Work done to slide a circle over another circle assuming a horizontal force and no friction

Question

There are two ways that I think that this problem may be able to be solved, however, they give different answers, therefore at least one of them is incorrect.

Let's start with a simpler problem, the classic example of lifting a weight $m g $ with a vertical force $F$, a distance $\Delta z$ gives a work done $W_F$ of $mg \Delta z$. This can be shown by knowing that the net work done $W_{net}$ equates to the change in kinetic energy $\Delta K$, which, assuming that the velocity at the beginning and the velocity at the end of the movement is the same, will equal $0$, hence:

$W_{net}= \Delta K=0$

Now, without going into too much other detail it can be shown that $W_{net}$ is made up of the work done from the vertical force $W_F$ and the work done by gravity $W_g$ and therefore $W_F= mg \Delta z$, as previously stated. However, let's consider the slightly more complex problem:

Let's assume that we want to slide (yet assume zero friction) a circle over another circle with lever arm $L$, assuming zero initial velocity, zero final velocity, initial angle $\theta_0$, final angle $\frac{\pi}{2}$ and a horizontal force $F$ that acts on the moving circle (see diagram below).

How do we solve this problem? Approach 1 follows as above for the vertical lifting case:

$W_{net}=0$ and $W_{net}= W_g + W_F$ therefore,

$-W_g= W_F$ so, considering the component of $mg$ in the direction of movement and integrating:

$-W_g= - \int_{\theta_0}^\frac{\pi}{2} mg \cos(\theta) ds = -\int_{\theta_0}^\frac{\pi}{2} mg \cos(\theta) L d\theta = -mg L (1-\sin(\theta_0)$

Where $ds$ is the infinitesimal chunk of the path the circle follows and equals $L d\theta$. Therefore...

$W_F=-mg L (1-\sin(\theta_0)$. However, this seems strange as it doesn't seem to account for the fact that $F$ is in the horizontal direction and would give the same answer if $F$ were applied vertically.

Approach 2 is as follows: can we assume that the forces are equal the whole way up the path of the centroid? In other words, the equation of motion in the tangential direction with no acceleration is:

$0= F \sin(\theta)- mg \cos(\theta) $, rearranging gives:

$F= mg \cot(\theta)$, now let's integrate along the path with infinitesimal length $ds$:

$\int_{\theta_0}^\frac{\pi}{2} F ds = \int_{\theta_0}^\frac{\pi}{2} mg \cot(\theta) ds= \int_{\theta_0}^\frac{\pi}{2} mg \cot(\theta) L d\theta= -mgL [\ln(\sin(\theta_0))]$

Therefore, that final expression is equal to $\int_{\theta_0}^\frac{\pi}{2} F ds$ which also equals $W_F$. Therefore, we have solved the problem. BUT, why does the first approach not work and how would you solve this problem using approach one. Or is there a fault in approach 2?

Sorry this is such a long question. If one of these approaches is correct and one is incorrect, then please explain what is wrong with the incorrect one. If they are both incorrect then please explain the correct approach to this problem.

Thanks a lot for any help you can provide.

Answer 1

Ok, I think i've solved it and now understand where my mistake was.

Approach 1 is the correct approach. Approach 2 is wrong; the formula $\int_{\theta_0}^\frac{\pi}{2}F ds$ is not the work done, because work done would actually be defined by $\int_{\theta_0}^{\pi/2}\mathbf{F}. d\mathbf{s}$, where $\mathbf{F}$ and $\mathbf{s}$ are both vectors, hence $\mathbf{F}$ .$d\mathbf{s}$ is the dot product of the vectors. Therefore, the component of $\mathbf{F}$ in the direction of $d\mathbf{s}$ is small (this being the definition of a dot product), even though $\mathbf{F}$ may be quite large in the horizontal direction. So, the work done would be the same if I applied the force such that it followed the movement of the circle, rather than always being from a horizontal direction, because in the former case F can be small the whole way up the path of the circle, whereas in the horizontal case F must be relatively large, but the component of it in the direction of $d\mathbf{s}$ would be equal to the case where it follows the direction.

This is kind of unintuitive, because you would have thought that applying the force not in the direction of movement would cause the work done to go up, but actually it doesn't and i'm pretty sure that that would be true even if you accounted for friction as well. Any comments/ corrections are still appreciated.