JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2020
- Hard
- MCQ
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 \(0.1 \;kg - m ^{2}\) and \(10\; rad \,s^{-1}\) respectively while those for the second one are \(0.2 \;kg - m ^{2}\) and \(5\; rad \,s ^{-1}\) 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 ...........\(J\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(6.67\)
Step-by-step Solution
Detailed explanation
Both discs are rotating in same sense
Angular momentum conserved for the system
i.e. \(\quad L _{1}+ L _{2}= L _{\text {final }}\)
\(I _{1} \omega_{1}+ I _{2} \omega_{2}=\left( I _{1}+ I _{2}\right) \omega_{ f }\)
\(0.1 \times 10+0.2 \times 5=(0.1+0.2) \times \omega_{f}\)
\(\omega_{ f }=\frac{20}{3}\)
Kinetic energy of combined disc system
\(\Rightarrow \frac{1}{2}\left( I _{1}+ I _{2}\right) \omega_{ f }^{2}\)
\(=\frac{1}{2}(0.1+0.2) \cdot\left(\frac{20}{3}\right)^{2}\)
\(=\frac{0.3}{2} \times \frac{400}{9}=\frac{120}{18}=\frac{20}{3} J\)
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