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Rotation of a uniform rigid disc about a fixed smooth axis

  1. Mar 4, 2013 #1
    1. The problem statement, all variables and given/known data
    A uniform circular disc has mass M and diameter AB of length 4a. The disc rotates in a vertical plane about a fixed smooth axis perpendicular to the disc through the point D of AB where AD=a. The disc is released from rest with AB horizontal. (See attached diagram)

    (a) Calculate the component parallel to AB of the force on the axis when AB makes an angle θ with the downward vertical.

    (b) Calculate the component perpendicular to AB of the force on the axis when AB makes an angle θ with the downward vertical

    2. Relevant equations
    KE = (1/2)Iω2
    I of a disc about axis perpendicular to disc through centre is (1/2)mr2
    Transverse component of acc. = r(dω/dt)
    Radial component of acc. = rω2

    3. The attempt at a solution
    I have done (a) okay with no problems and got the correct answer. I used a similar approach for part (b) but got the incorrect answer. For part (b):

    The moment of inertia is (1/2)M(2a)2 + Ma2 = 3Ma2
    (using the parallel axis theorem).

    Using conservation of energy:

    (1/2)(3Ma22 = Mgacosθ
    (3/2)aω2 = gcosθ

    Differentiating w.r.t. θ gives:

    3aω(dω/dθ) = -gsinθ
    3aω(dω/dt)(dt/dθ) = -gsinθ
    (dω/dt) = -(g/3a)sinθ *

    Resolving forces: Mgsinθ - Y = Ma(dω/dt)

    Substituting in *: Mgsinθ - Y = -(Mag/3a)sinθ

    Y = Mgsinθ + (1/3)Mgsinθ = (4/3)Mgsinθ

    The answer is supposed to be (2/3)Mgsinθ. I can't see what I have done wrong. Any help would be greatly appreciated!

    Attached Files:

  2. jcsd
  3. Mar 4, 2013 #2
    Resolving forces: Mgsinθ - Y = Ma(dω/dt)

    I don't like what I see here. You have the angular acceleration, not the linear acceleration.
    Maybe rather use

    [itex]\tau[/itex] = I [itex]\alpha[/itex] ?
  4. Mar 4, 2013 #3
    I don't understand what is wrong with resolving forces, I have used the transverse component of acceleration for circular motion which is radius*(dω/dt) and the forces acting on the disc which are parallel to that acceleration vector (as in the diagram)?

    However if I use τ = I(dω/dt) = 3Ma2(dω/dt)

    aMgsinθ = 3Ma2(dω/dt)
    gsinθ = 3a(dω/dt)
    (dω/dt) = (g/3a)sinθ

    Before, I got the negative of this - if I substitute this into my force equation however I do get the correct answer of (2/3)Mgsinθ.

    But you said that that equation is wrong - how else can I get an expression involving Y though?
  5. Mar 4, 2013 #4
    Ok, Y does not have a torque.
    Then rather try using the tangential acceleration in your equation:
    Resolving forces: Mgsinθ - Y = Ma(dω/dt)
    You can see that this equation of yours is dimensionally wrong.
  6. Mar 4, 2013 #5

    LHS: (kg*m*s-2) - (N) = N
    RHS: kg*m*(s-1)(s-1) = kgms-2 = N
  7. Mar 5, 2013 #6
    Sorry, yes, you did use the transverse component of the acceleration.
    You could setup

    [itex]\sum \tau[/itex] = I [itex]\alpha[/itex]

    about another point along AB.
    Last edited: Mar 5, 2013
  8. Mar 5, 2013 #7
    I believe the radian is not included in dimensional analysis.

    For example in τ = Iα, if you include the radian you would have:

    LHS = Nm
    RHS = kg*m2*rad*s-2 = (kg*m*s-2)*m*rad = Nm*rad

    so you must ignore the 'unit' radian.
  9. Mar 5, 2013 #8
    Yes, I edited my previous post.
  10. Mar 5, 2013 #9


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    Because of the way theta is measured, dω/dt is the acceleration anticlockwise (in your diagram). So the above equation should read Y - Mgsinθ = Ma(dω/dt).
  11. Mar 6, 2013 #10
    Oh, and by the same logic for τ = Iα,

    aMgsinθ = -3Ma2(dω/dt)

    which, along with the force equation, gives the correct answer.

    Thank you for your help!
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