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)

Moment of inertia of disc $D_{1}$ about an axis passing through its centre and normal to its plane is
$I_{1}=\frac{M R^{2}}{2}=\frac{(2 k g)(0.2 m)^{2}}{2}=0.04\, kg \, m ^{2}$
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 $L_{f}=I_{\omega}=\left(I_{1}+I_{2}\right) \omega$
According to law of conservation of angular momentum, we get
$L_{i}=L_{f}$
$I_{1}+I_{2}=\left(I_{1}+I_{2}\right) \, \omega$
$\omega=\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}$
$\frac{\left(0.04\, kg\, m { }^{2}\right)\left(50\, rad \, s ^{-1}\right)+\left(0.02\, kg\, m ^{2}\right)\left(200\, rad\, s ^{-1}\right)}{(0.04+0.02) \, kg\, m ^{2}}$
$=\frac{(2+4)}{0.06} \, rad \, s^{-1}=100 rad\, s ^{-1}$