Force, Work, and Energy

 February 17

After studying this some more, I found that on a quarter of a circle, the total friction is:

E= -μmg*cos(π/4)*r*sqrt(2)

The reason is that at π/4, the curve can be split into two parts. The upper eighth of the circle has a lower amount of normal force. The bottom eighth of the circle has a greater amount of normal force. When each point along the arc is added up, they cancel out and reduce to the normal force at π/2. A argument that I devised is below:

 Spoiler

Consider the two triangles. If each of the smaller triangles are 30-60-90s and the outer larger triangle is 45-45-90, then it is easy to show that if a particle travels along the hypotenuses of the two smaller triangles, it will experience the same amount of friction as traveling the hypotenuse as the larger triangle.

To simplify things, each of the 30-60-90 triangles have sides 1, sqrt(3), and 2. Therefore, the outer triangle has two sides 1+sqrt(3) and hypotenuse sqrt(2)*(1+sqrt(3)).

Friction is defined to be μmg*cos(θ, so a particle traveling along the inside path will lose the following amount of heat in total:

2*μmg*cos(π/3) + 2*μmg*cos(π/6)

2*μmg*(1/2) + 2*μmg*(sqrt(3)/2)

μmg*(1+sqrt(3))

The amount of friction lost to the outer path is:

sqrt(2)*(1+sqrt(3))*μmg*cos(π/4)

sqrt(2)*(1+sqrt(3))*μmg*(1/sqrt(2))

μmg*(1+sqrt(3))

Lo and behold, they're the same. If these are the same, then I can generalize this to a circle, because a circle can be constructed by drawing a tangent to all the possible subdivisions that form 90 degree angles. Therefore, the heat lost due to friction is μmg, multiplied by the angle between its endpoints, or cos(π/4), multiplied by the length of the upper arc:

E= -μmg*cos(π/4)*r*sqrt(2)

E= -μmg*r

My new derivation seems correct.

 Spoiler

Assuming these two blue lines are completely the same, only symmetric about the midpoint of the arc, then the magnitudes of the normal forces will add up to be the same.

Also, This seems to be in agreement with the formula I derived earlier in the day (see my previous post):

E = -∫ μmg (dy/dx) dx = -μmg y(x)

If this process could be applied to more general curves, then it vastly simplifies things. My attempt to do this is to subdivide the curve along the inflexion points and the local minima and maxima.

 Spoiler

Is there a way to show that the total friction above or below each of these curves is the following?

E= μmg*cos(angle between the endpoints)*(length of the upper line)

If this is exact, I'll be darned happy.

February 22

I met with my TA and he told me that my process seems true, and that he had never thought of friction in the way I described. However, I did make one error, and that was using Giancoli's E= ∫ F cos(θ dl. My TA told me that this formula is designed for horizontal forces, and the force I wanted was vertical. Therefore, it'd be more logical to substitute sin(θ in the place of cos(θ. Working it out:

E = - ∫ μmg sin(arctan(-dx/dy)) dl

dl = sqrt(1+(dy/dx)^2) dx

sin(arctan(-dx/dy)) = (-dx/dy) /sqrt(1+(dx/dy)^2)

Therefore:

E = -∫ μmg (-dx/dy) sqrt(1+(dy/dx)^2)/sqrt(1+(dx/dy)^2) dx

E = -∫ μmg (-dx/dy) (dy/dx) dx = ∫ μmg dx

E = μmgx

What this means is that the shape of the curve doesn't matter. The sliding friction is purely dependent on the horizontal displacement, which makes sense based on the above. Therefore, my textbook is wrong, and I am happy. Lol. Being happy when my textbook is wrong... is wrong. ._.