A horizontal plywood disk with mass 7.00 kg and diameter 1.00 m pivots on frictionless bearings about a vertical axis through its center. You attach a circular model-railroad track of negligible mass and average diameter 0.95 m to the disk. A 1.20-kg, battery-driven model train rests on the tracks. To demonstrate conservation of angular momentum, you switch on the train’s engine. The train moves counterclockwise, soon attaining a constant speed of 0.600 m/s relative to the tracks. Find the magnitude and direction of the angular velocity of the disk relative to the earth.

Solution 97P The expression for angular momentum is given by, L = I , where I is the moment of inertia of the object concerned and is its angular velocity. Angular momentum of the system is given to be conserved. This means there is no external torque on it. Now, the mass of the disk m = 7.00 kg d Its diameter = 1.00 m Radius r = d.50 m Therefore, the angular momentum of the disk is L = I d d d L = m r 2 d 2 d d d 1 2 2 L d 2× 7.00 kg × (0.50) m × d 2 L d 0.875 kg.m …..(1)d Mass of the train m = 1.20 kg t The diameter of the railroad track = 0.95 m Radius of the railroad track r track= 0.475 m v = 0.600 m/s t Therefore, angular momentum of the train L = m r 2 , is the angular velocity t t track t t of the train. 2 2 L t 1.20 kg × (0.475) m × t L = 0.270 kg.m …..(2) t t Now, given linear speed v = 0.600 m/s Radius of the track r track= 0.475 m Angular speed of the train with respect to the track is , = v/r = (0.600 m/s)/0.475 m = 1.26 rad/s T track Now, T = t d 1.26 rad/s = t d +d1.26 rad/s = …..(3t Given that, angular momentum of the system is conserved. Therefore, the summation of the angular momentum of the train and the disk will be zero. Adding equations (1) and (2) and substituting the value of in equation (3), t L +dL = 0t 0.875 + 0.270( + 1.26) = 0 d d 1.145 = 0.34 d =d 0.30 rad/s Therefore, the angular velocity of the disk is 0.30 rad/s. The negative sign indicates that the disk will move clockwise.