Question

A hollow ball has mass M=2.0kg, radius R=0.35m, and moment of inertia about the center of mass I=(2/3)MR². The ball is thrown without bouncing, to the right with an initial speed 2.0m/s and backspin. The hoop moves across the rough floor (coefficient of sliding friction = 0.25) and returns to its original position with a speed of 0.5 m/s. All surfaces and the hoop may be treated as ideally rigid. Develop an expression for angular velocity of the hoop as a function of time while it is rolling with slipping. Determine the backspin with which the hoop must be released as well as the furthest distance travelled to the right.

Answer 1

co-efficient of friction   = 0.25

frictional force f = 0.25 *2*9.8 = 4.9 N

only frictional force provides torque about center and causes the ball to roll

torque = f R = I

MI of the hollow sphere I = 2/3 MR²

angular acceleration   = 3f /2MR = -3*4.9/2*2*0.35 = 4.33 rads/s/s

angular speed of the ball = _i + t = _i + 4.33t rads/s

_i - initial angular speed.

Initially the ball slides and rolls across the floor friction opposes the motion and it comes to rest finally , then the ball rolls back without slipping.

initial velocity of the ball = 2 m/s

KE = 1/2 Mv² = f  s , work done against friction.

distance moved s = 0.5*2* 2 = 2m

linear acceleration a= v²/2s = 4/2*2 = 1 m/s/s

final linear vel.=0

duration of slipping t = v_i /a = 2 s

ball returns with the speed of 0.5 cm without slipping.

final angular speed _f = v_cm/R = -0.5/0.35 = -10/7 rads /s , back spin

initial angular speed

_i = -10/7 - 4.33*2 = -10.09 rads/s - back spin required for the ball.