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Rotational kinematics

  1. Jan 12, 2005 #1
    1. A uniform rod of length 3m is suspended at one end so that it can move about an axis perpendicular to its length. The moment of inertia about the end is 6kgm^2 and the mass of the rod is 2kg. If the rod is initially horizontal and then released, find the angular velocity of the rod when
    i) it is inclined at 30 degrees to the horizontal,
    ii) reaches the vertical.

    oh no!! I don't have a clue of how to figure out this problem....

    2.) A flywheel with an axis 1.0m in diameter is mounted in frictionless bearings set in motion by applying a steadsy tension of 2N to a thin thread wound tightly round the axle. The moment of inertia of the system about its axis of rotation is 5.0 x10^-4 kgm^2. Calculate
    i) the angular accleration of the flywheel when 1m of thread has been pulled off the axle,
    ii) the constant retarding couple which must then be applied to bring the flywheel to rest in one complete turn, the tension in the thread having been completely removed.

    For i), I set: Tr = I alpha, where T is the tension, and r is the radius of axis
    2 x 0.5 x 10^-2 = 5 x10^-4 alpha
    angular accleration = 20 rad/s^2
    ummm....it is correct, but I haven't considered the length of thread being "1m", so what should I do if 2m of thread has been pulled off the axle. Is the answer the same?

    For ii), I know that for calculating couple, the equation is force x distance, i.e. 2 x 0.5 x10^-2....but it just doesn't make sense in this question
     
  2. jcsd
  3. Jan 13, 2005 #2
    1: use conservation of energy.... can you see how energy related to the angle[tex] \theta [/tex] ?
    2:a
    if you didn't consider the length 1m, where is your o.5 came from :confused:
    2:b what the hell is COUPLE?
     
  4. Jan 13, 2005 #3

    cepheid

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    If I remember right, a couple or "moment couple"... refers to forces that occur in opposing pairs. Any moment (ie torque) can be replaced by a pair of opposing forces, equal in magnitude, and opposite in direction. Obviously the forces lie in the plane of rotation, each equidistant from (and on opposite sides of) the point about which the torque is being applied. Vice versa...the replacement can happen in reverse too (any couple can be represented by the corresponding moment) => They are interchangeable.
     
  5. Jan 13, 2005 #4
    Vincent, I can solve 1.) now....Thank you for your help......well, I really haven't consider the 1m of thread, for 0.5 x 10^-2, it is the flywheel axis diameter


    For no.2, I am still stuck in it....>_< I try to use the equation like w^2 = 2as...but I guess angular acceleration stays teh same...so I don't need to calculate it again......
     
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