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One end of a uniform rod of mass m and length $l$ is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity . The force exerted by the clamp on the rod has a horizontal component.

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m $ω$ ²l

Zero

1/2 m $ω$ ²l

answer is D.

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Detailed Solution

In rotatory motion the angular velocity is constant for all points.

Consider a small piece of mass dm then force is dF.

$dF = (dm) x ω^{2}$

We know dm = M/L dx
$F = \int_{0}^{L} (\frac{M}{L} dx) {xω}^{2}$

$= \frac{M}{L} . ω^{2} \int_{0}^{L} xdx = \frac{M}{L} ω^{2} {(\frac{x^{2}}{2})}_{0}^{L}$

$F = \frac{1}{2} {mω}^{2} L$

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