# 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

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

ehild

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$^{2}$

ehild
Homework Helper
What do you get for the frequency?

ehild

What do you get for the frequency?

ehild

I'm getting 0.1465 for the frequency

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

ehild

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

ehild

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.

ehild

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?

ehild
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.

ehild

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