Recent content by jabgaunt92

Yes I originally reversed the angles as you said. This has cleared up my problem now. Thank you for the help.

This is what I originally did and got: V^2= L^2 ∅(dot)^2 +R^2 θ(dot)^2 +2LR∅(dot)θ(dot)cos(∅-θ) However, I was doubtful to this as when ∅=θ should the result not reduce to 1/2 I ∅(dot)^2? Also, is the moment of inertia about the COM or at the point at which it is pivoted? Thanks

Well, I am sure that this system is analogous to a double pendulum. So I will use the generalised co-ordinates θ,∅. θ: angle the rod makes with the attached point on the disk. ∅:angle the disk makes from the attached point to its COM. Now, I do not think we use T=1/2 M v^2 as this does not...

Homework Statement The pendulum of a grandfather clock consists of a thin rod of length L (and negligible mass) attached at its upper end to a fixed point, and attached at its lower end to a point on the edge of a uniform disk of radius R, mass M, and negligible thickness. The disk is free...