Question

Two discs rotating about their respective axis of rotation with angular speeds $2rad{s}^{-1}$ and $5rad{s}^{-1}$ are brought into contact such that their axes of rotation coincide. Now, the angular speed of the system becomes $4rad{s}^{-1}$. If the moment of inertia of the second disc is $1\times 10^{-3}kg{m}^2$, then the moment of inertia of the first disc (in $kg{m}^2$) is

• A. $0.25\times 10^{-3}$
• B. $1.5\times 10^{-3}$
• C. $1.25\times 10^{-3}$
• D. $0.75\times 10^{-3}$
• E. $0.5\times 10^{-3}$

Answer 1

Answer: (E)

Relation between angular momentum and moment of inertia
$J=I\omega$
where $I_1=?$
$I_2=1\times 10^{-3}kg-m^2$
${\omega }_1=2$ rad / s
${\omega }_2=5$ rad / s
$\omega =4$ rad / s
So, $J=J_1+J_2$
$\left(I_1+1\times 10^{-3}\right)\times 4=I_1\times 2+5\times 1\times 10^{-3}$
$4I_1+4\times 10^{-3}=2I_1+5\times 10^{-3}$
$2I_1=1\times 10^{-3}$
$I_1=\frac{1}{2}\times 10^{-3}$
$=0.5\times 10^{-3}kg-m^2$