College-level centripetal force questions

[Tkennan]

Overall, I've done pretty well on this physics homework (which is due 10-13 @ 11PM Central), but these two questions are bugging the hell out of me.

1. An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away. The coeficient of static friction between the person and the wall is µ and the radius of the cylinder is R. Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimum necessary. Find the magnitude of the frictional force between the person and the wall.

2. A curve of radius r is banked at angle theta so that a car traveling with uniform speed v can round the curve without relying on friction to keep it from slipping to its left or right. Find the component of the net force parallel to the incline.

As you can see, neither question provides any numbers, such that the answers are formulas. They are multiple choice, but there are 15 friggin' choices so they might as well not be.

Any help is appreciated.

 

[Tkennan]

And to start off the discussion, I could have sworn the answer to number 1 was F=(µmv^2)/r, but that was wrong. That was the first choice too, just to tempt me into an easy answer. >_<

 

[Checkfate]

What forces are acting on the person in question 1? Is the person accelerating upwards or downwards? What does that say about the vertical forces? Try using the equations without numbers (With the assumption that the force holding him up equals the force pulling him down (What forces are these?), see if any of the variables cancel out.

 

[Tkennan]

What forces are acting on the person in question 1? Is the person accelerating upwards or downwards? What does that say about the vertical forces? Try using the equations without numbers (With the assumption that the force holding him up equals the force pulling him down (What forces are these?), see if any of the variables cancel out.

Well, that's the problem. I can do these problems when the circular rotation and the object are both on the horizontal plane, i.e. a car on pavement, but wouldn't the formula need to be altered to accomidate a horiztonal rotation and vertical contact/friction? There's nothing whatsoever in my textbook on this that I can find.

As far as forces, there'd be gravity, centripetal force, friction...I guess I get confused when I'm trying to rotate horizontally to keep an object up vertically, I can't work out logically how these three forces would interact, much less come up with a formula for it.

 

[QuantumCrash]

If worst comes to worst, always draw a free-body force diagram. You will see that reaction is the only force involved in centripetal acceleration. From there, you can use it to compare the vertical forces, friction and weight.

 

[Tkennan]

If worst comes to worst, always draw a free-body force diagram. You will see that reaction is the only force involved in centripetal acceleration. From there, you can use it to compare the vertical forces, friction and weight.

I've drawn a diagram, but that only helps in understanding the problem, not the question. For both of these I don't really know what they're asking. Is frictional force the coefficient of friction, or some component of the maximum frictional force? On number two, is the force parallel to the angle the normal force, or does some sort of trigonemtry have to be applied to the centripetal and vertical forces?

 

[Doc Al]

Both of these questions are probably much easier than you think they are.

Is frictional force the coefficient of friction, or some component of the maximum frictional force?

By "friction force" they mean the actual friction force. (D'oh!) Hint: Which direction does the friction force point? What's the net force in that direction?

On number two, is the force parallel to the angle the normal force, or does some sort of trigonemtry have to be applied to the centripetal and vertical forces?

You are asked to find a component of the net force, so start by finding the net force! Hint: What's the acceleration?

 

[Tkennan]

Both of these questions are probably much easier than you think they are.

By "friction force" they mean the actual friction force. (D'oh!) Hint: Which direction does the friction force point? What's the net force in that direction?

The net force in that direction would simply be the centripetal force, no? I've tried (mv^2)/r and that's not the answer.

 

[Doc Al]

The net force in that direction would simply be the centripetal force, no? I've tried (mv^2)/r and that's not the answer.

No. I asked two questions. First answer the first one.

 

[Tkennan]

I suppose it either points to the center of the circle or in the direction of gravity...I'm not sure which, friction tends to confuse me. Assuming then that it's just downward, would µmg be all that is at work for the frictional force? That would make sense. Thanks :)

Edit: nevermind, that was wrong too. 0 for 3 so far.

 

[Doc Al]

Friction is what's holding the person up! So the friction force must point upwards to counter gravity, which is pulling down. What's the net force in the vertical direction? What must the friction equal?

 

[Tkennan]

Okay, gotcha. F=mg *slaps self*. Muchos gracias.

So, onto number two...my first inclination was indeed to find the net force, but one thing that bothered me was that none of the answers had any variables other than theta, r, v, and m. I couldn't picture how to find the netforce without gravity, nor any way I'd use gravity on both sides such that it would cancel out...

 

[Doc Al]

Hint: Use Newton's 2nd law. Hint2: What's the acceleration?

 

[Tkennan]

So F=ma, and a=v^2/r (centripetal acceleration), correct? Now I know I need to use trig in some fashion, I assume to get this force, which is horizontal, in the direction parallel to the angle. That seems really weird though. I initially tried (mv^2)/r*cos(theta), which made the most sense to me, but that was wrong.

 

[Doc Al]

So F=ma, and a=v^2/r (centripetal acceleration), correct?

Yes, that will give you the net force, which acts horizontally.

Now I know I need to use trig in some fashion, I assume to get this force, which is horizontal, in the direction parallel to the angle.
You are asked to find the component parallel
That seems really weird though. I initially tried (mv^2)/r*cos(theta), which made the most sense to me, but that was wrong.

That's not wrong.

Of course, that's a pretty generic answer and it probably is not one of the choices. So analyze the forces acting on the car and figure out another expression for that net force. (One that takes full advantage of the condition that the angle is optimized for zero friction at speed v.)

 

[Tkennan]

Well that was one of the choices and it counted it wrong, so that's pretty intimidating. Not sure what you mean by taking full advantage of the angle.

Anyhow, I'm going to be out for 3 hours or so. Thanks for all your help, and if you have any more hints or ideas on that question that could account for my answer being wrong, by all means post 'em. :)

 

[Doc Al]

I don't see how it can be wrong (unless they what you to include a minus sign, since it acts down the incline). It's kind of like asking, What's the net force on an object? and having one of the choices be "ma". Well, yeah, that's right (albeit generic).

My test would not include that as a choice, since I would want you to do some analysis of the forces. (Hint: Find the angle that the road makes.)

 

[Tkennan]

Just a few more hours before I need to have it answered, so here's a pic of the problem and all the answer choices to see if I left something out. I'm personally out of ideas.

http://img124.imageshack.us/img124/8600/q14nj8.jpg
http://img133.imageshack.us/img133/2864/q14alc3.jpg

I tried answer choice 1 and got it wrong.

 

[Doc Al]

I tried answer choice 1 and got it wrong.

Yes, that answer is wrong! The cos(theta) term is on the bottom--that makes no sense. If the net force is mv^2/r horizontal, what's the component parallel to the ramp? (I guess I misread your answer before.)

 

[Tkennan]

Yes, that answer is wrong! The cos(theta) term is on the bottom--that makes no sense. If the net force is mv^2/r horizontal, what's the component parallel to the ramp? (I guess I misread your answer before.)

Parallel to the ramp, as I understand it, would be cos(theta)=((mv^2)/r)/F, where F is the force parallel to the ramp. Thus, F=(mv^2)/r*cos(theta). Obviously I'm missing something, but what?

 

[Doc Al]

You have it exactly reversed. If you have a force (or any vector) X in some direction, the component of that force at an angle theta from that direction is X cos(theta).

Try this example: Someone kicks a ball with speed 5 m/s at an angle of 30 degrees above the horizontal. What's the x-component of that initial velocity?

 

[Tkennan]

Then it would be 5cos(30). But if you only knew the x component, which is the case here, and IT was 5 m/s, to find the speed it travels at 30 degrees above the horizontal, it would be 5/cos(30), right?

I'm sure that you're right and I'm missing something (or I wouldn't have come here for help , but I'm just lost as to the logic here.

 

[Doc Al]

Then it would be 5cos(30). But if you only knew the x component, which is the case here, and IT was 5 m/s, to find the speed it travels at 30 degrees above the horizontal, it would be 5/cos(30), right?

Ah, but that's not the case here! mv^2/r is not just a component of the force, it is the entire force! You are given the full vector and have to find the component.

The analogous case with velocity would be: If something is moving at a velocity of 5 m/s horizontally, what is the velocity component 30 degrees above the horizontal?

 

[Tkennan]

I just can't grasp this. Isn't your analogy equivalent to the last part of my previous post, and therefore worked ala 5/cos(30)? And I see no difference between mv^2/r being the full force or the horizontal component, because either way, isn't all of the force pointing towards the center of the circle on the horizontal plane? *goes cross-eyed with frustration*

 

[Doc Al]

It makes all the difference in the world whether you are given the component (a part) and need to find the full vector (the whole), versus having the whole vector and only need a component of the whole. For one thing, the component is always smaller that the full vector. (Dividing by cos(theta) makes something bigger.)

Maybe this will help. Since mv^2/r is the full vector, find both components: the parallel component and the perpendicular component. Actually draw the x-y axis (x = parallel; y = perpendicular), the full vector, and then the components. Maybe then you'll see the proper triangles involved.

 

[Doc Al]

Just for the record, when you first wrote this in post #14:

I initially tried (mv^2)/r*cos(theta), which made the most sense to me, but that was wrong.

I assumed (D'oh!) that you meant it to be understood as:
[(mv^2)/r]*cos(theta)

not:
(mv^2)/[r*cos(theta)]

That's why I said it wasn't wrong. (But this last version is wrong, as we've discussed.)

Don't give up on understanding this. When it clicks, you'll slap yourself again.

 

[Tkennan]

Don't give up on understanding this. When it clicks, you'll slap yourself again.

Yeah, I probably will. I went ahead and answered with cos* rather than /cos even though I don't understand it yet (I haven't put a lot of thought into it though, now that I can work backwards from the answer it will be a hell of a lot easier). For now though, my brain demands I take a break and feed it some video games before it sees another physics problem.

Thanks a ton for your prompt help.