JEE Main 2024 · Physics · Rotational Motion · medium

A uniform disc A of mass 2 kg and radius 0.2 m spins freely about its vertical axis at 60 rad s^-1. It is gently lowered coaxially on to a stationary…

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A uniform disc

A

of mass

2\ \mathrm{kg}

and radius

0.2\ \mathrm{m}

spins freely about its vertical axis at

60\ \mathrm{rad\,s^{-1}}

. It is gently lowered coaxially on to a stationary disc

B

of mass

6\ \mathrm{kg}

and the same radius, and the two discs finally rotate together because of friction between them. The kinetic energy lost to friction is

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Answer: (A) $54\ \mathrm{J}$

Moments of inertia:

I_A=\tfrac12(2)(0.2)^2=0.04

I_B=\tfrac12(6)(0.2)^2=0.12\ \mathrm{kg\,m^2}

. No external torque, so

I_A\omega=(I_A+I_B)\omega'

and

\omega'=\dfrac{0.04\times60}{0.16}=15\ \mathrm{rad\,s^{-1}}

. Initial energy

\tfrac12(0.04)(60)^2=72\ \mathrm{J}

; final

\tfrac12(0.16)(15)^2=18\ \mathrm{J}

. Energy lost

=72-18=54\ \mathrm{J}

(equivalently

\tfrac{I_B}{I_A+I_B}=75\%

of the initial energy). The answer

18

J is the final energy, not the loss.

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