Question

Bubba (m=100kg) and Sally (m=60kg) are playing on a merry-go-round. Treat the merry-go-round as a disk with mass 100 kg and radius 1.3 m. Both Bubba and Salley are at the edge of the disk when the disk is rotating at .25 revolutions per second. Suppse Bubba moves inward so that he is now exactly at the center. Treat the two children as point masses. Assuming there is no external torques or forces,

A) What is the new rotational speed of the merry-go-round?

B) How much work is done? Who, what did this work?

Answer 1

here,

mass of Bubba , m1 = 100 kg

mass of Sally , m2 = 60 kg

radius , r = 1.3 m

mass of merry-go-round , M = 100 kg

initial angular speed , w0 = 0.25 rev/s = 1.57 rad/s

let the final angular speed be w

using conservation of angular momentum

I1 * w1 = I2 * w2

(m1 * r^2 + m2 * r^2 + 0.5 * M * r^2) * 1.57 = ( m1 * 0^2 + m2 * r^2 + 0.5 * M * r^2 ) * w

1.3^2 * ( 100 + 60 + 0.5 * 100) * 1.57 = ( 0 + 60 * 1.3^2 + 0.5 * 100 * 1.3^2) * w

solving for w

w = 3 rad/s

A)

the new rotational speed is 3 rad/s

B)

the work done , W = the kinetic energy gained

W = 0.5 * I2 * w2^2 - 0.5 * I1 * w1^2

W = 0.5 * (- (m1 * r^2 + m2 * r^2 + 0.5 * M * r^2) * 1.57^2 + (( m1 * 0^2 + m2 * r^2 + 0.5 * M * r^2 ) * w^2) )

W = 0.5 * (- 1.3^2 * ( 100 + 60 + 0.5 * 100) * 1.57^2 + ( 0 + 60 * 1.3^2 + 0.5 * 100 * 1.3^2) * 3^2 )

W = 399.2 J

the work done is 399.2 J

theis is the work done done by the friction force when Bubba moves fromt he edge to center