In Fig. 11.16 the lower disk, of mass 370 g and radius 3.5 cm, is rotating at 18
ID: 1262140 • Letter: I
Question
In Fig. 11.16 the lower disk, of mass 370 g and radius 3.5 cm, is rotating at 180 rpm on a frictionless shaft of negligible radius. The upper disk, of mass 270 g and radius 2.3 cm, is initially not rotating. It drops freely down the shaft onto the lower disk, and frictional forces act to bring the two disks to a common rotational speed.
(a) What is that speed? RPM
(b) What fraction of the initial kinetic energy is lost to friction? %
In Fig. 11.16 the lower disk, of mass 370 g and radius 3.5 cm, is rotating at 180 rpm on a frictionless shaft of negligible radius. The upper disk, of mass 270 g and radius 2.3 cm, is initially not rotating. It drops freely down the shaft onto the lower disk, and frictional forces act to bring the two disks to a common rotational speed. (a) What is that speed? RPM (b) What fraction of the initial kinetic energy is lost to friction? %Explanation / Answer
Given,
mass of lower disk = 0.37 kg
mass of upper disk = 0.270 kg
radius of lower disk = 0.035 m
radius of upper disk = 0.023 m
speed of lower disk = 180 rpm
by consrvation of angular momentum
initial angular momentum = final angular momentum
initial angular momentum = moment of inertia * angular speed
initial angular momentum = 0.5 * Ml * Rl^2 * 180
initial angular momentum = 0.5 * 0.37 * 0.035^2 * 180
initial angular momentum = 0.0407
final angular momentum = 0.5 * Ml * Rl^2 * speed + 0.5 * Mu * Ru^2 * speed
final angular momentum = 0.5 * speed (0.37 * 0.035^2 + 0.27 * 0.023^2)
final angular momentum = 0.000298 * speed
0.0407 = 0.000298 * speed
speed = 136.57 rpm
initial kinetic energy = 0.5 * 0.5 * 0.37 * 0.035^2 * 180^2
initial kinetic energy = 3.671
final kinetic energy = 0.5 * 0.5 * 136.57^2 * (0.37 * 0.035^2 + 0.27 * 0.023^2)
final kinetic energy = 2.77
lost KE = (3.671 - 2.77)/3.671 * 100
fraction of the initial kinetic energy is lost to friction = 24.54 %