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Predict the position of a particle on a rigid body

  1. Nov 26, 2015 #1
    1. Problem Statement
    Assume there is an rigid object with mass m in 2D space, an impulse J = FΔt is applied at time t1 at the particle Pimp and Pimp is on the exterior boundary of the object. The impulse cause a free plane motion of the object and the object is only affected by the force of the gravity. Here I am interested in what is the position of a particular particle Pi of the object at a later time point t2.
    2-3. Relevant Equations and The Attempt at a Solution
    I developed a equation for this but not sure whether it is correct:
    \begin{equation}p_i(t_2) = p_i(t_1) + \int_{t_1}^{t_2} v^\prime + \omega^\prime \times (p_i(t) - p_r(t)) + \mathbf{a} t \,dt\end{equation}
    where:
    \begin{equation} v^\prime = v_0 + J/ m
    \\
    \omega^\prime = \omega_0 + I^{-1} \tau
    \\
    \tau = (p_{imp}(t_1) - p_{r}(t_1)) \times J
    \\
    \mathbf{a} = F/m + \tau / I \times (p_i(t) - p_r(t)) - \omega_0^2(p_i(t) - p_r(t))
    \end{equation}
    v0 and ω0 are the initial linear velocity and angular velocity respectively.
    pr refers to the position of the particle around which the object rotates.

    is the equation correct? It is not a homework but related to my research. I am from computer science and do not have enough physics. Please help..
     
    Last edited: Nov 26, 2015
  2. jcsd
  3. Nov 26, 2015 #2

    andrewkirk

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    I don't think that's possible, unless the initial net torque is zero. That's because a nonzero net torque will make the body rotate so that, if the direction of the force remains constant, the angle of the force to the direction from point of application to the body's centre of mass will change, thereby changing the torque. On the other hand, if the direction of the force rotates to keep the torque constant, the force vector is not constant.
     
  4. Nov 26, 2015 #3
    Thank you for your reply.
    I changed the assumption to " the object is only affected by the force of gravity". does the equation make sense?
     
  5. Nov 26, 2015 #4

    andrewkirk

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    An impulse is the result of applying a force over a period. What is the period over which the force is applied, and what is the relationship of that period to the period ##[t_1,t_2]##?

    In the 3rd equation in your part 2, it looks like you are still assuming that the force is always applied at right angles to the line connecting the point of application to the centre of mass. For the reason outlined in post 2, that is unlikely to be correct.
     
  6. Nov 26, 2015 #5
    Sorry it made you confused. what i want to say is, there is an impulse J applied at a time point t1 and lasts for Δt, and I am interested in the position of a particle at a later time point t2 after the impulse is applied. Initially I thought Δt is very short (i.e. t1 + Δt ≈ t1) so I ignored the Δt in the equation. Now I updated the equation:
    \begin{equation}p_i(t_2) = p_i(t_1 + \Delta t) + \int_{t_1 + \Delta t}^{t_2} v^\prime + \omega^\prime \times (p_i(t) - p_r(t)) + \mathbf{a} t \,dt\end{equation}

    Thank you for pointing it out.


    In the 3rd equation, × refers to a cross product of vectors. I am not assuming "the force is always applied at right angles to the line connecting the point of application to the centre of mass."
     
  7. Nov 26, 2015 #6

    andrewkirk

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    That makes it clearer.

    With that explanation, the term ##\mathbf{a}t## can be removed from the integral, as there is no acceleration in the time period ##[t_1+\Delta t,t_2]##.
    Accordingly, the last equation in (2) can be discarded.

    ##\tau## is the angular impulse that was imparted in the period ##[t_1,t_1+\Delta t]##. It is not equal to ##(p_{imp}(t_1) - p_{r}(t_1)) \times J## but rather to

    $$
    \int_{t_1}^{t_1+\Delta t}(\vec p_{imp}(t) - \vec p_{r}(t)) \times \frac{d\vec J(t)}{dt}dt
    $$

    which is equal to

    $$
    \int_{t_1}^{t_1+\Delta t}(\vec p_{imp}(t) - \vec p_{r}(t)) \times \vec F(t)dt
    $$

    where ##\vec F(t)## is the force applied at time ##t##.
     
  8. Nov 26, 2015 #7

    looks great ! thanks.
    I am just wondering what will happen if the object is affected by a constant force through the motion, In this case, we have to include a. is it like:
    \begin{equation}
    \mathbf{a} = \vec{F_{net}}/m + \tau / I \times (p_i(t) - p_r(t)) - \omega_0^2(p_i(t) - p_r(t))
    \\
    \tau =(p_{mc}(t) - p_r(t)) \times \vec{F_{net}}(t) dt
    \end{equation}
    where pmc is the position of the mass centre and Fnet is the constant net force that acted upon the mass centre.
     
  9. Nov 27, 2015 #8

    andrewkirk

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    The new formula for ##\vec a(t)## uses ##\tau## to represent a torque, whereas it is already in use as representing an angular impulse.
    Is ##\vec F_{net}(t)## intended to be the new force that applies in the interval ##[t_1+\Delta t,t_2]##?
     
  10. Nov 27, 2015 #9
    yes, correct.

    sorry, i should use a new notation for this torque (which is different from the torque generated by the impulse): ## \tau_{new} ##. and this torque is created by ## \vec{F}_{net} ##. Here I update the equation using the new notation:

    \begin{equation}
    \mathbf{a} = \vec{F_{net}}/m + \tau_{new} / I \times (p_i(t) - p_r(t)) - \omega_0^2(p_i(t) - p_r(t))
    \\
    \tau_{new} =(p_{mc}(t) - p_r(t)) \times \vec{F_{net}}(t) dt
    \end{equation}

    is it ok now? thanks
     
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