A disc of mass $m_{1}$ is freely rotating with constant angular speed $ω$ . Another disc of mass $m_{2}$ & same radius is gently kept on the first disc. If the surfaces in contact are rough, the fractional decrease in kinetic energy of the system will be

\frac{m_{1}}{m_{2}}

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\frac{m_{2}}{m_{1} + m_{2}}

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\frac{m_{2}}{m_{1}}

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\frac{m_{1}}{m_{1} + m_{2}}

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Solution

The correct option is B $\frac{m_{2}}{m_{1} + m_{2}}$
As no external torque, so angular momentum is conserved.
$m_{1} ω = (m_{1} + m_{2}) ω_{f}$
$ω_{f} = (\frac{m_{1}}{m_{1} + m_{2}}) ω_{0}$
$△ K E = K_{i} - K_{f}$
$= \frac{1}{2} m_{1} r_{1}^{2} (ω_{0})^{2} - \frac{1}{2} (m_{1} + m_{2}) r_{1}^{2} [(\frac{m_{1}}{m_{1} + m_{2}}) ω]^{2}$
$= \frac{1}{2} r_{1}^{2} [m_{1} ω^{2} - \frac{m_{1}^{2} ω^{2}}{m_{1} + m_{2}}]$
$= \frac{1}{2} r^{2} [\frac{m_{1}^{2} ω^{2} + m_{2} ω^{2} - m_{1}^{2} ω^{2}}{m_{1} + m_{2}}]$
$△ K E = (\frac{m_{2}}{m_{1} + m_{2}}) \times \frac{1}{2} ω^{2} r^{2}$
Fractional Decrease $= \frac{m_{2}}{m_{1} + m_{2}}$

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A disc of mass m1 is freely rotating with constant angular speed ω. Another disc of mass m2 & same radius is gently kept on the first disc. If the surfaces in contact are rough, the fractional decrease in kinetic energy of the system will be

The correct option is B m2m1+m2As no external torque, so angular momentum is conserved.m1ω=(m1+m2)ωfωf=(m1m1+m2)ω0△KE=Ki−Kf=12m1r21(ω0)2−12(m1+m2)r21[(m1m1+m2)ω]2=12r21[m1ω2−m21ω2m1+m2]=12r2[m21ω2+m2ω2−m21ω2m1+m2]△KE=(m2m1+m2)×12ω2r2Fractional Decrease=m2m1+m2

A disc of mass $m_{1}$ is freely rotating with constant angular speed $ω$ . Another disc of mass $m_{2}$ & same radius is gently kept on the first disc. If the surfaces in contact are rough, the fractional decrease in kinetic energy of the system will be