Rotational Motion

JEE Main level, single correct, medium

210 of 280

Question 210 of 280

Two discs have moments of intertia I1{I}_{1} and I2{I}_{2} about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, ω1{\omega }_{1} and ω2{\omega }_{2} respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:

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Answer: (C) I1I22(I1+I2)(ω1−ω2)2\dfrac{{I}_{1}{I}_{2}}{2\left({I}_{1}+{I}_{2}\right)}{\left({\omega }_{1}-{\omega }_{2}\right)}^{2}

Angular momentum conservation
I1ω1+I2ω2=(I1+I2)ω{I}_{1}{\omega }_{1}+{I}_{2}{\omega }_{2}=\left({I}_{1}+{I}_{2}\right)\omega
ω=I1ω1+I2ω2I1+I2\omega =\dfrac{{I}_{1}{\omega }_{1}+{I}_{2}{\omega }_{2}}{{I}_{1}+{I}_{2}}
Loss
=12I1ω12+12I2ω22−12(I1+I2)ω2=\dfrac{1}{2}{I}_{1}{\omega }_{1}^{2}+\dfrac{1}{2}{I}_{2}{\omega }_{2}^{2}-\dfrac{1}{2}\left({I}_{1}+{I}_{2}\right){\omega }^{2}
=12I1ω12+12I2ω22−12(I1+I2)(I1ω1+I2ω2I1+I2)2=\dfrac{1}{2}{I}_{1}{\omega }_{1}^{2}+\dfrac{1}{2}{I}_{2}{\omega }_{2}^{2}-\dfrac{1}{2}\left({I}_{1}+{I}_{2}\right){\left(\dfrac{{I}_{1}{\omega }_{1}+{I}_{2}{\omega }_{2}}{{I}_{1}+{I}_{2}}\right)}^{2}
=12I1I2(I1+I2)(ω1−ω2)2=\dfrac{1}{2}\dfrac{{I}_{1}{I}_{2}}{\left({I}_{1}+{I}_{2}\right)}{\left({\omega }_{1}-{\omega }_{2}\right)}^{2}
Ei−Ef=I1I2(ω1−ω2)22(I1+I2){E}_{i}-{E}_{f}=\dfrac{{I}_{1}{I}_{2}{\left({\omega }_{1}-{\omega }_{2}\right)}^{2}}{2\left({I}_{1}+{I}_{2}\right)}

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