Homework Help: Centripetal Acceleration of amusement ride

1. Sep 23, 2013

hsphysics2

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

a$_{c}$= 4∏$^{2}$r∫$^{2}$

3. 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

2. Sep 23, 2013

ehild

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

ehild

3. Sep 23, 2013

hsphysics2

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

4. Sep 23, 2013

ehild

What do you get for the frequency?

ehild

5. Sep 23, 2013

hsphysics2

I'm getting 0.1465 for the frequency

6. Sep 23, 2013

ehild

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

ehild

7. Sep 24, 2013

hsphysics2

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*

8. Sep 24, 2013

ehild

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

9. Sep 24, 2013

hsphysics2

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?

10. Sep 24, 2013

ehild

fg
∫$^{2}$=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