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

Answer: (B)

100

Answer 2

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
$L_i = L_f$
$I_1 +I_2 = (I_1 + I_2) \omega$
$\omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}$
$\frac{(0.04\,kg\,m^2)(50\,rad\,s^{-1}) + (0.02\,kg\,m^2)(200\,rad\,s^{-1})}{(0.04 + 0.02) \,kg\, m^2}$
$= \frac{(2+4)}{0.06} rad \,s^{-1} = 100 \,rad\, s^{-1}$