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A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is : (Moment of inertia of rod about A is (ml2/3))
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3g/2l 2l/3g 3g/2l2 mg(l/2) |
3g/2l |
\(\tau = I \alpha\)
mg x \(\frac{l}{2}\) = \(\frac{ml^2}{3}\)(α) α = \(\frac{3g}{2l}\) |
