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College-level centripetal force questions

  1. Oct 12, 2006 #1
    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.
     
  2. jcsd
  3. Oct 12, 2006 #2
    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. >_<
     
  4. Oct 12, 2006 #3
    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.
     
    Last edited: Oct 12, 2006
  5. Oct 13, 2006 #4
    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.
     
  6. Oct 13, 2006 #5
    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.
     
  7. Oct 13, 2006 #6
    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?
     
  8. Oct 13, 2006 #7

    Doc Al

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    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?

    You are asked to find a component of the net force, so start by finding the net force! Hint: What's the acceleration?
     
  9. Oct 13, 2006 #8
    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. :frown:
     
  10. Oct 13, 2006 #9

    Doc Al

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    No. I asked two questions. First answer the first one.
     
  11. Oct 13, 2006 #10
    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.
     
    Last edited: Oct 13, 2006
  12. Oct 13, 2006 #11

    Doc Al

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    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?
     
  13. Oct 13, 2006 #12
    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...
     
  14. Oct 13, 2006 #13

    Doc Al

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    Hint: Use Newton's 2nd law. Hint2: What's the acceleration?
     
  15. Oct 13, 2006 #14
    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.
     
  16. Oct 13, 2006 #15

    Doc Al

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    Yes, that will give you the net force, which acts horizontally.
    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.)
     
  17. Oct 13, 2006 #16
    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 gonna 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. :)
     
  18. Oct 13, 2006 #17

    Doc Al

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    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.)
     
  19. Oct 13, 2006 #18
    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.

    [​IMG]
    [​IMG]

    I tried answer choice 1 and got it wrong.
     
  20. Oct 13, 2006 #19

    Doc Al

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    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.)
     
  21. Oct 13, 2006 #20
    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?
     
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