# Uniform Circular Motion and wheel revolutions

1. Sep 19, 2004

### Parth Dave

A clump of mud is attached to the outside of a bicycle wheel, radius 700mm.

how many revolutions per second does this wheel have to spin to fling off the mud if the coefficient of static friction attaching it to the wheel is 0.64.

The real trouble im having with this is find the normal force. I assume it depends on centripetal force, but is it correct to say centripetal force is equal to normal force.

2. Sep 19, 2004

### Christina De Rose

NO, I don't think you can say that. The normal force acts perpendicular to a surface, while the centripetal force acts toward the centre of the circle. I think the normal force is simply Fg, or mass x 9.81. This way, when you say that Fc = Fg, the masses cancel out.

3. Sep 19, 2004

### Christina De Rose

sorry, i made a mistake. The frictional force is equal to the centripetal force.

4. Sep 19, 2004

### Parth Dave

Another question i have is, is the wheel rolling? And if so, doesnt that mean the force of gravity is dynamic in the y-direction. And thus the normal force will constantly change. And if the clump of mud is at the very top of the wheel at one point. Than isnt the normal force Fc + Fg?

5. Sep 19, 2004

### Parth Dave

Why is that so?

6. Sep 19, 2004

### Christina De Rose

You're right, the normal force would constantly change...
I'm as confused as you are now...

7. Sep 19, 2004

### Parth Dave

Although yours is the only logical solution i can see out of it. Because there is no other way to cancel out the mass..........

8. Sep 19, 2004

### Pyrrhus

The normal force of the mud is a centrifugal force, and the friction force must be a tangential force. It seems like that.. from my interpretetion.

9. Sep 19, 2004

### Sirus

Do they give you the correct answer (like any good question should)?

10. Sep 19, 2004

### Parth Dave

I have no idea what a tangential force is....

11. Sep 19, 2004

### Parth Dave

Sirus... no.

And i now know what a tangential force is, but how does that help me solve it?

12. Sep 19, 2004

### Sirus

Centrifugal force does not exist, actually.

13. Sep 19, 2004

### Parth Dave

i assume he meant centripetal, but whatever.

14. Sep 19, 2004

### Pyrrhus

Hmm, maybe i'm reading it wrong, but the mud is on the outside of the wheel, so it will have a normal force pushing always in the opposite direction of the center of the wheel.

15. Sep 19, 2004

### christinono

correct me if I'm wrong, but the normal force is directed toward the center of the wheel.

Do U have the answer to this problem?

16. Sep 19, 2004

### Pyrrhus

I meant to the force of the mud with surface of the wheel, but yes i was wrong, because it will be assuming the mud has a mass, it seems the problem doens't give it.

17. Sep 19, 2004

### Sirus

I think I figured it out, although I get a weird answer. Give me a second to Latex it (still learning).

18. Sep 19, 2004

### Sirus

Ok. The mud will need the smallest speed to make it fling off at the bottom of it's trajectory (when it's under the wheel, if it's rolling). yes, I know it won't come off if it's between the wheel and the pavement, but we can use the data from that moment if we say it flies off the very instant it comes out from under the wheel.

$$F_{c}=F_{g}$$ at that point, so:

$$\frac{mv^2}{r}=mg\\v=\sqrt{gr}$$

Now relate this to revolutions per second:

$$\sqrt{gr}=x2\pi\mbox{r}$$

Couldn't get the r to work, there. Anyways, now solve for x, which is the revolutions per second. The frictional force is the centripetal force, so you don't even have to include it. Answer is weird though. I think this is right.

19. Sep 19, 2004

### Sirus

I get 0.1884... revolutions per second. That seems extremely slow. Too bad you don't have the answer.

20. Sep 19, 2004

### Sirus

Oops, noticed a mistake in the latex, there is a supposed to be a new line starting after mg where it says mgv, so it is supposed to be v = square root g times r. Also remember that the revolutions would have to be just faster since we assumed that centripetal force equals gravitational force.