A rigid bar of mass 11 kg and length 82 m is released from r

A rigid bar of mass 1.1 kg and length 8.2 m is released from rest in the horizontal position. The rod’s angular velocity when it has rotated through 90° is Answer... rad/s

Solution

>> Let angular velocity after rotation = \"w\"

and, linear velocity = \"V\"

Also, V = rw

, where, r = L/2 = 4.1 m

>> Now, Applying Energy Conservation

Energy before rotation = Energy after rotation

NOTE :dUE TO UNCLEARITY IN FIGURE, I AM ASSUMING THAT ONLY ROTATION IS HAPPENING, THUS NOT CONSIDERING LINEAR KINETIC ENERGY IN AFTERWARDS ENERGY.

=> mgh = (1/2)Iw²

As, I = Moment of Inertia of rod = ML²/3 = 1.1*8.2*8.2/3 = 24.655 Kg-m²

=> 1.1*9.81*4.1 = 0.5*24.655*w²

=> w = 3.589 RAD/S ......REQUIRED ANGULAR VELOCITY ........