- #1

neonjr

- 6

- 0

**Centripetal or CoEf ?**

## Homework Statement

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...

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- Thread starter neonjr
- Start date

- #1

neonjr

- 6

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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

Oomair

- 36

- 0

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

- #3

neonjr

- 6

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

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

- #4

- 15,450

- 687

These are the variables you are trying to solve for. You don't really need a mass as it will disappear in the end.and how do i do that without having any mass to work with, any velocity to work with, any acceleration to work with.

- #5

Oomair

- 36

- 0

and you know that friction is a force...

- #6

Tentothe

- 13

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

- #7

neonjr

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And after drawing a FBD I am 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

- 15,450

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- #9

neonjr

- 6

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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

- 15,450

- 687

- #11

neonjr

- 6

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m. What is the maximum horizontal force the person can exert with his shoes?

would this not be 9.8 N ?

- #12

Tentothe

- 13

- 0

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 let's 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

- 15,450

- 687

would this not be 9.8 N ?

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.

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