# Uniform Circular Motion

• September 16th 2008, 09:13 PM
Elite_Guard89
Uniform Circular Motion
A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t(1)=4.00s, it is at point (5.00m, 6.00m) with a velocity (3.00 m/s)j and acceleration in the position x direction. At time t(2) = 10.0s, it has a velocity of (-3.00 m/s)i and acceleration in the positive y direction. What are the x and y coordinates of the center of the circular path if t(2)-t(1) is less than one period?

I have tried a bunch of stuff on this problem and I keep getting the wrong answer....help please.
• September 17th 2008, 11:58 AM
icemanfan
Quote:

Originally Posted by topsquark
What about r? We are stuck here. I cannot find a way to find r without more information.

-Dan

From the first statement you know that k = 6, because the particle at (5,6) is directly to the left of the center of the circle. Also, the equations should be:

$x = h + r \sin(\omega{t} + b)$

$y = 6 + r \cos(\omega{t} + b)$

The trig functions are reversed because the particle is moving clockwise.

Furthermore, $6\omega = \frac{3}{4} \cdot 2\pi$.
• September 17th 2008, 01:11 PM
Ima agree with iceman here, I think you solved for the angular velocity wrong top.

I think maybe we can use a relation relating the constant velocity around the circular track to the distance traveled to find r

we know velocity of a particle in uniform circular motion, v = omega*r... we have omega, we have v... vooooooila make sure your units are rad/s
• September 17th 2008, 01:32 PM
icemanfan
I would suggest this method of solving the problem:

$\frac{dx}{dt} = \frac{\pi}{4}r\cos{(\frac{\pi}{4}t + b)}$

$\frac{dy}{dt} = -\frac{\pi}{4}r\sin{(\frac{\pi}{4}t + b)}$

$3 = \sqrt{\frac{\pi^2r^2}{16}} = \frac{\pi{r}}{4}$

$12 = \pi{r}$

$\frac{12}{\pi} = r$
• September 17th 2008, 02:35 PM
iceman,

I'm not sure you can make the assumption that this person is experienced w/ calculus.... it may or may not be the case

To Elite, if you can follow the calculus...

Just note that be broke the velocity vector into x and y components (dx/dt) and (dy/dt). Same way you solve for the magnitude of any right triangle you would take sqrt(dx/dt^2 + dy/dt^2) sorry for not using the script.

Assuming you don't know calculus, I would

1) Solve for omega = (change in angle)/(change in time) = 3pi/2 / 6 = pi/4 rad/s

2) Note that v = omega*r => r = v/omega = 3/pi/4

Same solution, different method depending on your background
• September 17th 2008, 02:37 PM
topsquark
Thanks guys. Don't know where my head was at. (Doh)

-Dan
• September 18th 2008, 03:18 PM
Elite_Guard89
Quote:

Originally Posted by icemanfan
From the first statement you know that k = 6, because the particle at (5,6) is directly to the left of the center of the circle. Also, the equations should be:

$x = h + r \sin(\omega{t} + b)$

$y = 6 + r \cos(\omega{t} + b)$

The trig functions are reversed because the particle is moving clockwise.

Furthermore, $6\omega = \frac{3}{4} \cdot 2\pi$.

Where did you guys get these equations exactly?
• September 18th 2008, 03:30 PM
icemanfan
Quote:

Originally Posted by Elite_Guard89
Where did you guys get these equations exactly?

The equations for a particle in circular motion about the center (h, k) are either going to be

$x = h + sin(\omega{t} + b)$
$y = k + cos(\omega{t} + b)$
if the particle is moving clockwise

or

$x = h + cos(\omega{t} + b)$
$y = k + sin(\omega{t} + b)$
if the particle is moving counterclockwise.

We know from the direction of acceleration, the given fact that between t = 4 and t = 10 we do not have a full cycle, and the directions of velocity at t = 4 and t = 10 that we have moved 3/4 of a revolution clockwise. In radians, a full revolution is $2\pi$ radians, so 3/4 of a revolution is $\frac{3}{4} \cdot 2\pi$ radians. This is equal to the total change in angle, which is $\omega(t_2 - t_1)$ = (rate of change in angle per time)x(time taken), or in this case $\omega(10 - 4) = 6\omega$