DC Pandey Mechanics Part 2
Q. A uniform bar of length l stands vertically touching a wall OA, when slightly displaced, its lower end begins to slide along the floor. obtain an expression for the angular velocity 
of the bar as a function of 
. neglect friction everywhere.
[ D C PANDEY page 27, introductory exercise 9.5, problem 2 ]
the point of zero velocity lies at a distance of l/2 from the centre of mass of the rod.
moment of inertia of the rod about that point i=(ml^2/12)+m(l/2)^2=(ml^2)/3
we know that
decrease inpotential energy=increase in kinetic energy
(mgl/2)(1-cos )=I^2
(mgl/2)(1-cos )=(ml^2)/3 * ^2
(3g/2l)(1-cos)=^2
hence =sqrt{3g(1-cos)/2l}
please approve my answer!!!!!!!!
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