A uniform thin rod of length 8m is released from rest from the position shown. The acceleration of point at x = Image not present m just after release is

Dear Student,Taking Moment about the hinge of the rod , mg× l2   =(M.I)×α , ( Where (M. I)= ml23   (Moment of Inertia))[ m=mass of rod, l= length of rod =8m, g = gravity, α = angular acceleration] α= 3g2l=3×102×8=158Acceleration at point x = ax = xα =15x8 = 83(Given)Thus, x = 6445mRegards

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