Uniform Circular Motion and wheel revolutions

[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 I am 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.

 

[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.

 

[Christina De Rose]

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

 

[Parth Dave]

Another question i have is, is the wheel rolling? And if so, doesn't 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 isn't the normal force Fc + Fg?

 

[Parth Dave]

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

Why is that so?

 

[Christina De Rose]

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

 

[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...

 

[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.

 

[Sirus]

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

 

[Parth Dave]

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

 

[Parth Dave]

Sirus... no.

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

 

[Sirus]

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.

Centrifugal force does not exist, actually.

 

[Parth Dave]

i assume he meant centripetal, but whatever.

 

[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.

 

[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?

 

[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.

 

[Sirus]

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

 

[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.

 

[Sirus]

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

 

[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.

 

[Parth Dave]

But then where does static friction come into play? I have a hard time believing it is negligable.

 

[Sirus]

It is included when we use centripetal force, since this force exists here only due to static friction (the mud is not connected to the wheel in any other way).

 

[Parth Dave]

The only thing that makes some sense to me is that:
Fn = Fg
Fc = Fs

Thus, Fs = mg0.64 = mv^2/r
v = sqrt(g*0.64/0.7)

so the rev/sec is v/2pir or

sqrt(g*0.64/0.7) / 2pi(0.7) = 0.681 rev/sec

Seems a little more reasonable, but it is not entirely convincing to myself.

 

[Sirus]

Fn only equals Fg at the top of the wheel. I think we can say, however, that $F_{n}=F_{static friction}$, since those point in opposite directions at all times. This doesn't help us though. What don't you understand about my method?

 

[Parth Dave]

Fn <> Fs, because Fs is based on Fn...

Also, what about normal force in your situation. Because centripetal force doesn't necessarily have to be the only force holding the mud up. If it is touching the ground than the ground is holding it up too.

 

[Sirus]

I don't really understand the first part of your post. I don't think a normal force really exists when the mud is below the halfway point of the wheel, since static frictional force is a two-way electrical attraction between the wheel and mud (no normal force required since mud pulls on wheel as much as wheel pulls on mud). Normal force exists above halfway point because their is now something between the ground and the mud. Does this make sense?

Yes, the ground pushes on the mud at one moment, but remember what I said earlier, that we only use the data from that moment, and say the mud is flung off the instant it is no longer between the pavement and the wheel (so the ground no longer pushes on it).