So I have the Ek's as equal:
Eqn1: 1/6 ML2ωF2= 1/2mωi2R2
Equation 2 is the conservation of angular momentum;
Eqn 2: mω2iR^2 = 1/3ML2ωf
Eliminating ωf I get
mR^2 = 1/3 ML^2EDIT: So final answer as R= √(1/3)M/m *L
Which is still leaving me with the length in terms of the masses, is this...
If I let the ball have a given angular velocity I can write conservation of angular momentum? I can give the Rod an angular velocity after the collision of ωf.
Kinetic energy is conserved so ½mvi2= ½Mvf2; where Where v=ωr so;
½mωi^2 R^2 = ½M ωf^2 L^2
I could use angular momentum of the ball...
Homework Statement
A uniform thin rod of Length L and mass M can freely rotate about a point 0 and is at rest in at the vertical. A ball of mass m on a light string of length R, which is also attached about the pivot is deflected by a small angle from the vertical and let go of.
If the...