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One end of a long 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 with a uniform angular velocity w. The force exerted by the clamp on the rod has a horizontal component.

\[\begin {array} {1 1} (a)\;mw^2l & \quad (b)\;zero \\ (c)\;mg & \quad  (d)\;\frac{1}{2}mw^2l \end {array}\]
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Horizontal component Force= Total centrifugal force acting on the rod.
Consider an element at a distance x of width dx.
$dm=\large\frac{M}{L}\;$$dx$
centrifugal force $dF=\large\frac{dmv^2}{x}$
$\qquad= \large\frac{dmx^2 w^2}{x}$
$\qquad= \large\frac{M}{l}$$ dx.xw^2$
$\therefore$ Total force $= \int \limits^l_0 \large\frac{M}{l} w^2 x dx =\large\frac{1}{2} $$Mw^2l$
answered Nov 27, 2013 by meena.p
edited Jun 21, 2014 by lmohan717
 

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