A solid sphere of mass M and radius 2 R rolls down an inclined plane of height h without slipping. The speed of its centre of mass when it reaches the bottom is

$\sqrt{\frac{6}{7} gh}$

$\sqrt{3 gh}$

$\sqrt{\frac{10}{7} gh}$

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

Solution

$\sqrt{\frac{10}{7} gh}$

When solid sphere rolls down on an inclined plane, then it has both rotational and translational kinetic energy

K = K_rot + K_trans

K = $\frac{1}{2} {Iω}^{2} + \frac{1}{2} M v^{2}$

where,

I= moment of inertia of solid sphere = $\frac{2}{5}$ MR²

∴ K = $\frac{1}{2} (\frac{2}{5} M {(2 R)}^{2}) ω^{2} + \frac{1}{2} {Mv}^{2}$ [ $∵$ R =2R ]

= $\frac{4}{5} {MR}^{2} (\frac{v}{2 R}) + \frac{1}{2} {Mv}^{2}$ [ $∵$ v = Rω and R = 2R ]

K = $\frac{1}{5} {Mv}^{2} + \frac{1}{2} {Mv}^{2}$

K = $\frac{7}{10}$ Mv²

Now, gain in KE= loss in PE

$\frac{7}{10}$ Mv² = Mgh

⇒ v = $\sqrt{\frac{10}{7} gh}$

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