- #1
stunner5000pt
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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!
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!
