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AÂ rod of length L is hinged from one end.It is brougt to horizontal position and then released. the angular velocity of the rod when it is in vertical positon
wrt initial position of rod , when roD is verticle then center of mass of rod has fallen a height of L/2..... now , applying conservation of energy change in (kE) = difference in potential energy IW2/2 = mgdh (mL2/3)(W2/2) = mg(L/2) W =(3g/L)1/2 Irod = mL2/3 , dh = L/2
wrt initial position of rod , when roD is verticle then center of mass of rod has fallen a height of L/2.....
now , applying conservation of energy
change in (kE) = difference in potential energy
IW2/2 = mgdh
(mL2/3)(W2/2) = mg(L/2)
W =(3g/L)1/2
Irod = mL2/3 , dh = L/2
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