Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres.

Question:

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are $0.1 \mathrm{~kg}-\mathrm{m}^{2}$ and $10 \mathrm{rad} \mathrm{s}^{-1}$ respectively while those for the second one are $0.2 \mathrm{~kg}-\mathrm{m}^{2}$ and $5 \mathrm{rad} \mathrm{s}^{-1}$ respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is :

  1. $\frac{10}{3} \mathrm{~J}$

  2. $\frac{20}{3} \mathrm{~J}$

  3. $\frac{5}{3} \mathrm{~J}$

  4. $\frac{2}{3} \mathrm{~J}$


Correct Option: , 2

Solution:

(2) Initial angular momentum $=I_{1} \omega_{1}+I_{2} \omega_{2}$

Let $\omega$ be angular speed of the combined system.

Final angular momentum $=I_{1} \omega+I_{2} \omega$

According to conservation of angular momentum

$\left(I_{1}+I_{2}\right) \omega=I_{1} \omega_{1}+I_{2} \omega_{2}$

$\Rightarrow \omega=\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}=\frac{0.1 \times 10+0.2 \times 5}{0.1+0.2}=\frac{20}{3}$

Final rotational kinetic energy

$K_{f}=\frac{1}{2} I_{1} \omega^{2}+\frac{1}{2} I_{2} \omega^{2}=\frac{1}{2}(0.1+0.2) \times\left(\frac{20}{3}\right)^{2}$

$\Rightarrow K_{f}=\frac{20}{3} \mathrm{~J}$

Leave a comment