Determining possible trajectories (Classical Mechanics)

AI Thread Summary
The discussion focuses on determining the possible trajectories of a particle with constant velocity and angular momentum. It is established that since the angular momentum is constant, the net torque and force acting on the particle must be zero. This leads to the conclusion that the particle can move in a linear path, but the constant speed suggests circular trajectories are likely. Participants emphasize the importance of using polar coordinates for analysis and the need to differentiate angular momentum to gain further insights. The conversation highlights the interplay between velocity, angular momentum, and the resulting motion of the particle.
MrCreamer
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Homework Statement



Determine possible trajectories for particle with constant magnitude of velocity |\dot{\vec{r}}| = v0 and constant angular momentum \vec{L} = \vec{L}0

Homework Equations



|\dot{\vec{r}}| = v0
\vec{L} = \vec{L}0

The Attempt at a Solution



I know that L dot is zero and thereby the torque is zero. My intuition tells me that the possibly trajectories would be circles but mathematically, I am not sure where to start.
 
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Look at he definition of L. Perhaps the time derivative of that tells you something
 
Well, as I said, if \dot{\vec{L}} = 0, then the net torque and net force are zero. The constant velocity magnitude implies that the direction of the velocity vector can change but the speed would remain constant.

If F is zero, then you have

m\ddot{x} = 0

Which implies that x(t) is of some linear form:

x(t) = at + b, where a = v_{0} and b = x_{0}.

I'm assuming the question requires the graphing of the trajectories in phase space and hence would require some form of mathematical development in terms of the energy of the system.
 
I think BvU meant for you to differentiate ##\vec{L} = \vec{r}\times\vec{F}## and to interpret the result.

By the way, your inference that ##\dot{\vec{L}} = 0## implies ##\vec{F}=0## is not correct.
 
I meant: differentiate ##\vec L = \vec r \times \vec p## wrt time, knowing ##\vec p = m\dot{\vec r}##. Don't venture into x and y because you have everything you need in polar coordinates.
 
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