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A uniform rod of length l is held vertically on a horizontal floor fixing its lower end, the rod is allowed to fall onto the ground. Find (i) its angular velocity at that instant of reaching the ground (ii) The linear velocity with which the tip of rod falls to floor. |
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Answer» SOLUTION :(i) The rod rotates about an axis through one END. From the principle of conservation of mechanical energy. Loss of P.E. of the rod is EQUAL to its gain in rotational K.E. `therefore mg.(1)/(2)=(1)/(2) I omega^(2) RARR mg.(1)/(2).=(1)/(2).(ML^(2))/(3)omega^(2)` on solving `omega = sqrt((3g)/(l))` (ii) `V = r omega` or `V = l omega = l sqrt(3g//l)=sqrt(3gl)`
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