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A cylinderical 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 horizontally 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 accelerations of the cylinder as it falls. |
Answer» Solution :The rod rotates and TRANSLATES while falling down.![]() `MG-2T=Ma` (translational motion) `2TR = I alpha` (Rotation motion) SINCE `I=(MR^(2))/(2)` and `alpha = (a)/(R )` on solving the aboveequation we get `u=(2g)/(3)` and `T=(Mg)/(6)` |
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