# Centripetal or CoEf ?

1. Apr 19, 2008

### neonjr

Centripetal or CoEf ???

1. The problem statement, all variables and given/known data
If the Coefficient of friction between the gym floor and your running shoes is 0.80, how fast could you run around a circle of radius 2.0m?

I don't know even where to start......

2. Relevant equations

3. The attempt at a solution

2. Apr 19, 2008

### Oomair

you should draw a free body diagram and label the forces that are occuring

and apply F=ma, hence a = (v^2)/R

3. Apr 19, 2008

### neonjr

and how do i do that without having any mass to work with, any velocity to work with, any acceleration to work with.

4. Apr 19, 2008

### D H

Staff Emeritus
These are the variables you are trying to solve for. You don't really need a mass as it will disappear in the end.

5. Apr 19, 2008

### Oomair

centripetal acceleration is where a = (v^2)/R, so you know that F=ma, since the problem involves motion around a circle, then the sum of forces is F=m(V^2)/(R)

and you know that friction is a force...

6. Apr 19, 2008

### Tentothe

Drawing a free-body diagram shows that there are three different types of forces acting on you in this situation: friction, normal, and gravitational. One of them is going to be unbalanced and cause the acceleration required to keep you running in a circle. Add up the forces acting on you to find the net force, which you should find is only this unbalanced force. Using Newton's second law, F_net = ma, where F_net is this unbalanced force and a is v^2/R, you can solve for the velocity. The mass does end up dropping out, the acceleration was substituted for in terms of velocity, and obviously you won't need the velocity because this is what you are trying to solve for.

Hint: Recall that the friction force is going to be $$\mu_{s}n$$, or the coefficient of friction multiplied by the normal force.

7. Apr 20, 2008

### neonjr

still getting no where with this, i'm really not understanding without a complete equation. The statement "You don't really need a mass as it will disappear in the end." makes absolutely no sense to me.

And after drawing a FBD im still not getting what I need to use here. I draw the circle figure out my distance, because i know the radius of the circle, so TTr^2. But to find out velocity in any situation, and GET an answer, not a theoretical answer like (xy2) you need more than one piece of info out of an equation. because V=d/t, all i have is distance, Gravity, and CoEf.....I've talked to three people at home here and they are all lost.

8. Apr 20, 2008

### D H

Staff Emeritus
There is an upper limit on the centripedal force the shoes can exert. This upper limit maps directly to the maximum speed someone can run around the circle. A person can accelerate less than this upper limit, but this means the person is not running around the circle as fast as they can. What is this upper limit? (HINT: Use the coefficient of friction.)

9. Apr 20, 2008

### neonjr

i'm sorry to be abrupt, but perhaps i need to say this. but i'm not that bright. i have not got a clue what you are talking about. you need to explain this in terms a 5 year old could understand, and start from the beginning. please.

10. Apr 20, 2008

### D H

Staff Emeritus
Suppose the person has a mass m. What is the maximum horizontal force the person can exert with his shoes?

11. Apr 20, 2008

### neonjr

would this not be 9.8 N ?

12. Apr 20, 2008

### Tentothe

Break this into dimensions first. What are your vertical forces and what are your horizontal forces?

Since the runner is not going to be moving in the vertical direction (not coming up off the ground or sinking through it), we know the acceleration in this direction is going to be 0. By Newton's second law, F=ma, where a is zero, the net force in the vertical direction must be therefore be zero. What does this say about the forces acting on the runner in the vertical direction? How can you use this information to analyze the horizontal direction?

In the horizontal direction, you do have an acceleration. It is what lets you run in a circle and is directed toward the center of the circle. Its magnitude is v^2/R. If you have an acceleration in the horizontal direction, you must therefore have a net force acting in the horizontal direction by Newton's second law.

Determine which forces you have in each dimension and set them equal to the mass times the acceleration in that dimension by Newton's second law. If the algebra is done correctly, the mass terms will drop out of the equation, allowing you to solve for the unknown you want: velocity.

You will be working with two equations here: one for each dimension.

13. Apr 20, 2008

### D H

Staff Emeritus
No. You are out-and-out guessing here. HINT: The maximum horizontal force is the coefficient of (static) friction times the normal force. Any larger force will result in slipping.