Question

Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc $D_1$ has $2kg$ mass and $0.2m$ radius and initial angular velocity of $50rad{s}^{-1}$. Disc $D_2$ has $4kg$ mass, $0.1m$ radius and initial angular velocity of $200rad{s}^{-1}$. 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
$I_1=\frac{M{R}^2}{2}=\frac{(2kg)(0.2m{)}^2}{2}=0.04kg{m}^2$
Initial angular velocity of disc $D_1,{\omega }_1=50rad{s}^{-1}$
Moment of inertia of disc $D_2$ about an axis passing through its centre and normal to its plane is
$I_2=\frac{(4kg)(0.1m{)}^2}{2}=0.02kg{m}^2$
Initial angular velocity of disc $D_2,{\omega }_2=200rad{s}^{-1}$
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
$L_i=L_f$
$I_1+I_2=\left(I_1+I_2\right)\omega$
$\omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}$
$\frac{\left(0.04kgm{}^2\right)\left(50rad{s}^{-1}\right)+\left(0.02kg{m}^2\right)\left(200rad{s}^{-1}\right)}{(0.04+0.02)kg{m}^2}$
$=\frac{(2+4)}{0.06}rad{s}^{-1}=100rad{s}^{-1}$