CP? A thin, light wire is wrapped around the rim of a wheel (?Fig. E9.45?). The wheel rotates without friction about a stationary horizontal axis that passes through the center of the wheel. The wheel is a uniform disk with radius R = 0.280 m. An object of mass ?m? = 4.20 kg is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration. If the suspended object moves downward a distance of 3.00 m in 2.00 s, what is the mass of the wheel?

Solution 49E We shall have to calculate the acceleration of the suspended object in order to proceed the other calculations. Distance moved by the object s = 3.00 m Time t = 2.00 s 1 2 From the equation s = at 2 we can calculate the acceleration as follows. a = 2s/t2 2 2 a = 2 × 3.00/(2.00) m/s 2 a = 1.5 m/s Let the tension in the wire be T . Since mass is moving downward with an acceleration a, hence the net acceleration on it is = g a Thus, the tension, T = 4.20 × (g a) N T = 4.20 × (9.8 1.5) N = 34.86 N The radius of the wheel R = 0.280 m Torque on the wheel = TR = 34.86 × 0.280 N.m = 9.76 N.m Let M be the mass of the wheel. 1 2 1 2 2 Its moment of inertia is I = MR2= 2 × M × (0.280) kg.m I = 0.0392M kg.m 2 Angular acceleration = a/R = 1.5/0.280 rad/s = 5.35 rad/s 2 From the relation, = I and substituting the calculated values, 2 2 9.76 N.m = 0.0392M kg.m × 5.35 rad/s M = 46.5 kg The mass of the wheel is 46.5 kg .