Rotational Motion

JEE Main level, numerical answer, medium

203 of 280

Question 203 of 280

Consider a disc of mass 5kg5kg, radius 2m2m, rotating with angular velocity of 10rads−110rad{s}^{-1} about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is _________JJ.

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Answer: 250

Initial angular momentum, Li=Iωi=MR22⋅ω=100kgm2s−1{L}_{i}=I{\omega }_{i}=\dfrac{M{R}^{2}}{2}\cdot \omega =100kg{m}^{2}{s}^{-1}
Initial kinetic energy, Ei=12⋅MR22⋅ω2=500J{E}_{i}=\dfrac{1}{2}\cdot \dfrac{M{R}^{2}}{2}\cdot {\omega }^{2}=500J
From conservation of angular momentum, we can write
L⃗i=L⃗f⇒100=2Iωf{\vec{L}}_{i}={\vec{L}}_{f} \Rightarrow 100=2I{\omega }_{f}
⇒ωf=5rads−1\Rightarrow {\omega }_{f}=5rad{s}^{-1}
Ef=2×12⋅5(2)22⋅(5)2=250J{E}_{f}=2\times \dfrac{1}{2}\cdot \dfrac{5(2{)}^{2}}{2}\cdot (5{)}^{2}=250J
Therefore, required energy loss ΔE=250J\Delta E=250J

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