# What is the centripetal acceleration of the child?

• Physics_0101
In summary, Tracy is a new member on a site and asks for help with a physics problem involving a child riding a Ferris wheel. The question asks for the centripetal acceleration, magnitude and direction of the force at different points of the ride, and the value of each variable in the equation for centripetal force. Another member, Whozum, provides guidance and equations to help Tracy solve the problem. Ultimately, Tracy is able to solve for the centripetal acceleration, magnitude of force, and direction of force at the lowest point of the ride, with the help of Whozum's explanation and sample solution.
Physics_0101
Hey guys,
I'm new here and my name's Tracy. I just found out about this site and i think its fabulous.

Anyway i had a question and did'nt exactly know where to post it so if this is the wrong forum to post it I'm sorry.

Could somebody help me out and solve this question for me.

Q. A child weighing 40kg takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 18 m.

a) What is the centripetal acceleration of the child?

b) What is the magnitude and direction of the force that the seat exerts on the child at the lowest point of the ride?

c) What force does the seat exert on the child at the highest point of the ride?

d) What force does the seat exert on the child when he is midway between the top and the bottom?

Thanks a lot! And hi again to everyone!

Why don't you show us your working so we can show you where you went wrong?
The first thing you need to do is find out how fast she's going, so find angular velocity, and linear velocity.

whozum said:
Why don't you show us your working so we can show you where you went wrong?
The first thing you need to do is find out how fast she's going, so find angular velocity, and linear velocity.
Hey,
I can't really show you my working since i have'nt even tried the problem because i don't know where to start. Thats why i need help.

Simple problems like such always confuse me .

I have a test tommorow and while studying i stumbled across this problem. Help would be greatly appreciated.

Ok tell me,
what the centripetal force is and what it does
the equation for centripetal force
tell me what each variable means, and what its value is in this problem.

EDIT: Ok. Maybe my hint was too revealing.

Take it away, Whozum.

Last edited:
whozum said:
Ok tell me,
what the centripetal force is and what it does
the equation for centripetal force
tell me what each variable means, and what its value is in this problem.
The centripetal force is that force pulling an object towards the center of a circular path as the object goes around the circle. An object can travel in a circle only if there is a centripetal force on it.

The equation I'm not sure of.

An object can travel in a circle only if there is a centripetal force on it.
This is correct but sounds better if you rephrase it: An object's net force is centripetal if the consequent motion is circular.

Does F = mv^2/r ring any bells?

whozum said:
This is correct but sounds better if you rephrase it: An object's net force is centripetal if the consequent motion is circular.

Does F = mv^2/r ring any bells?
Yeah!

So once i find the centripetal force using that formula i can calculate the centripetal acceleration using a=v²/r ?

Also in the first equation what would be the value of 'v' ?

V is the linear velocity of the object in motion. In the problem, they don't give you this directly, however you can figure it out with what theyve given you.

A child weighing 40kg takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 18 m.

The bolded material gives you a time, and a rotation rate. Can you figure out the linear velocity?

Hint: Figure out the distance the child travels per rotation, then how many rotations per minute, then rotations per second.

Thanks for the help Whozum!

What about the rest of the question .

If you don't mind can you eplain the answer too.

have you figured out the velocity in terms of the period yet?

You can use $v = r\omega$.

I'll post you my solution to #1:

Step One: Write down all that you know:
Child's weight = m = 40kg
Ferris Wheel Diameter = d = 18m
Ferris Wheel Radius = r = 9m
Rotation rate = $\omega$ = ?
Centripetal Acceleration = a = ?
Centripetal Force = F = ?

Step Two: Identify what you need.
Ultimately we want the centripetal force, which is just the mass times the centripetal acceleration. The rotation rate needs to be solved for from the given information.

$$\omega = \frac{\Delta \theta}{\Delta t}$$

$$v = r \omega$$

$$a = \frac{v^2}{r}$$

$$F = ma = m \frac{v^2}{r}$$

They are all solved for their unknowns.

Step Five: Plug in your numbers:

$$\omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi}{15s} = \frac{0.42 rad}{1 sec}$$

$$v = r\omega = (9)(0.42) = 3.77m/s$$

$$a = \frac{v^2}{r} = \frac{(3.77)^2}{(9)} = 1.58 m/s^2$$

$$F = m \frac{v^2}{r} = ma = (40kg)(1.58m/s^2) = 63.2N$$

The child experiences a centripetal acceleration of 1.58m/s^2
The child experiences a centripetal force of magnitude 63.2N

Can you follow the same procedure for the other two, and fully answer #2?

Last edited:
would it be the centripidal force + the gravitational force for when the kid is at the lowest point of the ride?

You need to use the formula ac = v^2 / r
where a sub c = the centripetal acceleration
v^2 = the velocity squared of the object in question
and r is ofcourse the radius of a circle around which an object is moving
Try it out and see what you get!

## 1. What is centripetal acceleration?

Centripetal acceleration is the acceleration towards the center of a circular path that an object experiences when moving along a curved path. It is always directed towards the center of the circle and its magnitude is equal to the square of the speed of the object divided by the radius of the circle.

## 2. How is centripetal acceleration calculated?

Centripetal acceleration can be calculated using the formula a = v^2/r, where a is the centripetal acceleration, v is the speed of the object, and r is the radius of the circular path.

## 3. What factors affect the centripetal acceleration of the child?

The factors that affect the centripetal acceleration of the child include the speed at which they are moving, the radius of the circular path they are on, and the mass of the child. The greater the speed or the smaller the radius, the greater the centripetal acceleration will be. Additionally, a greater mass will result in a greater centripetal force needed to keep the child moving in a circular path.

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

No, centripetal acceleration is the acceleration towards the center of a circular path, while centrifugal acceleration is the apparent outward acceleration experienced by an object in circular motion. Centrifugal acceleration is caused by inertia and is equal in magnitude but opposite in direction to the centripetal acceleration.

## 5. How does centripetal acceleration relate to the child on a merry-go-round?

On a merry-go-round, the child experiences centripetal acceleration as they are constantly changing direction while moving in a circular path. This acceleration is necessary to keep the child from flying off the ride. The faster the merry-go-round spins or the closer the child is to the center, the greater the centripetal acceleration will be.

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