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Energy - Rotation

  1. Jan 14, 2006 #1
    Hi, I'm having trouble with a question. Can someone please help me out?

    The scenario is I've got a uniform rod off mass M and a ball (particle) of mass m. They are connected together (so there's a rod with a ball at one end) to form a link called "2." The link 2 is of length L.

    The link 2 is hinged to a circular platform (the shape of the platform seems irrelevant, I think it's just to make the situation easier to visualise) at the other end (the end of the rod without the particle attached).

    The circular platform rotates at a constant angular velocity of w (omega). The angular motion of the link 2 is determined by the angular displacement [itex]\beta(t)[/itex], where beta is the angle between the link 2 and the vertical.

    I need to determine the kinetic energy stored in the link 2.

    Answer: [tex]T = \frac{1}{2}m\left[ {\left( { - R\omega \sin \left( {\omega t} \right) + L\mathop \beta \limits^ \bullet \cos \left( {\omega t} \right)} \right)^2 + \left( {R\omega \cos \left( {\omega t} \right) + L\mathop \beta \limits^ \bullet \sin \left( {\omega t} \right)} \right)^2 } \right]
    [/tex]

    In my first attempt I obtained the expression above except that in the terms with (d/dt)(beta) I had beta in the argument of the trig function instead of wt.

    Anyway, what I'm having problems with are the energy considerations. I am looking for an expression for the kinetic energy of the link 2. So I start off by isolating it from the rest of the set up and consider it's motion alone.

    From the given information I would say that there is only rotational kinetic energy stored in the link 2. It doesn't seem to be 'translating' so I'd say that there is no (1/2)mv^2 component in the kinetic energy, only a (1/2)Iw^2 component.

    That's about all I've got to work with right now. I'm fairly sure I've missed a lot of things because the expression given in the answer looks like something of the form (1/2)mv^2. Also, the the mass of the rod (M) in the link doesn't appear in the answer whereas the mass of the particle (m) does.

    I'm really unsure about what to do in this question. Can someone please help me out?

    Edit: I think that the kinetic energy T will be:

    [tex]
    T = \frac{1}{2}m\left| {\mathop {v_G }\limits^ \to } \right|^2 + \frac{1}{2}I_G \mathop \beta \limits^ \bullet ^2
    [/tex]

    So I need to determine the COM of the link 2 (which consists of a rod of mass M and a particle of mass m). This seems rather difficult because the particle is just a dot. It's not like having two blocks where I can determine their individual COMs and then find the common COM.
     
    Last edited: Jan 14, 2006
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  3. Jan 14, 2006 #2

    arildno

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    So, the disk is rotating in the horizontal plane, whereas the rod is hanging beneath it?
     
  4. Jan 14, 2006 #3
    Sorry I should've made it more clear.

    I'll try explaining the situation in the following way and hopefully it makes things more clear.

    Place a disk on the XY plane such that the disk is symmetric about X and Y axes. A rod is then attached (at one end) at a point "A" on the disk. The point A is a distance R from the the centre of the disk.

    The rod is free to rotate about the point A and it's angular displacement with respect to the vertical (a vertical line through point A) is given by b(t). The disk itself also rotates, but with angular displacement given by wt.

    I don't think that gravity is to be considered in this question. So for the purposes of the question it's probably acceptable to consider the disk as being placed on the ground or a table so that gravity is not a factor in any of the calculations.

    Note: The link that I referred to in my earlier post is composed of the rod (mass M) + particle (mass m).

    I think most the problem is just a bit of geometry. The main problem I'm having is determining the 'types' of energy that are involved in determining the kinetic energy "stored" in the link during the rotation of the entire system of disk + link.
     
  5. Jan 15, 2006 #4

    arildno

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    Now, is the rod constrained to oscillate in the plane defined by the disk's radial vector at A and the vertical, or is it constrained to oscillate in the plane defined by the vertical and the disk's tangential vector?

    Or, thirdly, is the hinge a ball hinge, so that the rod may also rotate in a plane parallell to the XY-plane (in addition to its rotation with respect to the vertical, b(t)) ?
     
  6. Jan 15, 2006 #5
    The vertical is just a reference for the angular displacement b(t). The disk lies on the XY plane and rotates on that plane. The rod is attached directly on top of the disk so that it rotates on a plane which is 'parallel' to the plane of rotation of the disk. The problem is a 2D question so all rotations are on planes which are parallel to the XY plane.

    Another way to look at the problem is to simply draw a circle symmetrically about the XY axes on a piece of paper. The circle is free to rotate. Next select a point near the circumference of the circle and draw a rod so that the lower end of the rod is such that the rod makes an angle b(t) with the vertical. b(t) is the angular displacement of the rod.
     
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