Find the Centripetal Acceleration at 2.5m from a Rotating Platform

[rafay233]

Homework Statement

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?

Homework Equations

?

The Attempt at a Solution

centripetal acceleration= \frac{v^2}{r}
=5.6=\frac{v^2}{4.3}
v=4.907m/scentripetal 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.

 

[rafay233]

uhh guys i don't know why it looks like that. Only read the first three stuff please.

Thank you

 

[rafay233]

uhh guys i don't know why it looks like that. Only read the first three stuff please.

Thank you

disregard that message please. i fixed the first post

 

[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

the answer is 3.3m/s

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

 

[Genoseeker]

ω

Homework Statement

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?

Homework Equations

?

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.

the RED needs your attention.
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

 

[Genoseeker]

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

the answer is 3.3m/s

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

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

 

[rafay233]

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

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