Rotational Motion

JEE Main level, numerical answer, medium

279 of 280

Question 279 of 280

Two discs of moment of inertia I1=4kgm2{I}_{1}=4kg{m}^{2} and I2=2kgm2{I}_{2}=2kg{m}^{2} about their central axes & normal to their planes, rotating with angular speeds 10rad{s}^{-1}&4rad{s}^{-1} respectively are brought into contact face to face with their axe of rotation coincident. The loss in kinetic energy of the system in the process is _________JJ.

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Answer: 24

Applying conservation of angular momentum, we get
I1ω1+I2ω2=(I1+I2)ω0⇒4×10+2×4=(4+2)ω0⇒ω0=8rads−1{I}_{1}{\omega }_{1}+{I}_{2}{\omega }_{2}=\left({I}_{1}+{I}_{2}\right){\omega }_{0} \Rightarrow 4\times 10+2\times 4=\left(4+2\right){\omega }_{0} \Rightarrow {\omega }_{0}=8rad{s}^{-1}
Now, initial rotational kinetic energy is E1=12I1ω12+12I2ω22=216J{E}_{1}=\dfrac{1}{2}{I}_{1}{\omega }_{1}^{2}+\dfrac{1}{2}{I}_{2}{\omega }_{2}^{2}=216J
and final rotational kinetic energy is
E2=12(I1+I2)ω02=192J{E}_{2}=\dfrac{1}{2}\left({I}_{1}+{I}_{2}\right){\omega }_{0}^{2}=192J
Required, energy loss ΔE=E1−E2=24J\Delta E={E}_{1}-{E}_{2}=24J

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