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Centripetal force and friction problem

  1. Oct 7, 2004 #1
    Hey guys/gals, I'm new here and new to physics, but I love it all. Ok, for some of you, this may be a walk in the park, but personally, I don't think this is an incredibly easy problem for someone who's never taken Physics before this semester. I tried many different approaches, and have come out with a few different answers and none seem reasonable. Heres the question;

    "You are at a baggage carousel at the airport. Your suitcase has not slid all the way down the slope and is going around at a constant speed on a circle (r=11m) as the carousel turns. The coefficient of static friction between the suitcase and the carousel is 0.760, and the angle (theta) that the luggage is resting at is 36 degrees. How much time is required for your suitcase to go around once.?"

    Now, I know the T=2PiR/V, but figuring out the velocity with the supplied info is keeping me stuck. I found something in the book that says that V=sqrt of [coefficient of static friction]*g*r. But that doesn't seem to give a reasonable answer, (something like 7 or 8 m/s :eek: ) Please help me.

    Last edited: Oct 8, 2004
  2. jcsd
  3. Oct 8, 2004 #2


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    I think the problem is a little misleading. It seems to me they are asking for the shortest possible time it would take for your luggage to go around. The basic idea is that friction would not be adequate to hold the luggage in place if the carousel is rotating too quickly.

    Do a force balance and compare the centripetal acceleration with the appropriate component of the frictional force.
  4. Oct 8, 2004 #3
    What does Force Balance mean? Do you mean the g/Sin(31Deg)=H or would it be the g/Tan(31)=Ha. Doing that to find the component force acting on it . But again, I'm not sure that works.

    Is there a way you can find out the mass? How can I find the normal Force? What exactly can I use in this problem to find the centripetal acceleration?

    Could I find the Velocity using Tan(theta)=V^2/rg?

    Please help just a wee bit more...
  5. Oct 8, 2004 #4

    I really need help on this!
  6. Oct 8, 2004 #5
    Ok, so here's what I did, tell me if this is close or right or flat out wrong please.

    V=SqRt(Us g r)=9.05m/s

    So the least amount of time needed to complete one rotation is 6.18 seconds.
    I think that's kinda fast. But I don't know what kind of friction .760 really means since I'm new at this. For all I know that number can be two pieces of hot treaded rubber touching each other.
  7. Oct 8, 2004 #6


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    The frictional force is

    [tex]F_f = \mu mg \cos \theta[/tex]

    and is directed parallel to the surface. To find the component of that force perpendicular to the axis of rotation multiply by [itex]\cos \theta[/itex]. Also, the normal force on the object due to the surface provides a component of force radially outward from the axis of rotation. Together, they give the centripetal force so

    [tex] \mu g \cos^2 \theta - g \sin^2 \theta = \frac {v^2}{R}[/tex]

    gives the critical speed. R is the radius of the circular path and v is the speed of the luggage.
    Last edited: Oct 8, 2004
  8. Oct 8, 2004 #7
    Why is \cos \theta squared?

    I see where you got this equation, and it's what I used, except for the squared cosine part....

    Thanks for this part though...:D
  9. Oct 9, 2004 #8


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    The first cosine factor is used in determining the normal force the luggage exerts on the conveyor which is necessary for determining the frictional force. The second one comes from the horizontal component of the frictional force which contributes to the centripetal acceleration (in the radial direction about the center of rotation).
  10. Oct 9, 2004 #9
    Alright, Where did the minus Sin squared com from?

    So far I got [tex] \mu g \cos \theta = \frac {v^2}{R} \cos \theta [/tex]

    So wouldn't the cosines cancel each other out? Instead of squaring them? Could you show me the algebra and why you did it?

  11. Oct 9, 2004 #10


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    There's an error in my equation. Let me explain.

    Ultimately, you want to calculate the centripetal force on the luggage. That's the force pulling the luggage toward the center of its "orbit." Part of that force is the frictional force tending to hold the lugguge to the conveyor and is proportional to the force the luggage exerts normally on the conveyor. However, the frictional force is directed parallel to the surface and, in this case, uphill. The magnitude of the normal force is

    [tex]F_n = mg\cos \theta[/tex]

    Multiply that by the coefficent of friction and you have the frictional force. As I said, ultimately we are interested in the component of that force along the radial direction which means multiplying it by an additional factor of [itex] \cos \theta[/itex].

    Now the conveyor also exerts a force on the luggage - it's the same normal force as above but directed oppositely. Again, we are interested in the component of that force directed radially outward so we multiply by [itex]\sin \theta[/itex].

    The radial component of those two forces contribute to the centripetal force so we have

    [tex]\mu g \cos^2 \theta - g \cos \theta \sin \theta = \frac {v^2}{R}[/tex]


    [tex]g \cos \theta \left( \mu \cos \theta - \sin \theta \right) = \frac {v^2}{R}[/tex]

    The frictional force tends to pull the luggage toward the center of rotation and the normal force of the incline tends to push luggage outward which is why they have opposite signs.
  12. Oct 9, 2004 #11
    alright, so doing all of that, I got 44.9 seconds. That seems more reasonable. I must say, you are awesome, but I have one final request for you so that I can more fully understand this, I will private message you this....Thanks!
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