noel thekkekara
Last Activity: 13 Years ago
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!!!!!!!!