Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r = 0.1 m. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is m/s. (Assume string does not slip and g = 10 m/s2)
Answer
2
Solution
Moment of inertia of the pulley (disc): I = 12(30m)r2 = 15mr2
No slipping both masses move with speed v and = vr.
Energy conservation: loss of P.E. = gain of K.E.
2mgh - mgh = 12(2m)v2 + 12mv2 + 12(15mr2) v2r2
mgh = 12(3m + 15m)v2 = 9mv2
v2 = gh9 = 10 3.69 = 4
v = 2 m/s
Answer: 2