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(C) As the body rolls down the inclined plane,it loses potential energy. In rolling,it acquires both linear and angular speeds,and hence,gains kinetic

Vedclass · Accepted Answer

$\sqrt{\frac{2gh}{\beta}}$

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(C) As the body rolls down the inclined plane,it loses potential energy. In rolling,it acquires both linear and angular speeds,and hence,gains kinetic energy of translation and rotation. By the law of conservation of mechanical energy:$Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} I \omega^2$Since the body rolls without slipping,$v = R\omega$,which implies $\omega = \frac{v}{R}$.Substituting this into the energy equation:$Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} I \left(\frac{v}{R}\right)^2$$Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} \frac{I}{R^2} v^2$$Mgh = \frac{1}{2} Mv^2 \left(1 + \frac{I}{MR^2}\right)$Given $\beta = 1 + \frac{I}{MR^2}$,we have:$Mgh = \frac{1}{2} Mv^2 \beta$$gh = \frac{1}{2} v^2 \beta$$v^2 = \frac{2gh}{\beta}$$v = \sqrt{\frac{2gh}{\beta}}$

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