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A board lies on two cylinders

  1. Jul 28, 2011 #1
    A board lies on top of 2 uniform cylinders that lie on a fixed plane inclined at an angle theta. The board has mass m and each of the cylinders has mass m/2. The system is released from rest. If there is no sipping between any of the surfaces, what is the acceleration of the board. Solve using force and torque.

    http://img192.imageshack.us/img192/4955/diagramwt.jpg [Broken]


    Since I have been having some problems using the torque method I figured ill use the Lagrange one. But the thing is, the answer I am getting through this method is wrong.
    I have tried using the Lagrange equations for that.
    where by, lets say the positions of the objects the left cylinder, right cylinder and board respectively as follows:
    (R[itex]\gamma[/itex]cos[itex]\theta[/itex] , H - R[itex]\gamma[/itex]sin[itex]\theta[/itex])
    (Scos[itex]\theta[/itex] + R[itex]\gamma[/itex]cos[itex]\theta[/itex] , H - Ssin[itex]\theta[/itex] - R[itex]\gamma[/itex]sin[itex]\theta[/itex])
    (xcos[itex]\theta[/itex], H - S/2 sin[itex]\theta[/itex] + Rsin[itex]\theta[/itex] -Xsin[itex]\theta[/itex])

    [itex]\gamma[/itex] being the angle of which the cylinder rotates

    after integration I get the velocities.
    then I plug into L=T - V, and from that to derive the equation of motion.

    Which in the end I get the acceleration to be a=3/2gsin[itex]\theta[/itex]
    Whereas it should be a = 6gsin[itex]\theta[/itex]

    I would very much like help with both of the methods.

    Thanks
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 28, 2011 #2

    kuruman

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    Your explanation of what you did confuses me. I am not sure exactly what you integrated to find the velocities. Can you be more specific?

    I hope you realize that the acceleration of the board as well as its instantaneous velocity is twice those of either cylinder.

    If you want to use forces and torques, you need to draw some free body diagrams.
    If you want to use the Lagrangian method, you need to explicitly (a) declare your generalized coordinates and (b) write down your constraints before proceeding to write T - V.
     
  4. Jul 28, 2011 #3
    Last edited by a moderator: May 5, 2017
  5. Jul 28, 2011 #4

    kuruman

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    I don't think you will get very far if you take derivatives "to get velocities and then plug in."

    How about drawing three separate FBDs, one for each cylinder and one for the board. Don't forget static friction; without it the cylinders and board will just slide and not roll!
     
  6. Jul 29, 2011 #5
    Hey Kuruman,
    Here is the diagram of the forces:
    http://img836.imageshack.us/img836/7909/96813713.jpg [Broken]

    so all in all, I have on the board (up being y positive, and right being x positive):

    X: maboard x = mgsin[itex]\theta[/itex] + 2f2
    Y: maboard y = 2N2 - mgcos[itex]\theta[/itex] = 0

    X: (m/2)acylinder x = (m/2)sin[itex]\theta[/itex] - f1 -f2
    Y: (m/2)acylinder y = N1 - N2 - (m/2)sin[itex]\theta[/itex] = 0

    and now I know that:
    I[itex]\omega[/itex] = R(f1-f2)
    and that:
    2acylinder x = aboard x (As you well reminded me :) )

    Am I good so far?
     
    Last edited by a moderator: May 5, 2017
  7. Jul 29, 2011 #6

    kuruman

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    You are sort of good so far. I have marked in red what needs fixing. The first (sinθ) doesn't really matter because you are concerned with the x-direction only.

    The torque equation (left side) needs serious fixing. What is Newton's second law for rotational motion?
     
  8. Jul 29, 2011 #7
    Whooops..... LOL
    Sorry for that, I did mean
    I[itex]\alpha[/itex] = R(f1-f2).

    I = 1/2[(m/2)R2]

    Since I am assuming positive rolling direction is clockwise then:
    acylinder x = - r[itex]\alpha[/itex]

    so my equation then becomes:
    -acylinder xm/4 = f1-f2

    correct?
     
  9. Jul 29, 2011 #8

    kuruman

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    Why the minus sign? Here acylinder x is the magnitude of the acceleration and is positive. The other side of the equation must also be positive.
     
  10. Jul 29, 2011 #9
    Hey,
    I am a bit confused with the signs, to tell the truth.
    I was taught that arel = R[itex]\alpha[/itex].
    So in relation to the ground, that is:
    acylinder - aground = R[itex]\alpha[/itex].
    But then if I try to solve in relation to the board, that is:

    acylinder - aboard it cannot be equal as well to R[itex]\alpha[/itex], because then I will get that aboard = 0.

    That is why I put the minus sign in the beginning to make sense a bit, that is to get
    2acylinder x = aboard x
     
  11. Jul 29, 2011 #10

    kuruman

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    Refer all accelerations relative to the ground, else you will be confused. The angular acceleration is into the screen and that defines the positive direction that yields increasing angular velocity in the clockwise direction. The net torque on the cylinder is also into the screen. It's gotta, by Newton's 2nd law the net torque and the angular acceleration are always in the same direction.
     
  12. Jul 29, 2011 #11
    but then, what is the equation with relation to the board?
    What is the correct way to look at it in order to indeed get that:
    2acylinder x = aboard x.
    or does it not even matter?
     
  13. Jul 29, 2011 #12

    kuruman

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    Look instantaneously at one cylinder as it rolls. There are infinitely many points on it. We are interested in only three.

    Point 1. The point in contact with the ground.
    It is at rest with respect to the ground (no slipping) therefore its acceleration is zero relative to the ground.

    Point 2. The center of the cylinder. It has acceleration acylinder relative to point 1. Let α be its angular acceleration. Relative to point 1, acylinder = α R.

    Point 3. The point in contact with the board.
    It too has angular acceleration α relative to point 1, same as the center because the cylinder is a rigid body. Its linear acceleration is a3 = α (2R). But notice that since the board does not slip on the cylinder, it must have the same acceleration as point 3. Therefore, the board's linear acceleration is also α (2R) which is twice the acceleration of the cylinder.
     
  14. Jul 29, 2011 #13
    WOW!
    Thanks for explaining it to me. It straightens out my head enormously.
    Alright, so getting back to the equations:

    I[itex]\omega[/itex] = acylinderI/R = macylinder/4 = f1 -f2
    giving :(1) macylinder + 4f2 = 4f1
    then since:
    (m/2)acylinder = (m/2)gsinθ - f1 -f2
    I will multiply it by 4 to get:
    2macylinder = 2mgsinθ - 4f1 -4f2
    and using
    2acylinder x = aboard x
    and plugging in (1)
    I then get:
    3maboard x = 2mgsinθ -8f2
    adding this equation to 4 times the initial force equation on the board I get:
    7maboard x = 6mgsinθ giving:
    aboard x = 6/7 gsinθ.
    Does it make sense to get this value?
     
  15. Jul 29, 2011 #14

    kuruman

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    I did it in two different ways and I didn't get (6/7) g sinθ for the acceleration of the cylinder. Check your algebra.
     
  16. Jul 29, 2011 #15
    I found my error.
    a = 12gsinθ/11.
    Correct?
     
  17. Jul 29, 2011 #16

    kuruman

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    Which acceleration is this, board or cylinder? The cylinder cannot have acceleration greater than g sinθ, but the board can.
     
  18. Jul 29, 2011 #17
    this is for the board
     
  19. Jul 29, 2011 #18

    kuruman

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    Then it is correct. :approve:
     
  20. Jul 29, 2011 #19
    AWESOME!!!!!
    Thank you SO much for your patience. I do have an upcoming test which most definitely will include this kind of question. I tried to understand this subject for like a month, and in a few hours doing this question have proven VERY valuable.

    Thank you!!!
     
  21. Jul 31, 2011 #20

    kuruman

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    Addendum: Understanding this problem conceptually

    The board's acceleration, (12/11)g sinθ is greater than g sinθ, what it would be if the board were sliding down a frictionless incline. This counterintuitive result may be understood if we consider three general observations:

    1. The force of static friction adjusts itself to provide the observed acceleration.
    2. Simple FBD analysis shows that, for any freely rolling object (no board) the linear acceleration of point "3", the point diametrically opposite to the point of contact, is given by
    [itex]a_{3}=\frac{2}{1+\frac{I_{CM}}{mR^{2}}}gsin \theta[/itex]

    where ICM is the moment of inertia of the roller about its CM.
    3. For rolling without slipping, the linear acceleration of a board placed on rollers must match a3.

    For a hoop, ICM = mR2 and a3 = g sinθ. If we place a board on rolling hoops, the acceleration provided by gravity to the board is g sinθ and already matches a3. Therefore, no static friction is required in this case.

    For any roller other than a hoop, ICM < mR2 and a3 > g sinθ. In such cases, some force of static friction is required to match the accelerations. The static friction force exerted by the board on the roller will be "uphill" to diminish a3. The reaction counterpart of this will act on the board downhill and will increase the acceleration g sinθ that is already provided by gravity.
     
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