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