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Homework Help: Torque and Moment of Inertia

  1. Apr 27, 2005 #1
    What would be the acceleration of a yo-yo if these are my datas:

    m= 10.456g
    r(small) = 0.7mm
    R= 1.9mm

    Now on I've had accelerations varying from 0.006m/s² to 930m/s² -yeah right...

    Worked on that prrety all of the time today and all I got left is a few so that I'm really in a hurry right now.
  2. jcsd
  3. Apr 27, 2005 #2
    You haven't enough data, as far as I can tell.
  4. Apr 28, 2005 #3

    Andrew Mason

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    Think of the torque of the centre of mass of the yo-yo about the point of tangential contact with the string:

    [tex]\tau = mgr_{small} = I\alpha[/tex]

    where I is the moment of inertia of the yo-yo about that small radius. Use the parallel axis theorem to work that out.

    Use the relationship between [itex]\alpha[/itex] and acceleration to find the acceleration of the centre of mass.

    Are you sure these dimensions are mm and not cm? That is one awfully small yo-yo!

  5. Apr 28, 2005 #4


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    {Yoyo Small Radius} = r = (0.7 mm) = (7e(-4) meters)
    {Yoyo Large Radius} = R = (1.9 mm) = (1.9e(-3) meters)
    {Yoyo Mass} = (10.456 g) = (1.0456e(-2) kg)
    {Yoyo Moment of Inertia about CM Axis} = I = (1/2)*m*(R^2)
    {Force Acting Thru CM} = F = m*a
    {Gravitational Force on Yoyo} = m*g
    {String Tension on Yoyo} = T
    {Torque from String Tension about CM Axis} = τ = r*T

    For Linear Motion of CM:
    F = m*a =
    = T - mg
    ::: ⇒ a = (T/m) - g ::: <---- Eq #1

    For Rotational Motion about CM axis:
    τ = I*α = (1/2)*m*(R^2)*{-a/r} =
    = r*T ::: <---- Torque from tension "T" acting at radius "r"
    ::: ⇒ T = -(1/2)*m*a*(R/r)^2 ::: <---- Eq #2

    Placing Eq #2 into Eq #1:
    a = -(1/2)*a*(R/r)^2 - g
    ::: ⇒ a*{1 + (1/2)*(R/r)^2} = -g

    Solve for "a" in terms of "R" and "r" (and "g"), values for which are given in the problem statement.

    Last edited: Apr 28, 2005
  6. Apr 28, 2005 #5
    something is wrong with your calculation
    I think T is confused with τ and R^2*-a/r is derived into a*(R/r)^2
    otherwise the answer makes sense either way (r² or r)
    lastly, you're corect sir Mason, I did translate my cm in mm AND add the mm sign therefore doing the process twice.
  7. Apr 28, 2005 #6

    Andrew Mason

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    Ok. So here is how I see it (which is equivalent to xanthym's approach):

    [tex]\tau = mgr = I\alpha = I(a/r) = (\frac{1}{2}mR^2 + mr^2)\frac{a}{r}[/tex]

    [tex]a = \frac{gr^2}{(\frac{1}{2}R^2 + r^2)} = \frac{2g}{(\frac{R^2}{r^2} + 2)}[/tex]

    I get a = 2.1 m/sec^2

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