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Need help with centripetal force problem.

  1. Mar 7, 2015 #1
    [Note from mentor: this thread was originally posted in one of the non-homework forums, therefore it does not use the standard homework template.]



    I am self-studying ap physics, and as I have no teacher to guide me, am seeking help online. I have a problem that I have been trying to solve for about an hour now. I had thought I had the answer, but when I saw the answer, it was very close, but it was the negative version of what I had. Here's the problem:

    The sides of a cone make an angle φ with the vertical. A small mass m is placed on the inside of the cone and the cone, with its point down, is revolved at an angular velocity ω (radians/sec) about its symmetry axis. If the coefficient of static friction is μ, at what positions on the cone can the mass be placed without sliding on the cone? (Give max and min distances, r, from the axis).

    The answer the book gives is:



    My attempt at a solution:

    I broke the centripetal force into its components: One component is inwardly perpendicular to the surface


    The other is pointing toward the tip(downward parallel to the surface). This component is


    The gravity perpendicular to the surface is outward from the cone


    The parallel component is toward the tip of the cone parallel to the surface of the cone


    The friction would be μ*FN≥Ff.

    The normal force here is

    FN=mω2rcosφ - mgsinφ

    I substitute the normal force into the friction inequality.

    I now applied the F=ma formula in the direction parallel to the surface, using the direction of the tip of the cone as positive. a=0, so it's just F=0.

    2rsinφ + mgcosφ - Ff = 0

    I substituted in the max static friction and solved for r.

    I got

    r = (g/ω2)((1 + μtanφ)/(μ - tanμ))

    This is off from the r max by a factor of -1. I have no idea why. The radius shouldn't be negative. Could someone please help me with that?

    Also, if I make the friction negative and solve for r, I would get

    r = (g/ω2)((μtanφ - 1)/(μ + tanφ)

    This is also off by a factor of -1 for the minimum r.

    Something I suspect is messing up my answer is that I have the centripetal force acting inward to the cone, perpendicular to the axis of revolution. If this were so, I would think that the object wouldn't fly out because all of the components are pointing toward the bottom tip. Could someone please provide a free-body diagram?

    Help would be greatly appreciated. I want to thoroughly understand this, so it would be great if you explained what I did wrong and didn't just tell me what to do.

    Last edited by a moderator: Mar 7, 2015
  2. jcsd
  3. Mar 7, 2015 #2


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    Staff: Mentor

    The centripetal force is purely horizontal - in the plane of the circular motion.
    The force from the cone has a vertical component, but this goes upwards not downwards.
  4. Mar 7, 2015 #3
    Ok, thanks.

    I'm still having trouble figuring out what the force actually is. Would it be possible for you to provide a free-body diagram, or guide me on how to figure out all of the forces?

  5. Mar 7, 2015 #4


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    Staff: Mentor

    Why don't you draw one?
  6. Mar 7, 2015 #5
  7. Mar 7, 2015 #6
    I have a diagram attached

    Attached Files:

  8. Mar 7, 2015 #7
    Could you please tell me if anything is wrong with the diagram I had attached a few minutes ago?
  9. Mar 7, 2015 #8
    You have some sign errors in your force balance equations.

  10. Mar 7, 2015 #9
    Was it the normal force? Should I make it FG(perpendicular) - FC(perpendicular)?
  11. Mar 7, 2015 #10
    FN - mgsinφ=mω2rcosφ

  12. Mar 7, 2015 #11
    Sorry, but I'm not understanding why one should add the centripetal and gravitational forces perpendicular to the surface rather than subtracting them to get the normal force. Aren't they in the opposite directions?
  13. Mar 7, 2015 #12
    The centripetal acceleration is directed radially inward. If the inwardly directed unit normal at the cone surface is ##\vec{i}_n##, then:

    Normal force on mass = ##F_n\vec{i}_n##

    Component of gravity acting on mass in normal direction = ##-mg\sin\phi\vec{i}_n##

    Mass times acceleration in normal direction = ##mω^2r\cos\phi\vec{i}_n##

    So, applying Newton's second law in the normal direction gives:


  14. Mar 7, 2015 #13
    Ok, thanks.

    I appreciate your help very much.

    I'll see if I can go on from there.
  15. Mar 7, 2015 #14
    Wait, but don't we want the acceleration to be zero?
  16. Mar 7, 2015 #15
    Never mind, I just figured it out.
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