Rotational Motion

JEE Main level, single correct, medium

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Question 222 of 280

Two coaxial discs, having moments of inertia I1{I}_{1} and I12\dfrac{{I}_{1}}{2} , are rotating with respective angular velocities ω1{\omega }_{1} and ω12\dfrac{{\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 Ef{E}_{f} and Ei{E}_{i} are the final and initial total energies, then (Ef−Ei)\left({E}_{f}-{E}_{i}\right) is:

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Answer: (D) −I1ω1224-\dfrac{{I}_{1}{\omega }_{1}^{2}}{24}

As external torque is zero, angular momentum remains conserved.
I1ω1+I12ω12=(I1+I12)ω{I}_{1}{\omega }_{1}+\dfrac{{I}_{1}}{2}\dfrac{{\omega }_{1}}{2}=\left({I}_{1}+\dfrac{{I}_{1}}{2}\right)\omega
∴∴ Common angular velocity ω=5ω16\omega =\dfrac{5{\omega }_{1}}{6}
∴∴ Ef−Ei=12×(I1+I12)ω2−12×I1ω12−12×Il2(ω12)2{E}_{f}-{E}_{i}=\dfrac{1}{2}\times \left({I}_{1}+\dfrac{{I}_{1}}{2}\right){\omega }^{2}-\dfrac{1}{2}\times {I}_{1}{\omega }_{1}{}^{2}-\dfrac{1}{2}\times \dfrac{{I}_{l}}{2}{\left(\dfrac{{\omega }_{1}}{2}\right)}^{2}
=12×3I12(56ω1)2−916I1ω12=\dfrac{1}{2}\times \dfrac{3{I}_{1}}{2}{\left(\dfrac{5}{6}{\omega }_{1}\right)}^{2}-\dfrac{9}{16}{I}_{1}{\omega }_{1}^{2}
=75I1ω12−81I1ω12144=\dfrac{75 {I}_{1}{\omega }_{1}^{2}-81 {I}_{1}{\omega }_{1}^{2}}{144}
=−6I1ω12144=-\dfrac{6{I}_{1}{\omega }_{1}^{2}}{144}
=−I1ω1224=-\dfrac{{I}_{1}{\omega }_{1}^{2}}{24}

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