Question

Two coaxial discs having moments of inertia $I_1$ and $\frac{I_1}{2}$ are rotating with respective angular velocities ${\omega }_1$ and $\frac{{\omega }_1}{2}$, about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If $E_f$ and $E_i$ are the final and initial total energies, then $E_f-E_i$ is

• A. $-\frac{I_1{\omega }_1^2}{12}$
• B. $\frac{I_1{\omega }_1^2}{6}$
• C. $\frac{3}{8}{I}_1{\omega }_1^2$
• D. $-\frac{I_1{\omega }_1^2}{24}$

Answer 1

The correct option is D $-\frac{I_1{\omega }_1^2}{24}$

Given:
Moments of inertia of first disc $=I_1$
Moments of inertia of second disc $I_2=\frac{I_1}{2}$
Angular velocity of first disc $={\omega }_1$
Angular velocity of second disc ${\omega }_2=\frac{{\omega }_1}{2}$
As there is no external torque is acting so angular momentum should be conserved

$\Rightarrow (I_1+I_2){\omega }_c=I_1{\omega }_1+I_2{\omega }_2$

[where, ${\omega }_c=$ common angular velocity of the system, when discs are in contact]
$\Rightarrow {\omega }_c=\frac{I_1{\omega }_1+(\frac{I_1}{2}\times \frac{{\omega }_1}{2})}{(I_1+\frac{I_1}{2})}$

$\Rightarrow {\omega }_c=(\frac{5}{4}\times \frac{2}{3}){\omega }_1$

$\Rightarrow {\omega }_c=\frac{5{\omega }_1}{6}$

Now differenc in the energies,

$E_f-E_i=\frac{1}{2}(I_1+I_2){\omega }_c^2-\frac{1}{2}{I}_1{\omega }_1^2-\frac{1}{2}{I}_2{\omega }_2^2$

Put $I_2=\frac{I_1}{2}$ and ${\omega }_c=\frac{5{\omega }_1}{6}$
We get
$E_f-E_i=-\frac{I_1{\omega }_1^2}{24}$

Hence option $\left(D\right)$ is correct.