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A cylindrical rod of mass .m.length L and radius .R. has two cords wound around it whose ends are attached to a rigid support. The rod is held horozontally with the two cords vertical. When the rod is released, the cords unwind and rod rotates. Find the tension in the cords as they unwound and determine the linear acceleration of the cylinder as it falls. |
Answer» Solution :The rod ROTATES and translates while falling down. `Mg-2T=Ma` (translational motion) `2TR=I ALPHA` (Rotational motion) Since `I=(MR^(2))/(2) and alpha=(a)/(R )` on solving the above equation we get `a=(2g)/(3) and T=(mg)/(6)` |
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