Question

Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc $D_1 $ has 2 kg mass and 0.2 m radius and initial angular velocity of $ 50\, rad \,s^{-1}.$ Disc $ D_2 $ has 4 kg mass, 0.1 m radius and initial angular velocity of $ 200\, rad\, s^{-1} $. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in rad $ s^{-1}$ ) of the system is

• A. 60
• B. 100
• C. 120
• D. 40

Answer 1

The Correct Option is B

Moment of inertia of disc $ D_1 $ about an axis passing through its centre and normal to its plane is
$ I_1 = \frac{MR^2}{2} = \frac{(2kg)(0.2m)^2}{2} = 0.04\,kg \,m^2 $
Initial angular velocity of disc $ D_1$, $ \omega_1 = 50\, rad \,s^ {-1} $
Moment of inertia of disc $ D_2 $ about an axis passing through its centre and normal to its plane is
$ I_2 = \frac{(4\,kg)(0.1\,m)^2}{2} = 0.02 \,kg \,m^2 $
Initial angular velocity of disc $ D_2 $, $ \omega_2 = 200\, rad\, s^{-1} $
Total initial angular momentum of the two discs is
$ L_i = I_1 \omega_1 + I_2 \omega_2 $
When two discs are brought in contact face to face (one on the top of the other) and their axes of rotation coincident, the moment of inertia $l$ of the system is equal to the sum of their individual moment of inertia.
$ I = I_1 +I_2 $
Let $ \omega $ be the final angular speed of the system.
The final angular momentum of the system is
$ L_f = I_ {\omega} = (I_1 + I_2) \omega $
According to law of conservation of angular momentum, we get
$ L_i = L_f $
$ I_1 +I_2 = (I_1 + I_2) \omega $
$ \omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} $
$ \frac{(0.04\,kg\,m^2)(50\,rad\,s^{-1}) + (0.02\,kg\,m^2)(200\,rad\,s^{-1})}{(0.04 + 0.02) \,kg\, m^2} $
$ = \frac{(2+4)}{0.06} rad \,s^{-1} = 100 \,rad\, s^{-1} $