- #1
KleZMeR
- 127
- 1
Hi All,
This is from a classical mechanics problem, and I already 'solved' the problem, but I'm interested in why a certain term is set to zero. I think I understand the concept but just want to clarify.
The problem is a table with a hole in it and two masses on a string, one mass is hanging through the hole with only a Z component, and the other is on the table with an X and Y component (Z plane).
When I take the cross product of R x mV, I get the angular momentum vector K which has only a 'vertical' component:
R x mV = [m*(r^2)*dθ + m*r*dr*sin(2θ)] K
But I am told that:
R x mV = m*(r^2)*dθ K
The sin(2θ) came from some trig identity work. So I am wondering is this because there is no effect on the K vector from a sin(2θ) factor which is only in the Z plane? Why is this term 0? Is dr = 0 ? I think r is fixed but the problem does say that gravity affects the hanging mass, so perhaps dr in the Z plane is not zero? Any help understanding this is appreciated.
This is from a classical mechanics problem, and I already 'solved' the problem, but I'm interested in why a certain term is set to zero. I think I understand the concept but just want to clarify.
The problem is a table with a hole in it and two masses on a string, one mass is hanging through the hole with only a Z component, and the other is on the table with an X and Y component (Z plane).
When I take the cross product of R x mV, I get the angular momentum vector K which has only a 'vertical' component:
R x mV = [m*(r^2)*dθ + m*r*dr*sin(2θ)] K
But I am told that:
R x mV = m*(r^2)*dθ K
The sin(2θ) came from some trig identity work. So I am wondering is this because there is no effect on the K vector from a sin(2θ) factor which is only in the Z plane? Why is this term 0? Is dr = 0 ? I think r is fixed but the problem does say that gravity affects the hanging mass, so perhaps dr in the Z plane is not zero? Any help understanding this is appreciated.