An object of mass 1 ~kg is allowed to hang tangentially from the rim of the wheel of radius R. When released from the rest, the block falls vertically through 4 ~m height in 2 seconds. The moment of inertia is 1 ~kg ~m ^2 . The radius of the wheel R is

Step-by-step solution

Let m = 1 kg be the mass of the block, I = 1 kg m ^2 be the moment of inertia of the wheel, R be the radius of the wheel, and h = 4 m be the distance fallen in time t = 2 s . The block starts from rest, so the distance fallen is given by h = 1 2 a t^2 , where a is the linear acceleration of the block. 4 = 1 2 a (2)^2 4 = 2a a = 2 m/s ^2 . The equation of motion for the block is mg - T = ma , where T is the tension in the string. 1 9.8 - T = 1 2 T = 9.8 - 2 = 7.8 N . The torque on the wheel is = T R = I , where is t