Questions
A uniform rod of length L (in between the supports) and mass m is placed on two supports A and B. The rod breaks suddenly at length L/10 from the support B. Find the reaction at support A immediately after the rod breaks.
detailed solution
Correct option is A
Torque =τ=910mg920L=Iα=m3910L2αα=3g2L Acceleration, aCM=α(AC) aCM=3g2L9L20=27g40 Now, 910mg−NA=maCM=m⋅27g40 or NA=940mgTalk to our academic expert!
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A uniform rood AB of length and mass is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is , the initial angular acceleration of the rod will be
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