Right so I defined the v in 1/4 to be x2' since it is moving with the block and then then v in 1/2(m3)x3'^2 so since x2' and x1' are equivalent I replaced them in the L to be x' and then left x3' so my L=(1/2)(x')^2 (m1+m2+(1/4)m3)+(1/2)m3(x3')+m2g(l-∏a-x2)
So for the added KE of cylinder due to the ground would be (1/2)mv^2?
Also, I was wondering if I need to deal with the fact that the cylinder is on m1 so it is a distance above the table which is equivalent to the height of that m1.
Yes, I did mean T not I. I guess I am just confused about what all I have to add to the KE. I thought that the ((1/4)mv^2) was the term due to the KE associated with the movement of the centre of mass.
Also, I do not understand conceptually how the rod can have angular momentum before it rotates. I thought to calculate angular momentum the object must be rotating.
Right, so L=Ʃri x (mi)(vi) and Ʃmi=m and v is the same for all the elements and Ʃr=d (length of the stick so why is L=mvd incorrect for before the rod hits the putty?
Homework Statement
A thin vertical rod of mass m and length d slides on a frictionless horizontal surface with velocity Vo to the right. It his and sticks to a small lump of putty and rotates about the putty at point O.
(a) Find the angular momentum of the rod about O just before and just...
Homework Statement
A mass m1 slides on a frictionless horizontal table. it is attached by a massless cord passing over a massless pulley to a mass, m2. A cylinder of mass m3, radius r, and moment of inertia 1/2(m3*r^2) rests on m1.
(a) Choose and specify generalized coordinates (two are...