Two masses m and 2 m are connected by a light string going over a pulley (disc) of mass 30 m with radius r=0.1 ~m . The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2 m 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 / s ² )

Step-by-step solution

Let the speed of the masses be v and the angular velocity of the pulley be = v/r since the string does not slip. The pulley is a disc of mass M = 30m and radius r . Its moment of inertia is I = 1 2 Mr^2 = 1 2 (30m)r^2 = 15mr^2 . Using the principle of conservation of mechanical energy, the loss in potential energy of the system equals the gain in kinetic energy. When the mass 2m descends by h = 3.6 m, the mass m ascends by h = 3.6 m. Loss in potential energy U = (2m)gh - (m)gh = mgh . Gain in kinetic energy K = 1 2