Question

A light, nonstretching cable is wrapped around a solid cylinder with mass 88 kg and radius 0.19 m. The cylinder rotates with negligible friction about a stationary horizontal axis. We attach the free end of the cable to a block of mass 48 kg, and release the block from rest at a distance 3.9 m above the floor. As the block falls, the cable unwinds without stretching or slipping. Find the angular speed of the cylinder (in radians/s) the moment before the block hits the floor. Use g=9.8 m/s2. Do not round during the calculation and keep at least two decimal places in your final answer.

Answer 1

let

M = 88 kg

R = 0.19 m

m = 48 kg

h = 3.9 m

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

Apply conservation of energy

rotational kinetic energy gained by cyllinder + linear kinetic energy gained by block = loss of potential energy of the block

(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*(R*w)^2 = m*g*h

(1/4)*88*0.19^2*w^2 + (1/2)*48*(0.19*w)^2 = 48*9.8*3.9

==> w = 33.24 rad/s <<<<<<<----------------Answer