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Stick on a table

  1. Nov 23, 2015 #1
    1. The problem statement, all variables and given/known data
    A stick of length L and mass m lies on a frictionless table.A force parallel to the table top,[itex] \vec{F} [/itex],is applied to one end of the stick for a very short time [itex]\int F \, dt[/itex][itex]=[/itex][itex]\vec{I}[/itex],the"impulse".Choose a convenient set of two generalized coordinates.Interms of your choices of generalized coordinates,what are the generalized forces when [itex]\vec{F}[/itex][itex]\neq[/itex] 0?What is the subsequent motion of the stick?

    2. Relevant equations
    [itex]\frac{d(\frac{\partial L}{\partial\dot{x}})}{dt}-\frac{\partial L}{\partial x}=0 [/itex]
    where L is the Lagrangian of the system L=T-V.
    [itex]\sum_{i=0}^M \vec F_{i}\cdot \frac{\partial \vec r_{i}}{\partial q_{k}}=f_{k}[/itex]
    where [itex]f_{k}[/itex] is the generalized force and [itex]\vec F_{i}[/itex] are the non-constraint forcesand[itex]q_{k}[/itex] is the generalized coordinate.


    3. The attempt at a solution
    Actually I couldn't obtain 2 degrees of freedom.I think they must be 3,which are θ,Φand z.
    But I supposed that θ is fixed.So the degrees of freedom are now "z" and "Φ".
    z isthe distance between the origin O and the lower end of the stick. Φ is the angle between the stick and z as shown in the figure below.
    [itex]d\vec r_{i}=(z-dl\ cos\phi)\vec e_{r}+{dl}\ sin\phi\vec e_{\theta}[/itex]
    and [itex]\vec {r}=(z-l\cos\phi)\vec e_{r}+l\sin\phi\vec e_{\theta}[/itex]
    I reached here, and then I didn't know how to proceed.
    Any hints?
     

    Attached Files:

  2. jcsd
  3. Nov 23, 2015 #2
    There are only two degrees of freedom.
    I suggest you consider the stick's center of gravity.
     
  4. Nov 24, 2015 #3
    Do you mean to consider the stick as a point mass which is the center of gravity, so the degrees of freedom will reduce to "z" and "θ" ?
    But in analytical mechanics we deal with the system as a whole, so we should consider all the parts of the system.
     
  5. Nov 24, 2015 #4

    DrClaude

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    Staff: Mentor

    Once the impulse is over, what will be the motion of the stick? Answering this will give you both what the origin should be (see .Scott's hint) and why there are only two relevant coordinates.
     
  6. Nov 24, 2015 #5
    Say the ends of the stick are at (x,y) = (-1,0) and (+1,0), the impulse is applied to (-1,0) and the center of gravity (COG) is at (0,0). Tell me what the impulse vector would need to be to sent the stick's COG in the +y direction.
     
  7. Nov 25, 2015 #6
    I guess I got the solution.
    Let [itex]O[/itex] be the the origin of the edge of the table and [itex]O_{1}[/itex] be the origin of the movable reference frame [itex]O_1\vec i_{1}j_{1}[/itex].
    [itex]O_{1}[/itex] is the initial position of COG of the stick,so [itex]\overrightarrow {OO}_{1}[/itex] is a fixed vector.
    The impulse acting on the stick is divided into two parts;
    ⇒ The impulse which is perpendicular to the end of the stick which is responsible for the rotation.
    ⇒The impulse which is along the direction of the stick and this impulse is responsible for the linear motion of the stick along its direction.

    Now let [itex]x_{1}[/itex] be the coordinate of COG along the direction [itex]\vec i_{1}[/itex] which is a movable coordinate rotating with angular velocity[itex]\dot{\phi}[/itex] which is the angular velocity of the stick and[itex]\phi[/itex] is the angle between the stick and [itex]\vec X[/itex].I note that O is the origin of the fixed reference frame [itex]O\vec X\vec Y[/itex].

    So [itex]\vec R[/itex] (the position vector of the cog of rod)=[itex]\overrightarrow {OO}_{1}+x_{1}cos\phi\vec X +x_{1}sin\phi\vec Y[/itex].
    so I can see here that the two coordinates [itex]x_{1}\thinspace and\thinspace\phi[/itex] are enough to determine the whole configuration of the system.
     

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    Last edited: Nov 25, 2015
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