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Derivation of Moment of Inertia: Why is centripetal acceleration ignored?

  1. Oct 13, 2009 #1
    Moment of inertia, I is the resistance of an object to rotational motion. It is the rotational analogue of mass and is given by the following integral, I = [tex]\int[/tex]miri2 . A simple derivation was explained to me as follows.

    A moment, τ is the product of force, F and lever arm, r.
    τ = F*r

    Newton's second law gives force, F as the product of mass, m and linear acceleration, a.
    F = m*a

    This can be substituted into the first equation as follows.
    τ = F*r = (m*a)*r

    If a particle in a solid body is rotating in a circle (because of an applied moment, τ) then we know it's tangetial acceleration, a_tang is given as below.
    a_tang = r*α

    Now, substitute this value into the previous equation to give.
    τ = (m*a)*r = (m*(r*α))*r = m*r2

    Hence, it is seen that the moment, τ on the particle is proportional to the angular acceleration, α and the constant of proportionality is the term m*r2. For a whole object this constant of proportionality would be the sum of this term over all it's particles, that is, the integral [tex]\int[/tex]miri2 . This constant of proportionality determines the magnitude of the resistance to rotational motion and is termed the moment of inertia, I.

    Now, I don't understand the following. For a particle rotating in a circle in a rigid solid, it's acceleration, a is a vector which can be decomposed into 2 components. If we use polar co-ordinates then the acceleration can be decomposed into the radial acceleration, a_radial (which I previously called the 'centripetal acceleration') and the tangential acceleration, a_tang. It is well known that for a point rotating in a circle these have the following values; a_tang = r*α and a_radial = r*ω2 = v2/r . So why do we only substitute the tangetial acceleration, a_tang into the term τ = m*a*r ? Why is the radial acceleration, a_radial ignored? I understand that the radial acceleration does no work because there is no displacement in the radial direction. However, I do not understand how this answers my question?


    Furthermore, if you can answer that question, I'm sure you know the answer to this related problem. Image a cylinder lying on a rough surface that is horizontal. Since the surface is horizontal the normal force, N is equal to the weight of the cylinder, m*g.
    N = m*g
    The horizontal surface is rough with an associated coeffecient of friction, μ. Now, a moment, τ is applied to the cylinder and it begins to rotate and move towards the right in pure roll. As such, there is a fricton force, F_fric which acts towards the left.
    F_fric = μ*N
    Again, this frictional force does no work, because the point in contact with the ground is always instantanesouly at rest (the centre of the cylinder has a velocity to the right of r*ω and the circumfrence point vertically above the point in contact with the surface has a velocity to the right of 2*r*ω). Now, what balances the frictional force? I can not find any other horizontal force so why doesn't the cylinder slide to the left? Instead, we observe it to roll to the right. Have I misunderstood the concept of the 'standard model of friction'?
     
  2. jcsd
  3. Oct 13, 2009 #2
    No displacement, - no acceleration. The centripetal force is completely compensated with an elastic force, so the resulting radial acceleration is taken into account as zero.
     
    Last edited: Oct 13, 2009
  4. Oct 13, 2009 #3

    Doc Al

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    :bugeye:
     
  5. Oct 13, 2009 #4
  6. Oct 13, 2009 #5

    Doc Al

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    Do you really think a point mass going in a circle has zero radial acceleration?
     
  7. Oct 13, 2009 #6
    Yes, I do.
     
  8. Oct 13, 2009 #7

    Doc Al

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    A better definition of moment is:
    [tex]\vec{\tau} = \vec{r} \times \vec{F}[/tex]

    Only the tangential component of the force on the particle/body contributes to the moment. (But a point mass going in a circle most certainly has a radial component of acceleration and thus a radial component of force acting on it!)
     
  9. Oct 13, 2009 #8

    Doc Al

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    OK.
    Don't assume that the static friction equals μ*N. That's the maximum value of static friction--the actual friction is just what is needed to prevent slipping between the surfaces.
    Can you be more specific in your example? You said you exerted a torque on the cylinder to start it rolling, so you must have exerted a force somewhere. As far as friction goes, what do you mean by "what balances the frictional force"?
     
  10. Oct 13, 2009 #9
    But why then it does not fly away or even move along radius?
     
  11. Oct 13, 2009 #10

    Doc Al

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    Uh... just because it's accelerating radially does not mean it's moving radially! You might want to review Newton's 2nd law and circular motion--pronto!
     
  12. Oct 13, 2009 #11
    I am studying the Newton's first law this year. The second one will be the next year.
     
  13. Oct 13, 2009 #12

    Doc Al

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    Good plan. Pace yourself. (Seriously--review what you actually wrote. Perhaps you meant to say something else.)
     
  14. Oct 13, 2009 #13
    It's ignored because centripetal acceleration does not contribute to torque. It acts in the same direction as the lever arm (r), so it has zero contribution to the torque. Tangential acceleration, however, acts 90 degrees to the lever arm, so it contributes to torque.
     
  15. Oct 13, 2009 #14

    Vanadium 50

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    I think the problem is one of language. Bob for Short is, I think, using the term radial acceleration to mean [itex]\ddot{r}[/itex]. Everyone else is using it to mean acceleration in the [itex]\hat{r}[/itex] direction.
     
  16. Oct 13, 2009 #15
    If one counts r from any place different from the center of masses, then there is a radial acceleration too but who cares?
     
  17. Oct 14, 2009 #16
    Ok, I understand the solution to the first problem because I forgot to use the correct definition of a moment which is, τ = r x F.

    For the second problem I have something to clarify. A torque, τ is exerted on a solid object on a horizontal and rough surface. The torque, τ is a couple. That is to say it is there because 2 equal and oppositely directed forces do not act co-linearly, instead they act on two parrellel lines of action seperated by a perpendicular distance, d. Thus, these 2 forces (one force is F, the other force is -F) create a moment, τ which is the same about any point in the plane and given by;
    τ = F x d
    Note, there is no net force created by these two forces but there is a net moment. Because, this moment is the same about any point in the plane it is given the special term 'couple'.

    Again, there is a normal force, N which equals the weight of the object, m*g. If the object is in pure roll, surely there must be a friction force, fr to prevent it sliding? If so, what balances this friction force?
     
  18. Oct 14, 2009 #17
    The observer who is stationary w.r.t that place
     
  19. Oct 14, 2009 #18

    Doc Al

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    I'm not sure why you are talking about couples. You can certainly have a torque without having a couple.

    If the cylinder is rolling without slipping on a horizontal surface at constant speed, no friction is required. If you are talking about a different situation, where you are exerting some kind of couple on the cylinder while its in contact with the ground, then a static friction force is required to prevent slipping. Nothing "balances" this friction force--the cylinder will accelerate.
     
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