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Physics-

General

Easy

Question

A hollow sphere is released from the top of a wedge, friction is sufficient for pure rolling of sphere on the wedge. There is no friction between the wedge and the ground. Radius of sphere is R. At the instant it leaves the wedge horizontally. Angular velocity of sphere $omega$ is

$\sqrt{\frac{12 g h}{7 R^{2}}}$
$\sqrt{\frac{27}{7} \frac{g h}{R^{2}}}$
$\sqrt{\frac{20 g h}{7 R^{2}}}$
$\sqrt{\frac{44 g h}{7 R^{2}}}$

The correct answer is: $square root of fraction numerator 12 g h over denominator 7 R to the power of 2 end exponent end fraction end root$

$omega equals fraction numerator 2 v over denominator R end fraction equals square root of fraction numerator 4 over denominator R to the power of 2 end exponent end fraction cross times fraction numerator 3 over denominator 7 end fraction g h end root equals square root of fraction numerator 12 g h over denominator 7 R to the power of 2 end exponent end fraction end root$

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