# Homework Help: Center of circular path?

1. Sep 12, 2004

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (7.00 m, 7.00 m) with a velocity of -5.00 i m/s and an acceleration of -12.0 j m/s2. What are the coordinates of the center of the circular path?

I figured you'd have to find the radius and then it would be a simple problem but I was wrong. How is this done.

2. Sep 12, 2004

### Janitor

Hints:

The center will be on a line perpendicular to the velocity. You are given that the velocity is in the -x direction, so the x coordinate of the center will be the same as the x coordinate of the particle.

To get the y coordinate of the center, note that the acceleration is in the -y direction, so the center will have a y coordinate less than the y coordinate of the particle. How much less than? Use the formula relating radius and speed and acceleration for uniform circular motion.

3. Sep 12, 2004

### e(ho0n3

Explain how "a particle moves horizontally in uniform circular motion" as well as "a horizontal xy plane". You could just say "A particle is moving in uniform circular motion on the xy-plane...". Anywho, finding the radius is the key to solving this problem. What exactly are you having trouble with?

4. Sep 12, 2004

For the y coordinate I get -2 and that is incorrect. I'm using a=v^2/r using a=-15 and v=-5. What am I doing wrong.

5. Sep 12, 2004

### 0aNoMaLi7

Again....i have the same problem as you for my class 'GingerBread27'....I know the people here are keen on YOU solving the problem...NOT someone else but I just figured it out and figured I'd give you a heads up if you hadn't yet.

This is a problem in visualization. With the help of "Janitor" and about 15 minutes of staring I was able to nail this problem. I've provided a diagram to get you on your way. It should help. Like "Janitor" suggested use the

a = (v^2)/ r ---- you have all the variables you need.....

good luck

here's the diagram to assist (below):

#### Attached Files:

• ###### uam_prob.jpg
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6. Sep 12, 2004

What in the world am I not seeing. I know the y-coordinate must be less than 7 but when I plug in the numbers and all I get -2 and it's not correct.

7. Sep 12, 2004

### 0aNoMaLi7

oh im sorry....the coordinate in my problem is (5,5) i also have different 'a' and 'v' values....one moment...ill spell it out for you

8. Sep 12, 2004

### 0aNoMaLi7

the end to your frustrations:

your coordinate is (7,7)
your velocity vector is -5
your acceleration vector is -12

using the relationship a = (v^2)/r
REARRANGE: r = (v^2)/a

lets call r 'VALUE'

the x-coordinate we know is 7

therefore, subtract 'VALUE' from 7 for y .......

now?

9. Sep 12, 2004

### 0aNoMaLi7

yes? no? maybe?

10. Sep 12, 2004