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Condition of rotational equilibrium in circular motion

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider a cyclist turning around a circular track with radius r and speed v as shown
    in attached fig.
    Let W be the weight of the system (cyclist + bicycle) (W = mg)
    N = normal force by ground on the system
    f = static friction by ground on the system
    θ = angle of inclination between cyclist and the vertical
    h = height of the centre of mass C of the system above the ground
    d = perpendicular distance from the point of contact between the ground and
    the tyres to the line of application of the weight

    Horizontally:
    f = m v2 / r

    Vertically , for equilibrium:
    N = mg

    for rotational equlibrium, taking moment about the centre of mass (C), we get:
    N d = f h

    My question is : For rotational equilibrium, can't we take moment about the point of contact of cycle with track? As far as I know, for rotational equilibrium, moments can be taken about any point and their summation must be zero.
    However, in this case, moment about the point of contact can't be zero as W produces a torque while torques due to N and f are zero. So the condition of rotational equilibrium gives two different conditions for two different points. Where is the fallacy?

    2. Relevant equations

    Moment about the bottom most point :
    moment = W x d
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 23, 2011 #2

    Doc Al

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    Staff: Mentor

    The problem with finding moments about the point of contact is that such a point is accelerating. The CM has the special property that the torque about it equals the rate of change of angular momentum in all cases. Not so for accelerating points in general.
     
  4. Oct 23, 2011 #3
    Thank you so much for the information sir. :smile:

    However, I have some doubts :
    1)Why can't the condition of rotational equilibrium be applied about an accelerating point and how do we take moments about such points?
    2)How does the special property of CM (i.e ζ = dL/dt) imply that the condition of rotational equilibrium can be applied about it?
    3)For translational motion, are there any such points from where the condition of translational equilibrium not be applied (i.e ƩF=0). My guess is the case of non-inertial frames.

    Also, can you please suggest some internet resource or textbook (preferably net resource) from where detailed theory related to this concept can be read?

    Thanks
     
  5. Oct 23, 2011 #4

    Doc Al

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    Staff: Mentor

    You cannot use your intuition based on solving problems without acceleration and apply it directly to accelerating cases. (If that's what you're thinking. The non-accelerating case is a special case of the the more general situation.)
    Without going into a detailed derivation, I'll just say that the CM is a special case. When you use an accelerating point as your reference for calculating torque and thus dL/dt, there's an extra term in your equation (essentially the centrifugal pseudoforce); that extra term goes to zero if the accelerating point is the CM. (Check out any intermediate Classical Mechanics book for a derivation.)
    Sure. If the object is accelerating, why would you expect ƩF=0?
    I don't know of any online reference (though I would be surprised if there weren't several, if you can find them). Look for lecture notes on classical mechanics.

    What textbooks do you have access to? (In case I have it also.)
     
  6. Oct 23, 2011 #5
    Do you mean that we calculate torque using r x F definition and then dL/dt. But in general, they aren't equal. However, in the non-accelerating case, they are.

    I have Goldstien and feynman lectures.

    Thanks a lot for helping..:smile:
     
  7. Oct 23, 2011 #6

    Doc Al

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    Staff: Mentor

    Exactly.
    I should have both of those, so I'll poke around a bit when I get the chance.
     
  8. Oct 23, 2011 #7
    ok..thanks again!
     
  9. Oct 23, 2011 #8

    Doc Al

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    Staff: Mentor

    I did check those two references (albeit quickly). Unfortunately, neither cover that issue. Goldstein is a bit too advanced and Feynman a bit too elementary. (That's why I suggest an intermediate level Classical Mechanics text.) David Morin's book covers the topic well. (But keep poking around online and you'll eventually hit upon a resource.)
     
  10. Oct 23, 2011 #9
    There is no fallacy. As Doc Al pointed out, the system is accelerating.
    In an accelerating system the moments about a point other than cm are not 0(zero), but M(effective).

    You can find M(eff) by integration. Take an element of mass dm and find its force(s). Sum all elemental masses dm to find the M(eff) and F(eff) . In this way you can takes the moments about any point. Nonetheless, such an approach can involve complicated calculations, depending upon the shape of the body. You may want to look up d'Alembert's principle.

    Note that when we use the cm to do calculations, the result is an approximation, for any body not symmentrical about the axis of rotation. The reason being is that the radius r of rotation changes for each element of the body. The effective force will not reduce to a vector attached to the cm rotating about the axis. In most cases the difference is small and we use the cm as the location for the effective force to simplify the calculations.

    As an example, consider an airplane making a banking turn. The airplane has an axis of symmetry: two elements of mass equally distant from the centre of the airplane will produce an effective force not equal to each other.
    For the higher up wing : F1 = dm r1 omega^2
    For the lower wing : F2 = dm r2 omega^2
    where r1 > r2

    Solving for the effective forces will produce a force and couple at cm of the plane, which can be replaced by an effective force off-centre from the cm.
     
  11. Oct 24, 2011 #10
    thanks for the help 256 bits.:smile:
     
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