Polar Coordinates

**bobak** · May 6th 2008, 09:15 AM

A particle P, moving in a plane, has polar coordinates $(r , \theta )$ referred to the fixed pole O and the initial line in the plane

Given that the particle moves under a force directed towards towards O, show that at time t $r^2 \frac{d \theta}{dt} = h$ .

Where h is a constant

I keep going around in circles with this. a little hint would be nice.

Many Thanks

Bobak

**topsquark** · May 6th 2008, 12:58 PM

Originally Posted by bobak

I keep going around in circles with this. a little hint would be nice.

Many Thanks

Bobak

Consider the equations of motion:
$m \frac{d^2x}{dt^2} = a_x$
and
$m \frac{d^2y}{dt^2} = a_y$

What do these equations look like in polar coordinates?
(Hint: The solution is
$m \left [ \frac{d^2r}{dt^2} - r \left ( \frac{d \theta}{dt} \right )^2 \right ] = f_r$
and
$m \left [ 2\frac{dr}{dt} \frac{d \theta}{dt} + r \frac{d^2 \theta}{dt^2} \right ] = 0$
Make sure you know how to derive these.)

Or alternately if you are in an advanced enough class, do this using Lagrangian methods. It's easier.

-Dan

**bobak** · May 6th 2008, 01:52 PM

Originally Posted by topsquark

Consider the equations of motion:
$m \frac{d^2x}{dt^2} = a_x$
and
$m \frac{d^2y}{dt^2} = a_y$

As the force is directed towards the origin that does that mean that $a_x \propto -x$ and similarly for $a_y$ or not ?

What do these equations look like in polar coordinates?
(Hint: The solution is
$m \left [ \frac{d^2r}{dt^2} - r \left ( \frac{d \theta}{dt} \right )^2 \right ] = f_r$
and
$m \left [ 2\frac{dr}{dt} \frac{d \theta}{dt} + r \frac{d^2 \theta}{dt^2} \right ] = 0$
Make sure you know how to derive these.)

This is all new to me, I can't find anything about this in my books, would you mind pointing me to some reading material so I can learn how these results are derived?

Many Thanks Dan

Bobak

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