A solid cylinder of mass M and radius R rolls | vedclass

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(C) By the law of conservation of energy,the potential energy at the top is equal to the total kinetic energy at the bottom.$PE = Mgh$$KE = \frac{1}{2

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$\frac{2}{R} \sqrt{\frac{g h}{3}}$

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(C) By the law of conservation of energy,the potential energy at the top is equal to the total kinetic energy at the bottom.$PE = Mgh$$KE = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the condition for pure rolling is $v = R\omega$.Substituting these into the kinetic energy equation:$KE = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$Equating potential energy and kinetic energy:$Mgh = \frac{3}{4} Mv^2$$v^2 = \frac{4gh}{3} \implies v = 2 \sqrt{\frac{gh}{3}}$Since $\omega = \frac{v}{R}$,the angular velocity is:$\omega = \frac{2}{R} \sqrt{\frac{gh}{3}}$

$A$ solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height $h$. The angular velocity of the cylinder when it reaches the bottom of the plane will be

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