is that supposed to be ½(\dot{x}^2+\dot{y}^2)? which would equal some constant.
And I'm still unsure as to what that then tells me. I am really struggling to understand where this is headed. I'm assuming in the end I want some equation for r(x,y;t) Sorry for being so difficult lol
so i took the derivative of x'y-y'x and set it equal to zero. (no change in angular momentum) and got x''y=y''x.
and the derivative of sqrt(x'^2+y'^2) and got ((x'x''+y'y'')/(sqrt(x'^2+y^2))=0
the problem is i don't really know what I'm looking for. I am not sure what the equation can tell me...
Magnitude of velocity is Sqrt(\dot{x}^2+\dot{y}^2)
the magnitude of the angular momentum would be (\dot{x}y- \dot{y}x)m
I got this just by taking the cross product of the velocity and position vectors
v={\dot{x},\dot{y},0}
r={x,y,0}
so angular momentum would not be constant in the helix.
If angular momentum is zero then the position and velocity vectors would either have to be parallel (or antiparallel) or one of them would have to be zero. doesn't make sense that a moving particles position or velocity vectors would be...
means that velocity vector and position vector always have to lie on the same 2 diminutional plane. which would mean angular momentum isn't constant in a helix.
hmm. well the parallelogram that the vectors of position and velocity make would have to remain a constant area...
if R*M*V*sin(theta) remains a constant that angular momentum is conserved. but wouldn't that mean that any of those 4 values could change if one of the other changes inversely to...
Well it would mean that the radius of the curve would either have to stay constant, or the velocity would need to change with it. So that combined with the constant velocity stipulation would mean that the particle would have to keep a constant radius from some origin.
But I don't this isn't...
I am thinking that implies that there is no torque, which would mean that the angular velocity remains constant? so that would mean that there either needs to me no external force or a force that is always in the direction of of the radius.
Homework Statement
A particle moves along a trajectory with constant magnitude of the velocity |\stackrel{→}{v}|=\stackrel{→}{v0} and constant angular momentum L⃗ = L⃗0. Determine the possible trajectories.
Homework Equations
d(L⃗)/(dt)=\stackrel{→}{N} where \stackrel{→}{N}=torque...