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)(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# Homework Help: Newton's equations

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