Question

We wrap a light, flexible cable around a solid cylinder with mass 0.95 kg and radius 0.23 m . The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to a block of mass 1.84 kg and release the object with no initial velocity at a distance 1.58 m above the floor. As the block falls, the cable unwinds without stretching or slipping, turning the cylinder. Suppose the falling mass is made of ideal rubber, so that no mechanical energy is lost when the mass hits the ground.

If the cylinder is originally not rotating and the block is released from rest at a height 1.58 m  above the ground, to what height will this mass rebound if it bounces straight back up from the floor?   

Answer 1

let
M = 0.95 kg
R = 0.23 m
m = 1.84 kg
h = 1.58 m

let v is the angular speed and v is the linear speed of the block just before hitting the ground.

Apply conservation of energy

Final kinetic energy = initial potential energy

(1/2)*I*w^2 + (1/2)*m*v^2 = m*g*h

(1/2)*(1/2)*M*R^2*w^2 + (1/2)*m*v^2 = m*g*h

(1/4)*M*v^2 + (1/2)*m*v^2 = m*g*h (since v = R*w)

v^2*(M/4 + m/2) = m*g*h

v = sqrt(m*g*h/(M/4 + m/2) )

= sqrt(1.84*9.8*1.58/(0.95/4 + 1.84/2) )

= 4.96 m/s

let H is the height reached after bouncing.

now again apply conservation of energy

m*g*H = (1/2)*v^2

H = v^2/(2*g)

= 4.96^2/(2*9.8)

= 1.26 m <<<<<<----------------Answer