NEET-XII-Physics
10: Rotational Mechanics
- #9One 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
ω. The force exerted by the clamp on the rod has a horizontal component
(a)
mω2l
(b) zero
(c) mg
(d)
12mω2t.digAnsr: dAns : (d) `` \frac{1}{2}m{\omega }^{2}l``.
Consider a small portion of rod at a distance x from the clamped end (as shown in fig.) with width dx and mass dm.
Centripetal force on this portion =`` {\omega }^{2}xdm``
`` dm=\left(\frac{m}{l}\right)ldx``
Force on the whole rod = F =`` {\int }_{0}^{l}{\omega }^{2}x\frac{m}{l}dx``
`` \therefore `` F = `` \frac{1}{2}m{\omega }^{2}l``
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