# Centripetal Acceleration of amusement ride

## Homework Statement

An amusement ride with a radius of 4.8m allows riders to experience 4.3g's of centripetal acceleration. What is the frequency of the ride?

## Homework Equations

a$_{c}$= 4∏$^{2}$r∫$^{2}$

## The Attempt at a Solution

a$_{c}$=4.3
r=4.8m
∫=?

I keep getting a weird answer when I rearrange that equation to solve for ∫, and sub the values in. It might be my algebra

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Homework Helper
What acceleration does 4.3 g mean in m/s2 units?

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What acceleration does 4.3 g mean in m/s2 units?

ehild
I think it's 1g=9.8m/s$^{2}$
therefore 4.3g= 42.14m/s$^{2}$

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Homework Helper
What do you get for the frequency?

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What do you get for the frequency?

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I'm getting 0.1465 for the frequency

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Homework Helper
It is not correct. How did you get it? Show your work.

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It is not correct. How did you get it? Show your work.

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a$_{c}$=4∏$^{2}$r∫$^{2}$
∫$^{2}$=a$_{c}$/4∏$^{2}$r
∫=±√(a$_{c}$/4∏$^{2}$r)
∫=±0.47157

*calculator mixup on the previous answer*

ehild
Homework Helper
You need to use parentheses. ac/4∏2r means ac/4 * ∏2 * r.
The numerical value is all right, but what is the unit? And give the final result with two significant digits.

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I think the unit would be hertz?

a$_{c}$=4∏$^{2}$r∫$^{2}$
∫$^{2}$=a$_{c}$/(4∏$^{2}$r∫$^{2}$)
∫=±√(a$_{c}$/(4∏$^{2}$r∫$^{2}$))
∫=±0.47hz

So it would be 0.47hz?

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Homework Helper
fg
I think the unit would be hertz?

a$_{c}$=4∏$^{2}$r∫$^{2}$
∫$^{2}$=a$_{c}$/(4∏$^{2}$r∫$^{2}$)
∫=±√(a$_{c}$/(4∏$^{2}$r∫$^{2}$))
∫=±0.47hz

So it would be 0.47hz?
∫$^{2}$=a$_{c}$/(4∏$^{2}$r∫$^{2}$)
You kept f2 on the right hand side...$$f^2=\frac{a_c}{4\pi^2 r}$$

Yes, it is Hz or 1/s.

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