Answer :
Answer:
[tex]\omega_2=2.304\ rev/s[/tex]
Explanation:
Given that
Mass of disc,M=3 kg
radius r= 65 cm
Mass of small mass ,m=0.07 kg
Initial speed = 2.2 rev/s
If external torque is zero then angular momentum of system will remain conserve.
Moment of inertia of disc at initial condition
[tex]I_1=\dfrac{Mr^2}{2}+mr^2[/tex]
[tex]I_1=\dfrac{3\times 0.65^2}{2}+0.07\times 0.65^2[/tex]
[tex]I_1=0.66\ kg.m^2[/tex]
Moment of inertia of disc at final condition
[tex]I_2=\dfrac{Mr^2}{2}[/tex]
[tex]I_2=\dfrac{3\times 0.65^2}{2}[/tex]
[tex]I_2=0.63\ kg.m^2[/tex]
So from conservation of angular momentum
[tex]I_1\omega_1=I_2\omega_2[/tex]
[tex]0.66\times 2.2=0.63\times \omega_2[/tex]
[tex]\omega_2=2.304\ rev/s[/tex]
This is final speed of disc after small mass flies off from the disc.