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Baggage Carousel Movement at an Angle

  1. Nov 2, 2008 #1
    1.Your suitcase is on a baggage carousel, Your suitcase has not slid all the way down the slope and is going around at a constant speed on a circle (r = 9.10 m) as the carousel turns. The coefficient of static friction between the suitcase and the carousel is 0.740, and the angle θ of the carousel is 7.16 °. How much time is required for your suitcase to go around once?

    2.Im not relaly sure what this question means?

    3. Im just looking for help about what this question means i dont relaly understand it?
    How can I know how long it takes if i dont know how fast its going?
  2. jcsd
  3. Nov 2, 2008 #2

    Doc Al

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    You're supposed to figure out how fast it's going. :wink:

    (Hint: Apply Newton's 2nd law.)
  4. Nov 3, 2008 #3
    hmmmm so f = ma, but in cicular motion that is f = ?

    If I found the normal force I think I would be able to find the centripetal acceleration, but Im not able to.

    Any other tip would be greatly appreciated
  5. Nov 3, 2008 #4
    This is my latest attempt,

    I put the x axis in the direction of the slope and y axis in the direction of the normal,

    I resolved the components on the x axis as follows,

    Weight of Baggage = Static Force + Centripetal force

    mgsinx = mgsinx(0.740) + ((mv^2)/r)cosx

    using this I get a velocity of 3.20 m/s

    and thus a time of 17.8 s

    I think Im on the right track but this answer is wrong :grumpy:
  6. Nov 3, 2008 #5

    Doc Al

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    You're on the right track, but you have the wrong value for the normal force and the sign of the centripetal force is off. Hint: Don't assume a value for the normal force, solve for it. (You'll need an equation for the y components as well.)

    (The problem will be slightly easier to solve if you use horizontal and vertical components.)
  7. Nov 4, 2008 #6
    Okay, firstly when you say the sign of the centripetal force is off, I assume you mean it should be negative. (Centripetal force acts towards the centre of rotation, so I thought this would be positive ) Any chance you could explain this?

    Normal Force. Im not sure how to solve this.

    im gonna try using the horizontal and vertical x and y axis now
  8. Nov 4, 2008 #7

    Doc Al

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    The component of the bag's weight acts down the slope, while the component of the centripetal acceleration will be up the slope. So a sign is off somewhere.
    Just call it N. It's one of the unknowns in your two equations.
  9. Nov 4, 2008 #8
    normal force something like Fn = -mg/cosx


    okay thats my crappy diagram and might help you explain to me where Im going wrong
  10. Nov 4, 2008 #9
    You shouldn't have the (-mg?) force. Part of the weight will be down the ramp... as you've indicated by the component (mg sin theta), and part of the weight (a component you are missing on your diagram) will press into the carrier and be supported by the normal force.

    In your diagram, you've left out the friction force, which relates to the normal force, which you could find by clear indication of components.

    Furthermore, if you're going to take a force (like the weight) in your free body diagram and replace it with components, be SURE you do this throughly, and then I suggest you mark it out in some noticeable way (texts tend to put a squiggle through it, but you could also "x" it out).
  11. Nov 4, 2008 #10

    Doc Al

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    You left out the friction force. Don't add extraneous stuff, like components. (Or -mg... what's that??)

    You should just have: Normal force; weight; friction force.

    Don't put Fc on a free body diagram. You'll get to that when you apply Newton's 2nd law.

    Edit: As physics girl phd says, be careful to distinguish components from the original vector if you want to put them on the diagram. Otherwise you might end up counting a force twice.
    Last edited: Nov 4, 2008
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