Answer to Question #108507 in Classical Mechanics for Chinmoy Kumar Bera

Question

the moment of inertia of a circular disc about an axis passing through it center and perpendicular to its plane is 10kg m^2it is rotating clockwise with an angular speed 60rpm.another didc with a rotational inertia of 5kgm^2 is rotating anticlockwise with an angular speed 30rpm.at some instant this disc is dropped on top of first disc.calculate final angular speed of the system.

Answer 1

As per the given question,

Moment of inertia of the first disc $(I_1)=10kg m^2$

Angular speed of the first disc$(\omega_1)=60 rpm$

Moment of inertia of the second disc $(I_2)=5kgm^2$

Angular speed of the second disc $(\omega_2)=30rpm$

When the smaller disc will drop on the larger, then net momentum always be conserve,

Let the final angular momentum be $\omega_3$

Hence, applying the conservation of the angular momentum

$I_1\omega_2-I_2\omega_2= (I_1+I_2) \omega_3$

$\Rightarrow \omega_3= \dfrac{I_1\omega_2-I_2\omega_2}{I_1+I_2}$

$\Rightarrow \omega_3= \dfrac{10\times 60-5\times 30}{10+5}$

$\Rightarrow \omega_3=\dfrac{600-300}{15}$

$\Rightarrow \omega_3=\dfrac{300}{15}=20rpm$