A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev/s. The merry-go-round is a solid disk, and the moment of inertia is = 1/2 MR2. . A 23.5 kg child is initially at rest, he then gets onto the merry-go-round by grabbing its outer edge. The child can be treated as a point mass located at the outer edge of the merry-go-round. His moment of inertia is mR2. After this rotational collision, the merry-go-round and the child have a commen final angular velocity (which is different from the merry-go-round's initital angular velocity). (e) What is the final angular velocity in rad/s after the child gets on?

where [tex]I_m[/tex] is the moment of inertia of the merry-go-round, [tex]W_m[/tex] is the initial angular velocity of the merry-go-round, [tex]I_s[/tex] is the moment of inertia of the merry-go-round and the child together and [tex]W_f[/tex] is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = [tex]\frac{1}{2}M_mR^2[/tex]

I = [tex]\frac{1}{2}(115 kg)(2.5m)^2[/tex]

I = 359.375 kg*m^2

Where [tex]M_m[/tex] is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2[tex]\pi[/tex] rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

[tex]I_s = \frac{1}{2}M_mR^2+mR^2[/tex]

[tex]I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2[/tex]

[tex]I_s = 506.25kg*m^2[/tex]

Finally we replace all the data:

[tex](359.375)(3.2672) = (506.25)W_f[/tex]

Solving for [tex]W_f[/tex]:

[tex]W_f = 2.319 rad/s[/tex]