Finding angular Momentum from a force

[fer Mnaj]

Homework Statement

Consider the force $\vec{F}(\vec{r}) = (−kq/r^2 +β^4/r^5) ˆr$

which affects an electron with load e. Calculate the torque that F exerts on e and the angular momentum
of the particle.

Relevant Equations

τ= r x F L= r x p

Is it correct to say that that τ=0 since r has the same directacion as F??
and for \vec{L} que need to find \vec{p}

So I thought solving this dif equation

$\int dp/dt =−kq/r^2 +β^4/r^5$

Do you agree in the path I am going?

 

[kuruman]

I do not agree with the direction you are going. The dif equation you have is incorrect: $dp/dt$ is a force; $m\int dp/dt$ is nonsense.

It looks like you have a central force here and the expression you provide is probably an electric field, not a force. What do you know about the torque that results from a central force or field for that matter?

To find the angular momentum, a velocity at some point is needed. You do not provide that information.

I do not understand "to which an electron is attached load e". Please explain what you mean.

 

[fer Mnaj]

Im studyin mechanics, so I don't think is related directly with electromagnetism
The torque in central force is zero.
Ive corrected m, I was thinking in order to get velocity from that force I need to integrate it

 

[haruspex]

Homework Statement:: Consider the force $\vec{F}(\vec{r}) = (−kq/r^2 +β^4/r^5) ˆr$
to which an electron is attached load e. Calculate the torque that F exerts on e and the angular momentum
of the particle.

That description is unintelligible.
Does it mean the force acts on an electron of charge -e? But if it is a force, we don't care about the charge, we need the mass.
A force acting directly on a particle would lead to a linear acceleration, not to an angular acceleration, nor to a velocity (unless we know how long it acts), nor an angular momentum.

Presumably this is a translation. You will need a better one.

 

[etotheipi]

I agree with the comments above, and the problem statement in its current form is quite baffling.

From what we have to go on, what you have here (as mentioned above) is a central force, i.e. something of the form $\vec{F} = F_r(r) \hat{r}$. Central forces have the special property that they constitute no torque about the origin of the coordinate system:$$\vec{\tau} = \vec{r} \times \vec{F} = r\hat{r} \times F_r(r) \hat{r} = rF_r(r) \hat{r} \times \hat{r} = \vec{0}$$And if you know that $\vec{\tau} = d\vec{L}/dt = \vec{0}$, then you can deduce that the angular momentum of the particle is a conserved quantity of motion.

But to find out what the angular momentum of the particle actually is, you need to specify the initial conditions in the form of $\vec{r}(0)$ and $\vec{v}(0)$.

 

[fer Mnaj]

changed the statement, and you are right in the first thing you said. In this case we are talking about a central force, so no angular acceleration as you say.
But we know that $\vec{F} = d \vec{p}/dt$ so i hoped we could get the velocity by integrating

 

[haruspex]

changed the statement, and you are right in the first thing you said. In this case we are talking about a central force, so no angular acceleration as you say.
But we know that $\vec{F} = d \vec{p}/dt$ so i hoped we could get the velocity by integrating

Yes, but your problem is that to get p by integrating F you'd need to integrate wrt t and the force is given as a function of r.
So what else can you do?