A uniform bar of length L and mass M hangs under the influence of gravity from a

Question

A uniform bar of length L and mass M hangs under the influence of gravity from a frictionless pivot. At t=0 the bar is given a horizontal instantaneous impulse J at its lower end. (a) What is the reaction impulse at the pivot at t=0? (b) What is the angular velocity about the center of mass immediately after the impulse is applied? (c) To what height above its initial position will the center of mass rise? Please use word with your equations. I need an explanation of the explanation reaction impulse=j j=m*(l/2)*w w=2j/ml 0.5*I*w^2=m*g*h I=m*l^2/3 h=(2/3)*j^2/m^2g

Explanation / Answer

eaction impulse=j because the net impulse balances at the pivot, more or less as newtons 3rd law. j=m*(l/2)*w ,as Centre of mass is at l/2. and impulse causes change in momentum. 0.5*I*w^2=m*g*h ; loss in KE = gain in PE I=m*(l^2)/3; moment of inertia about pivot. h=(2/3)*j^2/m^2g.

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