Centripetal Acceleration of amusement ride

[hsphysics2]

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]

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

ehild

 

[hsphysics2]

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]

What do you get for the frequency?

ehild

 

[hsphysics2]

What do you get for the frequency?

ehild

I'm getting 0.1465 for the frequency

 

[ehild]

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

 

[hsphysics2]

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]

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

ehild

 

[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?

 

[ehild]

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