Question

A 95.3-kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.77 rad/s. A monkey drops a 9.89-kg bunch of bananas vertically onto the platform. They hit the platform at 4/5 of its radius from the center, adhere to it there, and continue to rotate with it. Then the monkey, with a mass of 20.1 kg, drops vertically to the edge of the platform, grasps it, and continues to rotate with the platform. Find the angular velocity of the platform with its load. Model the platform as a disk of radius 1.61 m.

Answer 1

INITIAL ANGULAR MOMENTUM OF THE PLATFORM IS I1 = 0.5*m*r^2 = 0.5*95.3*r^2

Using law of conservation of angular momentum

I1*w1 = I2*w1

I2 = moment of inertia of platform + bananas + monkey

I2 = (0.5*m*r^2) + (mb*(4r/5)^2)+(mm*r^2)

I2 = (0.5*95.3*r^2)+(9.89*(4r/5)^2) + (20.1*r^2)

w1 = 1.77 rad/sec

then

I1*w1 = I2*w2

(0.5*95.3*r^2)*1.77 = ((0.5*95.3*r^2)+(9.89*(4r/5)^2) + (20.1*r^2))*w2

r^2 cancels
(0.5*95.3*1.77) = ((0.5*95.3)+(9.89*(4/5)^2)+(20.1))*w2

w2 = 1.13 rad/sec