A particle that feels an angular force only

[retro10x]

Homework Statement

"Consider a particle that feels an angular force only, of the form F=2mvw(theta direction). Show that r=Ae^theta + Be^-theta, where A and B are constants of integration."
v=dr/dt
w=d(theta)/dt

Homework Equations

F(radial)= 0 = m((dv/dt)-r*w^2)
F(theta)= 2mvw = m(r*(dw/dt)+2vw)

The Attempt at a Solution

So I can solve the F(theta) equation to find that r*(dw/dt)=0, hence dw/dt=0
and I can also solve the F(radial) equation to find that dv/dt=r*w^2
I also know that this question involves a separation of variables and then integrating
However the problem is that I don't know substitutions I can make to put the equations I have into a form that makes that possible. Any hints or ideas are appreciated ^-^

 

[voko]

dw/dt = 0 means w = W = const. So the first equation is then dv/dt = Cr, where C = W^2 = const. Are you saying you don't know how to solve this equation?

 

[retro10x]

Oh, thanks for pointing that out
dv/dt can written as v(dv/dr)
So 0.5v^2=0.5c*r^2+D ;where D is the constant of integration
v=sqrt(cr^2 +2D)=dr/dt
I end up with: t+E=intg[dr/sqrt(cr^2 +2D)] ;where E is the constant of integration
Wolfram Alpha can't even solve this integral :O
What am I missing?

 

[voko]

That integral can be converted to $$\int \frac {adx} {\sqrt {x^2 + 1}}$$.

 

[Nickie]

First, let's rewrite the given force equation in terms of the variables given in the problem: F = 2mvw(theta direction) = m(r*(dw/dt) + 2vw). This is similar to the F(theta) equation given in the problem, but we have substituted in v=dr/dt and w=d(theta)/dt.

Next, we can use the F(radial) equation to solve for dv/dt, which gives us dv/dt = r*w^2. We can then substitute this into the F(theta) equation to get m(r*w^2) = m(r*(dw/dt) + 2vw). This simplifies to r*(dw/dt) + 2vw = 0.

Now, we can use the separation of variables method to solve this differential equation. We can rearrange the equation to get dw/dt = -2vw/r. Then, we can separate the variables by multiplying both sides by dt and dividing by -2vr, giving us: (1/2v)*dv = (-1/r)*d(theta).

Integrating both sides with respect to their respective variables gives us: (1/2)*ln|v| = -ln|r| + C, where C is a constant of integration. Using properties of logarithms, we can rewrite this as ln|v| = -2ln|r| + C'. Then, we can exponentiate both sides to get v = e^C'/(r^2).

Now, we can substitute this expression for v into our original equation, dv/dt = r*w^2. This gives us e^C'/(r^2) = r*w^2. We can solve for r by multiplying both sides by r^2 and taking the square root, giving us r = (e^C')^(1/3) = A*e^(theta) + B*e^(-theta), where A and B are constants of integration.

Therefore, we have shown that r = A*e^(theta) + B*e^(-theta), as desired.