Centrifugal and Centripetal Forces Lesson 2 - A Sample Problem

There is really not much to say about Centripetal Forces on the introductory level past what was said in the last lesson, so I would like to take this opportunity to work a sample problem that encompasses several of the things you have learned in the past few lessons.

You are on the Barrel o' Fun at Astroworld ( you know, the round thing where you stick to the sides as it spins your brains out). The Barrel's diameter is 8.0m and spins at a rate of 1 revolution every 3 seconds.
a. What is the centripetal force on your 80kg body?
b. What is the minimum coefficient of friction needed to keep you from sliding down the wall?

First, we need to convert the angular velocity into rad/s.

1rev / 3s = .333 rev/s = 2.094rad/s

We know that 8.0m is the diameter, which means that 4.0m is the radius. Now we can figure out the centripetal acceleration, which leads us to the answer of part a.

aC = (2.094rad/s)2 * 4.0m
aC = 17.546m.s
FC = 80kg * 17.546m/s
FC = 1403.677N

For part b, we want to find out the coefficient of friction needed to keep you from slipping down the sides of the barrel. Since we don't want movement, we aren't looking for the coefficient of kinetic friction, but the coefficient of static friction. Therefore, the equation we will need to use is . Let's look at a picture to help us envision everything.

Picture of Barrel o' Fun

From this we can see two things, one is that the force of static friction needs to equal the force of gravity. The other is that the centripetal force is also the normal force. In an earlier lesson, we talked about how the normal force was always perpendicular to the point of contact. Well, in this case, your body is vertical and pushing on the side of the barrel. That means that the normal force must be horizontal and it must be pushing back at you. The centripetal force is always the force that is pulling inside the circle, and it just so happens that in this problem, they are the same thing.

So now all we need to do is find FS, put the variables into the equation and plug n' chug. Since FS is equal to FG, we can just use F = m * a to find FG.

FG = 80kg * 9.8m/s2
FG = 784N
FS = FG = 784N
784N = S * 1403.677N
S = .559

Isn't it great to see how much we've learned and what all we can do with it. Now let's continue to learn with the next few lessons on momentum.

Centripetal Force Sample Problems