- A cone of height h and base radius R is free to rotate around a fixed vertical axis. It has a thin groove cut in its surface. The cone is set rotating freely with angular speed ω0, and a small block of massm is released in the top of the frictionless groove and allowed to slide under gravity. Assume that the block stays in the groove. Take the moment of inertia of the cone around the vertical axis to beI0.
h is the height of the cone.
(a) What is the angular speed of the cone when the block reaches the bottom?
(b) Find the speed of the block in inertial space when it reaches the bottom.
Li = Lf
The Attempt at a Solution
OK on this question for part a I understand that angular momentum is conserved
Li = I0*w0
the final angular momentum should it not be that of the cone plus the marble as the marble is moving downwards with a velocity v which can be calculated from mgh=1/2 m v2 and so
Lf = (I0 + mR2)*wf + m*R* √(2*g*h)*h/√(h2+R2)
the solution does not take into account the speed of the marble at the bottom IE: m*R* √(2*g*h)*h/√(h2+R2)
Last part is sinθ = h/√(h2+R2)
The problems for this chapter are really subtle and confusing me..