How Does Adding Weight Affect the Angular Speed and Energy of a Merry-Go-Round?

[SelHype]

A 3.0-m-diameter merry-go-round with rotational inertia 120 kg*m² is spinning freely at 0.60 rev/s. Four 25-kg children sit suddenly on the edge of the merry-go-round. (a) Find the new angular speed in rev/s. (b) Determine the total energy lost to friction between the children and the merry-go-round in J.



L_i\omega_i=L_f\omega_f
I=.5mR²



Okay so I know that L_i\omega_i=L_f\omega_f. That means that (120 kg*m^2)(0.60 rev/s)=(mass of merry-go-round + 4(25kg))(1.5²)\omega_f.

To find the mass of the merry-go-round I took the Inertia and divided it by R² to get m = (120 kg*m²)/(1.5²) = 53.3.

When I plugged it into the equation I found the new angular speed to be .21 rev/s.


My problem is with part (b)...

So I know that J is (kg*m²)/s², but I am unsure as to how to solve for it...A little help, please?

 

[unscientific]

Find the angular velocity before the children get on, then find rotational energy use (1/2)I(omega^2).

Then find the angular velocity after the children get on. Take initial rotational energy - final rotational energy. The energy loss went to friction.

 

[Redbelly98]

While you got the correct answer for (a), you did make two errors in the process ... which fortunately canceled out.

First, let's clear up those equations:

L_i = L_f
and
I_merryω_i = (I_merry+I_children)ω_f

It's not necessary to find the mass of the merry-go-round, you can just use the I_merry value here to find ω_f.

To find the mass of the merry-go-round I took the Inertia and divided it by R² to get m = (120 kg*m²)/(1.5²) = 53.3.

Actually, this is 0.5*m, since I = 0.5 m R^2. But that's okay, since 0.5*m is what should have been used in your earlier expression,
(mass of merry-go-round + 4(25kg))

So the answer is 0.21 rev/s as you got.