Acceleration of a Particle in an Elliptical Path

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
LisaSpace
Messages
1
Reaction score
0

Homework Statement


Path of a particle attached to the end of a flexible bar is essentially an ellispe in the horizontal plane.

Determine the acceleration of the particle in an arbitrary position.

Homework Equations



The polar coordinates are:

R = ab/(b^2+(a^2-b^2)cos(theta)^2)^.5

theta(dot) = k/R^2

The Attempt at a Solution



I attempted to solve using the equation for acceleration in cylindrical coordinates, I'd write it out, but I'm not familiar with the code for math symbols.

I assumed z to be constant since motion was in a horizontal plane.

Is this the correct way to go about solving it? If it is then I'll will go and see where I messed up.

Thanks
 
Physics news on Phys.org
LisaSpace said:

Homework Statement


Path of a particle attached to the end of a flexible bar is essentially an ellispe in the horizontal plane.

Determine the acceleration of the particle in an arbitrary position.

Homework Equations



The polar coordinates are:

R = ab/(b^2+(a^2-b^2)cos(theta)^2)^.5

theta(dot) = k/R^2

The Attempt at a Solution



I attempted to solve using the equation for acceleration in cylindrical coordinates, I'd write it out, but I'm not familiar with the code for math symbols.

I assumed z to be constant since motion was in a horizontal plane.

Is this the correct way to go about solving it? If it is then I'll will go and see where I messed up.

Thanks

If you know the particle position, then you should be able to get the acceleration through differentiation and the chain rule. So yes, carry on and show us where you think you are messing up. You can use the buttons labeled X2 and X2 for super and sub scripts. There are also various symbols for copy and paste in my signature.