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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.

$\begin {array} {1 1} (a)\;mw^2l & \quad (b)\;zero \\ (c)\;mg & \quad (d)\;\frac{1}{2}mw^2l \end {array}$
Can you answer this question?

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
edited Jun 21, 2014