A 40kg boy running at 41 ms jumps tangentially onto a small

A 40-kg boy running at 4.1 m/s jumps tangentially onto a small stationary circular merry-go-round of radius 2.0 m and rotational inertia2.0×10² kgm2 pivoting on a frictionless bearing on its central shaft.

1. Determine the rotational speed of the merry-go-round after the boy jumps on it.

2. Find the change in kinetic energy of the system consisting of the boy and the merry-go-round.

3. Find the change in the boy\'s kinetic energy.

4. Find the change in the kinetic energy of the merry-go-round.

Solution

The child\'s angular velocity = v/r = 4.1m/s/2.0m = 2.05 rad/s
and his moment of inertia = m*r^2 = 40*2.0^2 = 160.0kg-m^2

After he jumps on the system\'s I = 200 + 160.0= 360.0 kg-m^2

Now = I*)initial/ I final = 160.0*2.05/360.0 = 0.9111 rad/s

So K = 1/2*I*^2 = 1/2*360.0*0.911^2 =149.42J