A solid uniform cylinder with mass 8.25 kg and diameter 15.0 cm is spinning at 220 rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.333. What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

Solution 61P Step 1 of 6: Given solid cylinder has mass M= 8.25 kg and diameter 15 cm, so the radius R=7.5cm= 0.075 m rotates with angular speed =220 rev/min and comes to rest =0 after 0 f undergoing = 5.25 rev angular displacement by the application of the brake on surface of cylinder with kinetic friction coefficient =0.333. Here need to calculate the normal k force that has to be applied to stop the cylinder. Step 2 of 6: Given data. Mass, M= 8.25 kg Radius, R=0.075 m kinetic friction coefficient =0.333 k Final angular speed, =0 f Initial angular speed, =220 rev/min 0 Using 1 rev =2 rad and 1 min= 60 sec = 220×2rad/s 0 60 =23.02 rad/sec 0 Angular displacement, = 5.25 rev Using 1 rev =2 rad = 5.25×2 rad =32.97 rad To find, Normal force , N= Step 3 of 6: In order to stop the rotating solid cylinder by the application of normal force, we need apply a normal force in such a way that the frictional force will be equal to the torque of the rotating cylinder. That is = N R k Where is torque, is coefficient of kinetic friction, N is normal force and R is radius. k Using = I I = kN R Solving for normal force, N= R …………...1 k Where I is moment of inertia and is angular acceleration.