Answer:
The final angular velocity is [tex]w_f = \frac{MR^2}{MR^2+ mr^2} w[/tex]
Explanation:
From the question we are told that
The mass of the first disk is m
The radius of the first disk is r
The mass of second disk is M
The radius of second disk is R
The speed of rotation is w
The moment of inertia of second disk is [tex]I = \frac{1}{2} MR^2[/tex]
Since the first disk is at rest initially
The initial angular momentum would be due to the second disk and this is mathematically represented as
[tex]L_i = Iw = \frac{1}{2} MR^2 w[/tex]
Now when the first disk is then dropped the angular momentum of the whole system now becomes
[tex]L_f = (I_1 + I_2 ) w_f= ( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f[/tex]
This above is because the formula for moment of inertia is the same for every disk
According to the law conservation of angular momentum
[tex]L_f = L_i[/tex]
[tex]( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f = \frac{1}{2} MR^2 w[/tex]
=> [tex]w_f = \frac{\frac{1}{2} MR^2 w }{( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2)}[/tex]
[tex]w_f = \frac{MR^2}{MR^2+ mr^2} w[/tex]
