A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rigid, massless rod. The rod and balls are rotating clockwise about their center of gravity at 20 rpm. What torque will bring the balls to a halt in 5.0 s?
SOLUTION: Two balls are connected by a rigid,massless rods are a rigid body rotating about an axis through center of gravity.we have placed origin of system on 1 kg ball. The net torque is expressed as = I .to calculate the torque we require the moment of inertia c.g and angular acceleration. The center of gravity is given by x cm = (m 1 1m x )2 2 +m 1 2 x = (1.0 kg)( 0 m)+(2.0 kg)(1.0 m)/(1.0 kg +2.0 kg) cm xcm = 0.667 m ycm = 0 m Given: w r 0 rad/s,w = 2i rpm = (20 rpm)(2/60) rad/s = 2/3 rad/s trt i 5 s So from kinematics relation angular velocity w = w +rt ti) r s In this eq all values are except angular acceleration ,then w r w +i(t t r s 0 rad/s = (2/3 ) rad/s+(5.0 s) = 2/15 rad/s 2 Torque that will brings balls to t = 5.0 s 2 2 = Icm = (2/3 kg. m )(2/15 rad/s ) =4/45 N/m =0.28 N/m Here negative sign indicates the torque acting in clockwise direction.
