# Vertical Circular Motion &amp; Centripetal Acceleration

• tem_osu
In summary, the first problem required finding the length of the circular track and then dividing it by the time taken to complete one lap to find the velocity. The second problem involved rewriting the equation using the correct values and then solving for the velocity.
tem_osu
[SOLVED] Vertical Circular Motion &amp; Centripetal Acceleration

## Homework Statement

1) A car travels at a constant speed around a circular track whose radius is 2.1 km. The car goes once around the track in 410 s. What is the magnitude of the centripetal acceleration of the car?

a=(v^2)/r

---

## Homework Statement

2)A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to five times her weight as she goes through the dip. If r = 20.5 m, how fast is the roller coaster traveling at the bottom of the dip?

F-mg = m(v^2/r)

## The Attempt at a Solution

v^2 = (20.5)(9.8)
v = 14.17 m/s - wrong answer

---

I believe you miscalculated the velocity in the first question. You probably want to give this another look.

In the second question, try rewriting

F -mg = m(v^2)/r all in terms of m, g, v, and r.

Then go through and see what you can cancel.

Last edited:
Thank you so much for replying.

For the 1st problem:
2.1 km = 2,100 m
velocity = 2,100 m / 410 s = 5.12 m/s (did anything went wrong in this step?)

a = (v^2)/r
a = (5.12^2)/2100 = 0.01248

---

For the 2nd problem:
F - mg = m(v^2)/r

Force is 5 times the weight, so F=5m

5m - mg = m(v^2)/r
m(5-g) = m(v^2)/r
rm(5-g)/m = m(v^2)/m
r(5-g) = v^2
(20.5)(5-9.8) = v^2
-98.4=v^2 (Do I ignore the negative sign?)

For the first question:

2.1 km is the radius of the track. You need to use this information to find the entire length of the track (more specifically, the circumference). Then use this information to find the velocity.

For the second question:

As you said, the force is 5 times the weight. However, this is not F = 5m. It should be F = 5mg because W = mg (the MASS is m, but the WEIGHT is mg).

I hope this makes sense; if not, I'll be up for a while and would be happy to try to help further.

I got the 2nd problem. The answer came out to be 28.35 m/s.

For the 1st problem:
a = (v^2)/r
r = 2100/(2*3.14) = 334.39

velocity = 334.39m/410s = 0.8156 m/s?
then the acceleration would be a= 0.8156/410 = 0.001989 m/s^2?

Congratulations on getting the answer to the second question.

Now, let's try to iron the wrinkles in the first problem.

The problem gives you the value for r. This is 2.1 km = 2,100 m. It also gives you the time it takes for the car to make one full lap around the track.

To get the velocity, you need to find the distance the car travels in the time it took to complete one lap. Thus, you need to find the length of the track and then divide this number by the time it took to traverse the track.

How do you think you will find the length of the track?

HINT: The circumference of a circle = 2*pi*r

So, the circumference is (2)(pi)(2100) = 13,194.69 m
the car traveled for 410 s, 13194.69/410 = 32.18 m/s?

EDIT: I got it! Thank you so much. You are a lifesaver!

Last edited:

## 1. What is vertical circular motion?

Vertical circular motion is a type of motion in which an object moves in a circular path while also moving up or down in a straight line. This type of motion is commonly seen in rides at amusement parks, such as a roller coaster.

## 2. What is centripetal acceleration?

Centripetal acceleration is the acceleration directed towards the center of a circular path. It is responsible for keeping an object moving in a circular path and is always perpendicular to the velocity of the object.

## 3. How is centripetal acceleration related to vertical circular motion?

In vertical circular motion, the centripetal acceleration is responsible for keeping the object moving in a circular path while the gravitational force is responsible for the object's vertical acceleration. These two forces work together to create a circular motion in a vertical plane.

## 4. How does the speed of an object in vertical circular motion affect the centripetal acceleration?

The centripetal acceleration is directly proportional to the square of the speed of an object in vertical circular motion. This means that as the speed increases, the centripetal acceleration also increases, and vice versa.

## 5. What is the difference between centripetal acceleration and centrifugal force?

Centripetal acceleration is the acceleration directed towards the center of a circular path, while centrifugal force is the outward force that appears to act on an object in circular motion. Centrifugal force is a fictitious force and only appears to exist due to the object's inertia.

• Introductory Physics Homework Help
Replies
55
Views
936
• Introductory Physics Homework Help
Replies
8
Views
322
• Introductory Physics Homework Help
Replies
28
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
731
• Introductory Physics Homework Help
Replies
9
Views
3K
• Introductory Physics Homework Help
Replies
9
Views
6K
• Introductory Physics Homework Help
Replies
2
Views
663