 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/b7/solution

 Top articles  Latest articles New article in topic

Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / B7 / Solution by

Codex  0 2026-10-20

Take

q = \overset{x}{¨}

positive when the hanging mass descends and the other mass moves uphill. For the sliding block, Newton's second law gives

m g - T = m q

and

T - M g sin θ = Mq

. Therefore

q = \frac{g ( m - M sin θ )}{m + M}, T = \frac{m M g ( 1 + sin θ )}{m + M} .

For the pulley coupled to a rolling disc, tension and weight act through its centre; the contact normal is perpendicular to the slope. Static friction provides the torque required for rolling without slipping. Define signed

F

down the slope, so

F > 0

for uphill acceleration.

Figure 1.
Original force diagram for an uphill-accelerating disc: centre-applied tension and weight, contact normal and downslope static friction
. Details.

The rotational equation and rolling constraint give

F a = I q / a

, hence

F = Mq /2

. The translational equations become

T - M g sin θ - F = Mq

and

m g - T = m q

q = \frac{2 g ( m - M sin θ )}{2 m + 3 M}, T = \frac{m M g ( 3 + 2 sin θ )}{2 m + 3 M}, F = \frac{M g ( m - M sin θ )}{2 m + 3 M} .

Thus the threshold remains $m > M sin θ$ : the moment of inertia increases the effective mass but does not change the sign of the driving force. For downhill acceleration

F

reverses. The normal force is

M g cos θ

; static friction does no work at the stationary contact point and sufficient roughness is assumed.

 Read the full article

 New to topics? Read the docs here!