Newton's Equations: Analyzing Constant Orbital Angular Momentum of Mass m

[stunner5000pt]

Consider a particle of mass m moving under the influence of a force $\vec{F} = -k \bullet \vec{r}$ , 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 $\vec{L} = \vec{r} \times \frac{d \vec{r}}{dt}$ 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.
$$\dot{\vec{L}} = \dot{\vec{r}} \times m \dot{\vec{r}} + \vec{r} \times m \ddot{\vec{r}}$$
first term is zero because the vectors are parallel to each other
second term is zero because $\vec{F} = m \ddot{\vec{r}}$ 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 $\dot{\vec{p}} = -k \vec{r}$ that is obtain the parametric equation of the trajectory given that $\vec{r} (t=0) = \vec{r_{0}}$ and $\dot{\vec{r}} (t=0) = \vec{v_{0}}$. Hint: write $m \ddot{\vec{r}} = -k \vec{r}$ in rectangular cartesian coordiantes

well ok ill do it for the X
let $$\vec{r} = (x,y,z)$$
let $$\vec{r_{0}} = (x_{0},y_{0},z_{0})$$
and $$\vec{v_{0}} = (v_{x},v_{y},v_{z})$$
from $$\dot{\vec{p}} = -k \vec{r}$$ it follows that
$$m \frac{d^2 x}{dt^2} = - kx$$
and has a solution
$$x(t) = C_{1} \cos(\sqrt{\frac{k}{m}} t) + C_{2} \sin(\sqrt{\frac{k}{m}} t)$$
using the intial conditions
$$x(t) = x_{0} \cos(\sqrt{\frac{k}{m}} t) + v_{x} \sqrt{\frac{m}{k}} \sin(\sqrt{\frac{k}{m}} t)$$
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 believe it would be a straight line and not a circle. So since it is moving in a plane shouldn't one of the parts (x,y,z) be zero?


Please help!

 

[stunner5000pt]

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 the z, y, x ? Thereby finding an expression relating z, ,y, and x?

 

[George Jones]

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.

$$x(t) = x_{0} \cos(\omega t) + \frac{v_{0x}}{\omega} \sin(\omega t)$$

$$y(t) = y_{0} \cos(\omega t) + \frac{v_{0y}}{\omega} \sin(\omega t)$$

To eliminate $t$:

1) mutliply both sides of the $x$ equation by $\omega y_{0}$;

2) mutliply both sides of the $y$ equation by $\omega x_{0}$;

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

4) mutliply both sides of the $x$ equation by $v_{0y}$;

5) mutliply both sides of the $y$ equation by $v_{0x}$;

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

7) square and add the equations in 3) and 6) to elliminate $t$.

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 $\vec{r}_{0}$ and $]\vec{v}_0$ are parallel or anti-parallel.

Regards,
George

 

[stunner5000pt]

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
$$\vec{F} \bullet d \vec{r} = dW = -dV( \vec{r})$$
$$k \bullet \vec{r} d \vec{r} = d(\frac{k\vec{r}^2}{2}) = -dV(\vec{r})$$
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 $$E = \frac{1}{2} m \dot{\vec{r}}^2 + V(\vec{r})$$ 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!

 

[stunner5000pt]

for the expression for dr/dt in the E expression
would $$\dot{\vec{r}} = (\dot{\vec{x}} , \dot{\vec{y}})$$
is htat correct? So the only 'variable' as such is going to be t?
as for $$V(r) = \frac{kr^2}{2}$$
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

 

[stunner5000pt]

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.

 

[George Jones]

Let $f$ be a function of (scalar) $r$, i.e., $f = f \left( r \right)$. Then

$$\vec{\nabla} f \left( r \right) = \frac{df}{dr} \left( r \right) \hat{r}.$$

Thus,

$$\begin{equation*}<br /> \begin{split}<br /> \vec{F} &= - k \vec{r}\\<br /> &= - k r \hat{r}\\<br /> &= - \vec{\nabla} \left( \frac{1}{2} k r^{2} \right)<br /> \end{split}<br /> \end{equation*}$$

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

$$E = \frac{1}{2} m v^{2} + \frac{1}{2} k r^{2} = \frac{1}{2} \frac{k}{\omega^2} \vec{v} \cdot \vec{v} + \frac{1}{2} k \vec{r} \cdot \vec{r}.$$

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

Don't use x and y coordinates.

Regards,
George