# Homework Help: Homework Help. Rotating platform. difference of acceleration when close to center

1. Apr 11, 2012

### rafay233

1. The problem statement, all variables and given/known data
A person is on a horizontal rotating platform at a distance of 4.3 m from its centre. This preson experiences a centripetal acceleration of 56m/s^2. What is the centripetal acceleration is experienced by another person who is at a distance of 2.5 m from the centre of the platform?

2. Relevant equations

?

3. The attempt at a solution
centripetal acceleration= $\frac{v^2}{r}$
=5.6=$\frac{v^2}{4.3}$
v=4.907m/s

centripetal acceleration= $\frac{v^2}{r}$
= $\frac{4.9^2}{2.5}$
= 9.6m/s^2
I know that is not the right answer because it should be lower when closer to the center.

Last edited: Apr 11, 2012
2. Apr 11, 2012

### rafay233

uhh guys i dont know why it looks like that. Only read the first three stuff please.

Thank you

3. Apr 11, 2012

### rafay233

disregard that message please. i fixed the first post

4. Apr 11, 2012

### rafay233

Okay guys I figured this out, but don't know why this works.

4.9m/s*1/4.3m = 1.14s

1.14*2.5= 2.84m/s

(2.84m/s)^2/2.5m = 3.246 = centripetal acceleration

Btw im not doing this to bump the thread, that is if it's possible.

Last edited: Apr 11, 2012
5. Apr 11, 2012

### Genoseeker

ω
Also

the linear velocity changes based on the radius. ANGULAR (ω) velocity remains constant though.

the first Centripetal Acc is correct the second one is wrong.

formula to use v = ω * r

use the initial v and r to find ω.

then use ω and new r to find new v.

then find the correct second Centripetal Acc

Last edited: Apr 11, 2012
6. Apr 11, 2012

### Genoseeker

its called the EDIT button. please learn to use it.

7. Apr 11, 2012

### rafay233

K sorry for being a noob, I thought it would be better if I made another post instead of editing the other one.