0, f = - T \frac{I + M r R}{I + M R ^{2}} .

Thus the requested magnitude of static friction is

F = ∣ f ∣ = T \frac{I + M r R}{I + M R ^{2}} .

It acts leftward. Released from rest, the spool accelerates rightward, toward the pulling hand, and rotates clockwise. There is no contradiction between leftward friction and rightward motion: static friction prevents slipping of the contact point, and the horizontal tension is larger than

∣ f ∣

because

r < R

.

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