Answer:
[tex]\frac{MR^{2}\omega _{i}}{(MR^{2}+mr^{2})}=\omega _{f}[/tex]
Explanation:
Mass of small disc = m
radius of small disc = r
Mass of large disc = M
Radius of large disc = R
Initial angular speed = ωi
Let the final angular speed is ωf
As no external torque of applied on the system so the angular system remains constant.
I1 ωi = I2 ωf
MR^2 ωi = (MR^2 + mr^2) ωf
[tex]MR^{2}\omega _{i}=(MR^{2}+mr^{2})\omega _{f}[/tex]
[tex]\frac{MR^{2}\omega _{i}}{(MR^{2}+mr^{2})}=\omega _{f}[/tex]
