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Static stick and cyclinder, did I approach this problem correctly?

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data
    A horizontal stick of mass m has its left end attached to a pivot on a plane inclined at an angle theta, while its right end rest on top of a cylinder also of mass m which in turn rests on the plane as shown. The coefficients of friction at both points is (mu).
    Sorry I couldn't add the symbols for some reason. I am always having problems attaching the drawing. Hopefully I can fix this after I post.

    a.) Assuming the system is at rest, what is the normal force from the plane on the cylinder?

    b.) What is the smallest value of (mu) in terms of theta for which the system does not slip?

    2. Relevant equations



    3. The attempt at a solution
    For a.) the force on the cylinder from the stick is just mg/2 and the force of gravity from the ball is just mg. Then the normal force on the ball from the plane must be (mg/2+mg)cos(theta) = 1.5mg*cos(theta). Hopefully I'm right up to that point.
    For part b.) do I need I need to use torques? I did not so I think I may be wrong. I just solved (mu)1.5mg*cos(theta) + (mu)0.5mg*cos(theta) >= 1.5mg*sin(theta) and got (mu) >= 3/4 * tan(theta)
    Can someone let me know if I did this correctly or if I need to implement torques? Any help would be great.
     
    Last edited: Sep 16, 2012
  2. jcsd
  3. Sep 17, 2012 #2

    CWatters

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    Part a) looks right.

    Part b) you have

    (mu)1.5mg*cos(theta) + (mu)0.5mg*cos(theta) >= 1.5mg*sin(theta)

    which looks correct if you are assuming the ball may slide but I do think you need to check that the ball won't roll out.
     
  4. Sep 17, 2012 #3
    Ok thats what I thought, but how would I do this. I would have the force of friction on the bottom of the cylinder = (mu)1.5mg*cos(theta) which points up the inclined plane. The force of friction on the top would = (mu)0.5mg which points towards the inclined plane along the stick. When I set these equal to each other to balance the torques the (mu)'s cancel out so I assume I'm doing something wrong.
     
  5. Sep 17, 2012 #4

    CWatters

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    I've edited this post because having thought about it a bit more I've decided your approach is correct.

    I believe it rolls down the hill if...

    (mu)1.5mg*cos(theta)*R > (mu)0.5mg*R

    (mu), mg and R cancel leaving

    cos(theta) > 1/3

    I think this is telling use that for cos(theta) > 1/3 it will roll. For cos (theta) < 1/3 use your answer for sliding..

    Would be nice to have input from another reader.
     
    Last edited: Sep 17, 2012
  6. Sep 17, 2012 #5
    I could't put this in the original, but here is the sketch
     

    Attached Files:

  7. Sep 18, 2012 #6
    It feels wrong :(

    Imagine that instead of a rod on top of the cylinder, the cylinder is held by a horizontal light string to the top of the cylinder instead. The additional mass of the rod is absorbed into the mass of the cylinder (which is now 3m/2). The only thing special about the string is that it breaks at the point where the rod would have slipped. (tension > μ mg/2)

    This system is now very like a block on an inclined plane as the cylinder can only move by slipping (or breaking the string). My memory has the minimum μ to avoid slip for a block on an inclined plane as tan(θ): so a result that is less than that feels wrong.
     
    Last edited: Sep 18, 2012
  8. Sep 18, 2012 #7
    There is the friction at the top of the cylinder though. So shouldn't that make the minimum (mu) less?
     
  9. Sep 18, 2012 #8
    I missed that :( so you are correct that μ can be less than tan(θ)

    ===

    Let the tension in the rod at the top be T
    Let the friction where the cylinder touches the inclined plane be F
    Let the normal reaction where the cylinder touches the inclined plane be R

    T=F (to avoid the cylinder rotating) <= this line is bothering me
    T ≤ μ 1/2 m g (limit for the rod)
    F ≤ μ R


    T + F cos(θ) = R sin(θ) (resolve horizontal forces)
    F ( 1 + cos(θ) ) = R sin(θ) (substitute T=F)
    F/R = sin(θ)/ ( 1 + cos(θ) )

    => μ ≥ sin(θ)/ ( 1 + cos(θ) )

    but that isn't a complete solution as the tension in the rod also restricts μ.

    F + T cos(θ) = 3/2 m g sin(θ) (resolve along the inclined plane)
    T (1 + cos(θ) ) = 3/2 m g sin(θ) (substitute T=F)
    T/(1/2 m g ) = 3 sin(θ) / ( 1 + cos(θ) )

    => μ ≥ 3 sin(θ)/ ( 1 + cos(θ) )

    which is always greater than the earlier result so is the required relationship for μ
     
    Last edited: Sep 18, 2012
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