Unsuccessful DIY - Seeking Help

[Brad Barker]

I tried to do this myself, but I was unsuccessful.

Thank you.

 

[Brad Barker]

Ok, I was able to do this after all.

My first method was to substitute in -bv in for F and try to show that the closed line integral of F with dr is not equal to 0. This did not get me anywhere, although I imagine that some masters of vector calculus could get the result this way.

What was successful for me was exploiting Stokes' Theorem, and also solving the differential equation

m dv/dt = -bv.

I got v from this, plugged it back into my equation for force, and then computed the curl, which is not equal to zero so long as the velocity is not equal to zero.

Could someone verify that this is correct?

Thank you.

 

[jdavel]

Brad,

No fair! You didn't really say that the force had to be in the same direction as the velocity! In that case it's obvious; the velocity of the mass is always increasing. So if it travels ina closed loop, it's going faster at the end than it was at the beginning. KE isn't conserved.

 

[Galileo]

Consider the work done on a particle in going from a to b.
Now consider going along the same path, but with twice the velocity.

 

[Andrew Mason]

Ok, I was able to do this after all.

My first method was to substitute in -bv in for F and try to show that the closed line integral of F with dr is not equal to 0. This did not get me anywhere, although I imagine that some masters of vector calculus could get the result this way.

What was successful for me was exploiting Stokes' Theorem, and also solving the differential equation

m dv/dt = -bv.

I got v from this, plugged it back into my equation for force, and then computed the curl, which is not equal to zero so long as the velocity is not equal to zero.

For the differential equation of motion:

\frac{dv}{dt} + \frac{b}{m}v = 0

the integrating factor is e^{\frac{b}{m}t

So:
e^{\frac{b}{m}t}\frac{dv}{dt} + e^{\frac{b}{m}t}\frac{b}{m}v = \frac{d}{dt}(ve^{\frac{b}{m}t}) = 0

This means that:

v = v_0e^{{\frac{-b}{m}t}

If the force is conservative, the work done is a function of position only, so the line integral of Fds along a closed path must always be 0:

\oint m\frac{dv}{dt}ds = - \oint bvds = W

Edit: oops - integration corrected as per subsequent posts:

W = -b \oint vds = -b\oint v^2dt = -b\int_{t_0}^{t_1} v_0^2e^{\frac{-2b}{m}t}dt + -b\int_{t_1}^{t_2} v_0^2e^{\frac{-2b}{m}t}dt

W = -b(\frac{-m}{2b}v_0^2(e^{\frac{-2b}{m}t_1}-e^{\frac{-2b}{m}t_0}) + (-b(\frac{-m}{2b}v_0^2(e^{\frac{-2b}{m}t_2}-e^{\frac{-2b}{m}t_1}))

W = \frac{1}{2}mv_0^2(e^{\frac{-2b}{m}t_1} - e^{\frac{-2b}{m}t_0}) + \frac{1}{2}mv_0^2(e^{\frac{-2b}{m}t_2} - e^{\frac{-2b}{m}t_1})

W = \frac{1}{2}mv_0^2(e^{\frac{-2b}{m}t_2} - e^{\frac{-2b}{m}t_0}) = 0 only if t_2 = t_0, which is impossible or if v_0 = 0.

In other words, the line integral along any path is time dependent, not position dependent and cannot be 0, so it is not a conservative force,

AM

 

[Brad Barker]

W = -b \oint vds = -b\int_A^B v_0e^{\frac{-b}{m}t} + -b\int_B^A v_0e^{\frac{-b}{m}t}

W = -b(\frac{-m}{b}v_0(e^{\frac{-b}{m}t_1}-e^{\frac{-b}{m}t_0}) + (-b(\frac{-m}{b}v_0(e^{\frac{-b}{m}t_2}-e^{\frac{-b}{m}t_1}))

W = mv_0(e^{\frac{-b}{m}t_1} - e^{\frac{-b}{m}t_0}) + mv_0(e^{\frac{-b}{m}t_2} - e^{\frac{-b}{m}t_1})

W = mv_0(e^{\frac{-b}{m}t_2} - e^{\frac{-b}{m}t_0}) = 0 only if t_2 = t_0, which is impossible or if v_0 = 0.

In other words, the line integral along any path is time dependent, not position dependent and cannot be 0, so it is not a conservative force,

AM

Thanks, Andrew.

It looks like you integrated with respect to time, although the line above, you have a "ds." I think this might change the answer, although the salient feature is the same.

 

[dextercioby]

Andrew,your analysis and computations are incorrect.

W_{1\rightarrow 2}=:\int_{1}^{2} \vec{F}\cdot d\vec{s}=-b\int_{1}^{2}\vec{v}\cdot d\vec{s}

Since \vec{v}\uparrow\uparrow \vec{r},then

W_{1\rightarrow 2}=-b\int_{1}^{2} v \ ds=-b\int_{t_{1}}^{t_{2}} \left(v\frac{ds}{dt}\right)dt=-b\int_{t_{1}}^{t_{2}} v^{2} dt

Do everything again.

Daniel.

 

[Andrew Mason]

Andrew,your analysis and computations are incorrect.

W_{1\rightarrow 2}=:\int_{1}^{2} \vec{F}\cdot d\vec{s}=-b\int_{1}^{2}\vec{v}\cdot d\vec{s}

Since \vec{v}\uparrow\uparrow \vec{r},then

W_{1\rightarrow 2}=-b\int_{1}^{2} v \ ds=-b\int_{t_{1}}^{t_{2}} \left(v\frac{ds}{dt}\right)dt=-b\int_{t_{1}}^{t_{2}} v^{2} dt

Do everything again.

Daniel.

Thanks for pointing that out - of course E_0 = mv^2/2 and not mv. See above correction. The bottom line is that W is not position dependent so it is not conservative.

AM

 

[PBRMEASAP]

Another way of saying it, entirely equivalent to AM's analysis, is that \vec{F} \cdot \vec{ds} has the same sign along the entire path of the particle. Further, this dot product is only zero where V is zero, the velocity has the same direction as \vec{ds}. Thus the only way a closed loop integral could be zero is if it has zero length--the particle doesn't move.

 

[vinter]

Consider the work done on a particle in going from a to b.
Now consider going along the same path, but with twice the velocity.

Why isn't anyone considering this elegant solution?

 

[Galileo]

Why isn't anyone considering this elegant solution?

Thank you vinter.