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A block of mass $m$ is kept on a horizontal ruler. The friction coefficient between the ruler and the block is $μ$ . The ruler is fixed at one end and the block is at a distance $L$ from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. What can the maximum constant angular speed be for which the block does not slip ?

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m g

5 m g

$\sqrt{\frac{μg}{L}}$

2 g

answer is B.

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Detailed Solution

The frictional force provides the necessary centripetal force.

$∴ {mLω}^{2} = μmg$

$ω = \sqrt{\frac{μg}{L}}$

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