Physics | VITEEE 2019 | Discussion Page

A solid cylinder of mass m and radius R rolls down the inclined plane without slipping. The speed of its C.M. when it reaches the bottom is

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$\sqrt{2gh}$

$\sqrt{\frac{4gh}{3}}$

$\sqrt{\frac{3}{4gh}}$

$\sqrt{4gh}$

Option B

By energy conservation

(K.E)_i+ (P.E)_i = (K.E)_f+ (P.E)_f

(K.E)_i =0,(P.E)_i = mgh, (P.E)_f = 0

(K.E)_f= $\frac{1}{2}I\omega^{2} + \frac{1}{2}mv^{2}$

For the solid cylinder, the moment of inertia,

I = $\frac{1}{2}mR^{2}$

So, mgh =

$\frac{1}{2}(\frac{1}{2}mR^{2})(\frac{v^{2}}{R^{2}})+\frac{1}{2}mv^{2}$

v = $\sqrt{\frac{4gh}{3}}$

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