Chetan Mandayam Nayakar
Last Activity: 14 Years ago
It can be shown(in case I am required to show, please ask me) that the moment of inertia about the axis of rotation is
(7/12)mR2, m is mass of cube and R is length of its side. Distance of centre of mass of cube from axis of
rotation =(R/√2)sin((π/4)+θ) from conservation of angular momentum, we get (7/12)mR2w = (mvR/√2)sin((π/4)+θ)
w = angular velocity = √2(6v/7R)sin((π/4)+θ)