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A uniform rod of mass M = 2 kg and length L = 1.5 m is attached to a wall with a frictionless pivot and a string as shown in the diagram above. The initial angle of the rod with respect to the wall, is theta = 39ï¿½. The string is then cut. The moment of inertia of a rod about an axis through one end is 1/3ML^2.

What is the angular acceleration of the rod, a, immediately after the string is cut?

What is the angular velocity av of the rod when it is horizontal (theta=90)?

Added Fri, 07 Aug '15

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for angular acceleration of the rod

net torque=mg*l/2*cos(51) =I*(angular acc)=ml^2/3 *(angular acc)

(angular acc) =3g cos(51)/2l =6.167 rad/sec^2 (ans)

2)Energy conservation

Potential energy change=rotational kinetic energy

mgl/2*(sin51) =0.5(ml^2/3)(av)^2

(av)^2 =3g/l *sin(51) =15.232

so av=3.90 rad/sec (ans)

Added Fri, 07 Aug '15

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