After thinking about it, I think that with the magnetic field from the dipole magnet pulling charge to one spot in the soft iron, that too would act like a dipole so I'd be looking for dipole-dipole forces.
F_{mag}=\frac{-3μ_0m_1m_2}{4πr^4}
Where m1 and m2 are the masses of the soft iron...
That makes a lot of sense. That equality seems to be messing things up quite a bit. Another approach came to me last night, though.
So if d_2 is increased by an amount of Δd_2 and is traveling at the same initial velocity, it will arrive at the initial collision point after m_1 has already got...
Homework Statement
A soft iron ball is fixed a distance d above the pole of a rectangular dipole magnet which is permanently magnetized. What is the force the iron ball feels due to the magnetic field?
The dimensions of the dipole magnet are a x a x b, where a < b
Homework Equations...
The farther away the object is, the smaller the Δv should be required to make the m2 mass avoid the m1 mass. I think the way it's set up now, it's saying that the farther away m2 is from the collision point, the more acceleration required to get m2 to the collision point "on time".
So if I...
A week later and I'm back at it. I still think my method might be somewhat flawed.
Just by thinking about it, the Δv should decrease as the distance of m2 from the collision point increases. Should it look exponential, approaching zero as Δv approaches infinite?
The way I have it set up...
Ah okay, thanks for the heads up :)
I figured that if I take the equation for acceleration that I found and use that to find a final velocity, the Δv would represent a minimum acceleration required to avoid collision, meaning any acceleration greater than that would be acceptable.
But that...
The issue indeed is that some acceleration is required to slow the object to a slower speed so that it avoids collision. In any realistic case any change in velocity wouldn't be in a negligible time. Could I use the acceleration I found a couple posts back in another kinematics equation to get...
So I was thinking a bit more and thought that maybe using the formula for uniform acceleration might help. Since both times should be the same:
d_2=v_2t+\frac{1}{2}a_2t^2
a_2=\frac{2(d_2-v_2t)}{t^2}
t=\frac{d_1+x}{v_1}
a=\frac{2(d_2-v_2(\frac{d_1+x}{v_1}))}{(\frac{d_1+x}{v_1})^2}...
Homework Statement
Two objects with masses m1 and m2 are traveling in a frictionless surface and will collide perpendicular to each other (m1 is moving on the +x-axis, m2 is moving on the +y-axis). The distances of objects m1 and m2 from the collision point are d1 and d2 respectively.
Q...