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F = MA Exam 2011 # 13 (Rotation/Rolling)

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data
    https://aapt.org/physicsteam/2012/upload/exam1-2011-1-3-answers_1.pdf
    Number 13

    2. Relevant equations
    Rolling objects: E = 1/2mv^2 + 1/2Iw^2
    I_cylinder = 1/2mR^2
    Torque = rf(perp)
    Torque = Iω = rfsinθ



    3. The attempt at a solution
    No idea how to proceed. Most likely starting with the torque equation.
    T = rF = .02F
     
  2. jcsd
  3. Jan 27, 2013 #2
    Think about what happens in non-rotational dynamics:
    In order for the apparatus to not move horizontally, the friction force must equal the horizontal component of the force (Fcosθ) caused by the pulling of the thread.

    Now, think in terms of rotational dynamics:
    The torque produced by the pulling of the thread must equal the torque produced by the friction in order for the apparatus to not rotate.

    What do you get from that?

    After you have solved this problem, here's something else for you to ponder: What happens when θ is less than 60 degrees? Try it yourself with an actual yo-yo!
     
  4. Jan 27, 2013 #3
    Ok here we go
    1) Tcosθ = Ff
    2) Torque by friction = Torque by string
    I'm confused about what the moment arm of friction is.
    If we look at the 2nd image of the cylinder/disk system (cross-section)
    I understand that the distance between point of rotation (the point touching ground) is √3 so the torque = √3Tsinθ
    But what is friction's moment arm - wouldn't it have a zero moment arm if the point of rotation is the one touching the ground, since that is where friction is applied?
     
  5. Jan 27, 2013 #4
    I guess my question is: what is the point of rotation?
     
  6. Jan 27, 2013 #5
    Nope. The lever arm is the perpendicular distance from the axis of rotation to the point of force application.

    Once again, the axis of rotation is the point you should be concerned about, not the point touching the ground.
     
  7. Jan 27, 2013 #6
    Ok, so the center of rotation is the point in middle.
    Net torque equation is:
    (.02)Ff = .01Tsinθ
    2Ff = Tsinθ
    Combining this with the first equation:
    Ff = Tcosθ gives:
    2Tcosθ = Tsinθ
    tanθ = 2
    θ = 63.5 degrees -- this is not a valid answer.
     
  8. Jan 27, 2013 #7
    You are almost there. ".01Tsinθ" is wrong. θ is not the angle between r and the force, T. What is the angle between r and T? Try drawing it out (look at the diagram on the F=MA test).
     
  9. Jan 27, 2013 #8
    Is the angle between r and the force 90 degrees?
    Thus:
    .02Ff = .01T
    2Ff = T
    2Tcostheta = T
    2costheta = 1
    costheta = 1/2
    theta = 60 - This is the answer?

    I think my understanding of torque/moment arms is poor. Could you explain how to find the moment arm given a non-perpendicular force in general please?
     
  10. Jan 27, 2013 #9
    Good. r is perpendicular to the force.

    Yeah, I think the main problem you were having was with moment arm angles.
    Draw a line from the axis of rotation to the point of force application. The angle between that line and the line of force is the angle you should used.

    Things will become much more clear when you study vectors and learn about cross products. rsin(angle) is perpendicular distance between the axis of rotation and the point of force application; it should always at a right angle to the force vector.

    Here's a good explanation with diagrams:
    http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html
     
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