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I am working on the following problem:

Given a mass $m$ at rest at the base of an incline surface shaped like 1/12th of a circle with radius $R$. The mass is pulled up the incline by a rope that forms angle $\alpha$ with the direction of velocity. The coefficient of friction between the mass and the surface is $\mu$. Find the minimum work required to pull the mass to the top of the incline.

http://imgur.com/bZMLUSx

I did the force calculations, split the force into angular and radial components (calling the angle between the normal and the vertical $\phi$), expressed the accelerations in polar form, manipulated the equations, integrated both sides from 0 to $\pi/6$, and got the following result:

$$ W_F = \frac{1}{1+\mu\tan\alpha}\left(mgR\left(1 - \frac{\sqrt{3}}{2} + \frac{\mu}{2}\right) + mR^2\int^{\pi/6}_{0}{\left(\ddot\phi + \mu\dot\phi^2\right)d\phi}\right). $$

The question asks for minimum work, and so as I understand it, the minimum work would be achieved if the second term in that sum, with the hairy integral, is equal to 0. However, I am struggling to make a coherent argument as to why that's the case. Could that integral ever go negative, resulting in an even smaller amount of work? If not, why not?

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  • $\begingroup$ I don't understand your reasoning. Are you pulling the mass along the curved part of the inclined plane? $\endgroup$ – Steven Mathey Sep 14 '14 at 2:06
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    $\begingroup$ Yes, the mass is being pulled up a curved incline, shaped like a 1/12th of a circle. Here's a diagram: dropbox.com/s/zucdaixfn05ovx7/diagram.png?dl=0 $\endgroup$ – allenrabinovich Sep 14 '14 at 2:21
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Integration by parts or using the identity $\frac{d}{dt}\dot{\phi}^2=\dot{\phi}\ddot{\phi}+\dot{\phi}\ddot{\phi}$ would be the trick here:

$\frac{d\phi}{dt}dt=d\phi$, so:

$\int \ddot{\phi} d\phi=\int \ddot{\phi} \dot{\phi} dt=\dot{\phi}^2-\int \ddot{\phi}\dot{\phi}dt$. Applying the usual trick (adding the integral to both sides and dividing by two), this shows $\int \ddot{\phi}d\phi=\frac{1}{2}\dot{\phi}^2$. Now it's easier to see what a positive/negative value there could mean. If $\dot{\phi}_0$ is some huge velocity and we slowed the box down by pushing (not pulling), we're really doing negative work (that we could use to generate power or lift another box or what have you). In fact, you could dominate all other terms with this negative work and find $W_F$ to be negative. So you have to restrict the initial/final velocities of the box. For example, if the angular velocity is zero at the start and end, then the $\ddot{\phi}$ integral must also be zero.

I'm considering only the $\ddot{\phi}$ term because the freedom that $\mu$ gives the other integral is enough to do lots of things. Like - by virtue of fiddling $\mu$ to be smaller - turn a case where the phi squared term dominates a negative $\ddot{\phi}$ term, into one where the negative $\ddot{\phi}$ term dominates the phi squared term.

(I'm really sleepy so sorry if there's a blatant mistake here)

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  • $\begingroup$ I think the problem only makes sense if the mass starts at rest. Otherwise, there is no "minimum" amount of work to be done -- one can arbitrarily have the mass moving arbitrarily quickly initially, and then work can be arbitrarily large in the negative direction. So, the mass starts at rest. Now, given your explanation, it makes sense that the first term has to be 0 -- that's the change in the kinetic energy. You don't want to transfer more energy to the mass, because then you are not minimizing work, and you can't receive any energy from the mass, because it started at rest. $\endgroup$ – allenrabinovich Sep 15 '14 at 21:03
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    $\begingroup$ Then, the second term is the velocity over time -- now, this can't be 0, because the mass has to move, but I think the argument here is that, if you take an infinite amount of time, you can make the velocity infinitesimal. So I think that's the totality of the argument for why this integral must be 0 (and can't be negative). $\endgroup$ – allenrabinovich Sep 15 '14 at 21:04

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