# Centripetal Acceleration of amusement ride

• hsphysics2
In summary, the conversation discusses finding the frequency of an amusement ride that allows riders to experience 4.3g's of centripetal acceleration with a radius of 4.8m. The formula used is a_{c}=4∏^{2}r∫^{2}, and after rearranging and solving for ∫, the correct frequency is found to be approximately 0.47Hz or 1/s. There was some confusion with the units and a calculator mixup, but the final result is verified by using parentheses and two significant digits.
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

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

ehild

ehild said:
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$^{2}$

What do you get for the frequency?

ehild

ehild said:
What do you get for the frequency?

ehild

I'm getting 0.1465 for the frequency

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

ehild said:
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*

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?

fg
hsphysics2 said:
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

1 person

## 1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It always points towards the center of the circle and is responsible for keeping the object moving in a curved path.

## 2. How is centripetal acceleration related to amusement rides?

Amusement rides, such as roller coasters and spinning rides, often involve circular motion. The centripetal acceleration experienced by riders is what makes them feel the sensation of being pushed outward and held in place.

## 3. How is centripetal acceleration calculated?

The formula for centripetal acceleration is a = v² / r, where a is the acceleration, v is the velocity of the object, and r is the radius of the circular path. This formula can be used to calculate the acceleration of amusement rides.

## 4. Is centripetal acceleration the same as centrifugal force?

No, centripetal acceleration and centrifugal force are not the same. Centripetal acceleration is a real force that acts towards the center of the circle, while centrifugal force is a fictitious force that appears to act outwards from the center of the circle. It is simply a result of the inertia of the object.

## 5. How does the speed of an amusement ride affect centripetal acceleration?

The speed of the ride directly affects the magnitude of the centripetal acceleration experienced by riders. The faster the ride moves, the greater the centripetal acceleration will be, resulting in a more intense ride experience.

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