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Urgent- a rotation question

  1. Apr 19, 2009 #1

    C.E

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    Hi I am really stuck on the following and have an exam on it in 2 days. Can somebody please help me. I have attempted it but think I have done it all wrong. (I have indicated how many marks each part of the question is worth at the side).

    a. A particle of mass m moves at constant velocity v relative to a point P with a closest distance of approach a. Find the angular momentum of the particle about P.[5]



    my attempt: mva, I don't think this is right as in my textbook it says to use mva if a is the closest distance of perpendicular approach which, it may not be.



    b. A belt drives the circumference of a cylindrical wheel of Radius R, mass M with no slipping. The tension in the belt is T. Friction in the wheel bearing causes an effective Torque N to act on the wheel.



    (i). If the velocity is constant what is T in terms of N.[5]

    my attempt: Friction= T (as acceleration=0)

    Therefore as N=TR, T=N/R.



    (ii). The bearing is lubricated removing all significant friction and the velocity of the belt increases with uniform acceleration a. Write T in terms of M, a and R.[15]



    my attempt: Force=Ma=T

    This is as far as I got and I am not even sure if this is right, can somebody please explain how to do this question?
     
  2. jcsd
  3. Apr 19, 2009 #2

    Doc Al

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    Staff: Mentor

    Sounds good to me. Since the velocity is constant, the point of closest approach must be where the velocity is perpendicular to the position vector.



    Sounds good.



    What's the torque on the cylinder? How does the acceleration of the belt relate to the angular acceleration of the cylinder?
     
  4. Apr 19, 2009 #3

    C.E

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    Is the acceleration of the belt (a)=R x acceleration of cylinder (c) so a=cR? And is the Torque just TR? If so, where do I go from here?
     
    Last edited: Apr 19, 2009
  5. Apr 19, 2009 #4

    Doc Al

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    Yes, since the belt doesn't slip, linear acceleration = angular acceleration x R.
    Yes.
    Apply Newton's 2nd law for rotation.
     
  6. Apr 19, 2009 #5

    C.E

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    Is this right then?
    Torque= moment of inertia x angular acceleration
    Hence TR=0.5MR^2 X a/R
    so T=0.5Ma
     
  7. Apr 19, 2009 #6

    Doc Al

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    Looks good.
     
  8. Apr 19, 2009 #7
    Dear C.E., please notice that Newton's second law is used for particles only. Hence, in problems 2, you should turn to dynamics of rotation and equilibrium law of mechanical moment.

    As to problem 1, I think it is not properly stated. If $m$ moves at constant velocity $v$ relative to the point $P$, the driving force must be perpendicular to the velocity of $m$ at any time, and there's no tangent component of the force, so the orbit must be a standard circle, rather than ellipse or anything else. Thus the radius of $m$ is always $a$. When we refer to the closest distance, it often means orbit like ellipse. I think you can tell your teacher this idea. On the other hand, even if for the ellipse, the velocity at the closest point is perpendicular to the position vector relative to $P$. Hence, probably the angular momentum of $m$ is $mva$.



    As to problem 2, since there's no graph for illustration, some confusion occurs, namely, the connecting point of the belt with the wheel, and the spatial angle of the belt (or the tension T) with the wheel. But, because the problem is easy in the whole, I think I can still give some constructive advice on the solution.

    The cylindrical wheel is not a particle, so "the velocity is constant" is to mean the the wheel advances via uniform rotation without slipping and the belt advances with constant linear speed. So, acceleration=0 leads to resultant of moments of forces being null, hence
    $$
    Friction \times arm of friction = Tension(T) \times arm of T
    $$
    the arm of force is relative to the contacting point (the tangent point) of the wheel with the ground. As stated, the friction is in the bearing, so the arm of friction is equal to the radius $R$. If the connecting point of the belt with the wheel is at the top, then the arm of T is $2R$, the diameter, twice that of friction. Hence, the friction is $T/2$ on this occasion. Hence
    $$
    N=(T/2) \times R
    $$
    $$
    T=2N/R
    $$
    This is just an example of formulating. Just make clear the contacting point of the forces and the arms of forces respectively, and using the method above.



    In b(ii), the acceleration of $a$ refers to the linear one measured by the belt. The acceleration of the belt is $a$, so the acceleration of the wheel is simply
    $$
    a'= a(wheel) = a(belt)/(2\pi R) = a/(2\pi R)
    $$
    The tension T is responsible for the acceleration of the wheel, so according to Newton's second law,
    $$
    a(wheel) = T/M
    $$
    Hence,
    $$
    a/(2\pi R) = T/M
    $$
    Hence,
    $$
    T= Ma/(2\pi R)
    $$



    I find that you know the concept of moment of inertia, so I would give some explanations.
    In the solution of b(ii) above, Newton's 2nd law is used. It is infact used for the centroid of the wheel.
     
    Last edited: Apr 20, 2009
  9. Apr 20, 2009 #8

    Doc Al

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    Staff: Mentor

    Newton's 2nd law can be applied to bodies (not just particles) for translation (∑F = ma) or for rotation (∑T = Iα, under suitable circumstances).

    There's nothing wrong with the statement of problem 1. It is given that the particle moves with constant velocity, thus there is no net force on it. It is certainly not moving in a circle.

    While a diagram would certainly be helpful, I think the simplest assumption is that the wheel rotates about a fixed axis. No reason to think that the wheel is rolling along the ground. C.E, please let us know if that is the case.
     
  10. Apr 21, 2009 #9
    --------------------------------------------------------------------------------

    Dear C.E., I would change my solution of problem b(ii). Sorry for my carelessness.

    case 1: As to problem 2(ii), if the wheel goes on the ground, then the acceleration of the belt is equal to the acceleration of the wheel (its centroid, to be exact). Hence, if the friction of theground is ignorable,

    Force=Ma=T.

    case 2: As Doc Al put it, the simplest assumption is that the wheel rotates about a fixed axis, its bearing. The acceleration of $a$ refers to the linear one measured by the belt. The acceleration of the belt is $a$, so the angular acceleration of the wheel is
    $$
    \beta= a(belt)/R = a/R
    $$
    The tension T is responsible for the anglar acceleration of the wheel, whose moment of inertia is
    $$
    L=(1/2)MR^2
    $$
    so according to the Generalized Newton's second law,
    $$
    T \times R = L \times \beta
    $$
    Hence,
    $$
    T \times R = (1/2)MaR
    $$
    Hence,
    $$
    T= Ma/2
    $$
     
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