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A uniform. rod of length l can rotate without friction about an axis passing through its upper end(Fig). The rod is deflected by an angle a_0and let go. Find the speed of the lower end of the rod as a function of the angle alpha |
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Answer» `K=U_0-U` where `K = 1/2 Iomega^2, U_0 =mgh_0 and U = mgh` The moment of inertia of a rod about an axis passing through one end is `I = 1/3ml^2` .In the course of oscillations the centre of gravity of the rod rises to a height `h_0=1/2l(1-cosalpha_0),h=1/2l(1-cosalpha)` Substituting the values obtained into the equation for the energy balance, we obtain `omega= sqrt((3G)/l(cos alpha-cosalpha_0))` the speed of the end of the rod is `v= omega l` . |
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