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Newton's equations

  1. Jan 6, 2006 #1
    Consider a particle of mass m moving under the influence of a force [itex] \vec{F} = -k \bullet \vec{r} [/itex] , k>0 is a constant where r is the position vector of mass m. (A fixed origin O in an inertial reference frame)
    Is the orbital angular momentum [itex] \vec{L} = \vec{r} \times \frac{d \vec{r}}{dt} [/itex] constant of the motion in this case? Explain what this implies about hte trajectory of the body?
    ok L is constnat if its derivative w.r.t. time is zero.
    [tex] \dot{\vec{L}} = \dot{\vec{r}} \times m \dot{\vec{r}} + \vec{r} \times m \ddot{\vec{r}} [/tex]
    first term is zero because the vectors are parallel to each other
    second term is zero because [itex] \vec{F} = m \ddot{\vec{r}} [/itex] thus the vecotrsa re parallel and thus dL/dt = 0 and thus L is constant.
    Since the angular momentum is a constnat, the torque (which is its derivative) is zero thus the mass m is going to move in a plane, and since F is a linear combination of r, the mass is going to move in a straight line
    Solve Newton's equation [itex] \dot{\vec{p}} = -k \vec{r} [/itex] that is obtain the parametric equation of the trajectory given that [itex] \vec{r} (t=0) = \vec{r_{0}} [/itex] and [itex] \dot{\vec{r}} (t=0) = \vec{v_{0}} [/itex]. Hint: write [itex] m \ddot{\vec{r}} = -k \vec{r} [/itex] in rectangular cartesian coordiantes

    well ok ill do it for the X
    let [tex] \vec{r} = (x,y,z) [/tex]
    let [tex] \vec{r_{0}} = (x_{0},y_{0},z_{0}) [/tex]
    and [tex] \vec{v_{0}} = (v_{x},v_{y},v_{z}) [/tex]
    from [tex] \dot{\vec{p}} = -k \vec{r} [/tex] it follows that
    [tex] m \frac{d^2 x}{dt^2} = - kx [/tex]
    and has a solution
    [tex] x(t) = C_{1} \cos(\sqrt{\frac{k}{m}} t) + C_{2} \sin(\sqrt{\frac{k}{m}} t) [/tex]
    using the intial conditions
    [tex] x(t) = x_{0} \cos(\sqrt{\frac{k}{m}} t) + v_{x} \sqrt{\frac{m}{k}} \sin(\sqrt{\frac{k}{m}} t) [/tex]
    i can imagine that hte answers for the y and z part are similar?
    Is this correct? They certainly do appear to be parametric equations...

    Eliminate the parameter t from your solution of b and identify the shape of the trajectory
    well it follows from the part b...
    i think it will move in a plane (from a). But as for the motion i belive it would be a straight line and not a circle. So since it is moving in a plane shouldnt one of the parts (x,y,z) be zero?


    Please help!
     
  2. jcsd
  3. Jan 7, 2006 #2
    can anyone help on this matter?
    i Solved the second order DEs correctly, right? But ist hte solution supposed to be so large?
    ALso how would one eliminate the the t variable? from teh z, y, x ? Thereby finding an expression relating z, ,y, and x?
     
  4. Jan 7, 2006 #3

    George Jones

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    Just like in your previous thread, all the motion takes place in a plane, andif a coordinate system is chosen with the z-axis perpendicular to this plane, ony x and y coordinates need be considrered. I think you solved the x equation correctly, and as you say, the y direction is similar.

    [tex]x(t) = x_{0} \cos(\omega t) + \frac{v_{0x}}{\omega} \sin(\omega t) [/tex]

    [tex]y(t) = y_{0} \cos(\omega t) + \frac{v_{0y}}{\omega} \sin(\omega t) [/tex]

    To eliminate [itex]t[/itex]:

    1) mutliply both sides of the [itex]x[/itex] equation by [itex]\omega y_{0}[/itex];

    2) mutliply both sides of the [itex]y[/itex] equation by [itex]\omega x_{0}[/itex];

    3) subtract the resulting equations to eliminate the cos terms;

    4) mutliply both sides of the [itex]x[/itex] equation by [itex]v_{0y}[/itex];

    5) mutliply both sides of the [itex]y[/itex] equation by [itex]v_{0x}[/itex];

    6) subtract the resulting equations to eliminate the sin terms;

    7) square and add the equations in 3) and 6) to elliminate [itex]t[/itex].

    This eliminates t, but is rather messy.There must be choice of x and y axes that exploits symmetry and shows that the motion is on an ellipse. Motion is along a straight line if [itex]\vec{r}_{0}[/itex] and [itex]]\vec{v}_0[/itex] are parallel or anti-parallel.

    Regards,
    George
     
    Last edited: Jan 7, 2006
  5. Jan 8, 2006 #4
    i see what you mean by total mess
    and clearly this is an ellipse
    also the z is zero

    th question aint over actaully...
    Verify that the force F = kr is conservative and derive the expression for hte corresponding potential energy V(r)

    Forcei s conservative if the expression for dW can be written as a differential of a scalar function
    so here
    [tex] \vec{F} \bullet d \vec{r} = dW = -dV( \vec{r}) [/tex]
    [tex] k \bullet \vec{r} d \vec{r} = d(\frac{k\vec{r}^2}{2}) = -dV(\vec{r}) [/tex]
    so since V is a scalar valued function thus the expression kr^2 /2 is a scalar valued term as well

    Calculate the total energy [tex] E = \frac{1}{2} m \dot{\vec{r}}^2 + V(\vec{r}) [/tex] for this system and verify that it can be written in terms of the constants of the problem

    so here would i be using the expressions R in component form and substituting for them? THat gets rid of the vector r but does bring about the parameter t. The question does ask for the expression for E to be written in terms of constants (not scalars then?) So then this wya would be the way to go?

    ALso thank you for the help so far George!
     
  6. Jan 10, 2006 #5
    for the expression for dr/dt in the E expression
    would [tex] \dot{\vec{r}} = (\dot{\vec{x}} , \dot{\vec{y}}) [/tex]
    is htat correct? So the only 'variable' as such is going to be t?
    as for [tex] V(r) = \frac{kr^2}{2} [/tex]
    then do i express r as x,y and multiply them by k as well?

    is this the the right wya to go for the second part of post #4 (the continued part of the origina lquestion)?

    also if it would not be too much trouble for you, could you help with this question of similar topic https://www.physicsforums.com/showthread.php?t=106174
     
    Last edited: Jan 10, 2006
  7. Jan 11, 2006 #6
    can anyone help with the last post
    i have done everything else and i am confident that most of the answers are correct

    just the last one with the energy formula that i need to figure out.
     
  8. Jan 12, 2006 #7

    George Jones

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    Let [itex]f[/itex] be a function of (scalar) [itex]r[/itex], i.e., [itex]f = f \left( r \right)[/itex]. Then

    [tex]\vec{\nabla} f \left( r \right) = \frac{df}{dr} \left( r \right) \hat{r}.[/tex]

    Thus,

    [tex]
    \begin{equation*}
    \begin{split}
    \vec{F} &= - k \vec{r}\\
    &= - k r \hat{r}\\
    &= - \vec{\nabla} \left( \frac{1}{2} k r^{2} \right)
    \end{split}
    \end{equation*}
    [/tex]

    Cosequently, the potential energy is [itex]V \left( r \right) = \frac{1}{2} k r^{2}[/itex], and the total energy is

    [tex]E = \frac{1}{2} m v^{2} + \frac{1}{2} k r^{2} = \frac{1}{2} \frac{k}{\omega^2} \vec{v} \cdot \vec{v} + \frac{1}{2} k \vec{r} \cdot \vec{r}.[/tex]

    Your solution for position can be written in vector form as [itex]\vec{r} = cos \left( \omega t \right) \vec{r}_{0} + \frac{1}{\omega} sin \left( \omega t \right) \vec{v}_{0}[/itex]. Differentiate this to get a vector equation for [itex]\vec{v}[/itex]. Use these vectors in the dot products in [itex]E[/itex].

    Don't use x and y coordinates.

    Regards,
    George
     
    Last edited: Jan 12, 2006
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