Could someone help me answer the following with complete solutions/free-body-dia

Question

Could someone help me answer the following with complete solutions/free-body-diagrams?

Thanks!

A bowling ball of mass m and radius R is released so that at the instant it touches the floor it is moving horizontally with a speed v_0 and is not rotating. It slides a distance s_1 in time t_1 before it begins to roll without slipping. If mu_k is the coefficient of kinetic friction between the ball and the floor, find s_1, t_1 and the final speed of the ball, v_1. What is the ratio of the final kinetic energy to the initial kinetic energy of the ball? Two boxes hang from a solid disk pulley that is free to rotate as the blocks rise and fall as shown below. The left box has a mass of m_1 = 4.7 kg and the right box has a mass m_2 = 1.9 kg. The pulley has mass m_3 = 3 kg and radius R = 0.12 m. Assuming that the pulley starts from rest, how many revolutions has the pulley made and what is the magnitude of the angular speed of the pulley after a time of t = 0.46 s? Consider a solid cylinder of mass M and radius R rolling down a ramp under the influence of gravity The coefficient of static friction of the ramp is mu_s = 0.2. At what angle theta will the ramp be too steep for the cylinder to roll, so that it slips instead?

Explanation / Answer

1. f = uk m g

torque = I alpha

uk m g R = ( 2 m R^2 / 5) (alpha)

alpha = 5 uk g / 2 R

and Fnet = m a

- uk m g = m a

a = - uk g

after time t1, w1 = v1/ r ( rolling without slipping)

w1 = w + alpha t = 0 + 5 uk g t /2 R

v1 = v0 - uk g t

w1 R = v1

5 uk g t1 / 2 = v0 - uk g t1

t1 = v0 / (3.5 uk g ) .............Ans

s1 = theta r = alpha t1^2 R / 2

= (5 uk g / 2 R ) ( v0 / 3.5 uk g )^2 R / 2

s1 = 0.102 v0^2 / uk g ...........Ans

v1 = v0 - uk g (v0 / 3.5 uk g) = 0.714 v0 .........Ans

Ki = m v0^2 /2

Kf = I w^2 /2 + m v^2 /2

= (2 m R^2 / 5) (5 v0 / 7 R)^2 /2 + m (0.714 v0)^2 / 2

= m v0^2 / 2

Kf / Ki = 1

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