Question

In the figure, a very light rope is wrapped around a wheel of radius R = 1.2 meters and does not slip. The wheel is mounted with frictionless bearings on an axle through its center. A block of mass 14 kg is suspended from the end of the rope. When the system is released from rest it is observed that the block descends 9 meters in 2.6 seconds. What is the moment of inertia of the wheel (give answer to the nearest 0.1 kg m2.)

Answer 1

given data is radius of wheel r = 1.2 m, block mass m = 14 kg, s = 9 m , t = 2.6 s

   linear accleration of the block is from equations of motion

   s = ut + 1/2 a t^2, u =0, => a = 2*s/t^2 = 2*9/2.6^2 =2.66 m/s2

   Tension in the rope is mg- T = ma ==> T = 14*9.8 -T = 14*2.66 = 99.96 N

now the torque acting on wheel is T = I*alpha= r*F Sin theta
                                

                   I = r*T*1/(a/r)

                   I = 1.2*99.96/(2.66/1.2)
                   I = 54.11 kg m^2