Uh, yeah.

What bothers me here (and makes this difficult to understand) is that I have no idea where the second equation comes from.

Well, I DO have an idea, but let me approach things this way.

To minimize energy loss you want to make sure the ball is not slipping as it rolls. This gives you a condition on the linear acceleration of the ball:

a = r*(alpha)

which in turn gives you a condition on the maximum angle you can roll the ball at (from rest, I presume.) Solve the relevent problem with this restriction and you should get your second equation, I presume.

Why are you including the spreadsheets? I think the question is looking for a derivation (calculations) not numbers.

How I would do this is as follows:

My sketch of the situation has a constant slope going downward and to the right at an angle (theta) with the horizontal. I'm going to define positive directions +x down the incline and +y perpendicular to and out of the plane of the incline.

The FBD has a weight (w) acting directly downward, a normal force (N) in the +y direction, and a (rolling) friction force (f) in the -x direction.

Call the mass of the ball m, the radius r, and the coefficient of (rolling) friction (mu). (Technically "rolling" friction is a subcategory of "static" friction, so (mu) = f(max)/N.)

Since the ball is not accelerating out of the plane of motion, we know that

SumFy = N - w*cos(theta) = 0

So N = mg*cos(theta)

We know that

SumFx = -f + w*sin(theta) = ma

-(mu)N + mg*sin(theta) = ma

-(mu)*mg*cos(theta) + mg*sin(theta) = ma

a = g*[sin(theta) - (mu)*cos(theta)]

This is the acceleration of the ball as it rolls down the plane. What we need to do now is have a way to state that the ball cannot slide. We know we MUST have the above acceleration if the ball is to roll without slipping (else the rolling friction equation is wrong, and hence our expression for a.)

So given an acceleration for the ball, can we say whether it is going to roll without slipping?

Sure! Let's take a look at the torque on the ball that causes it to rotate. Essentially the weight is acting on the center of the ball (at the CM, that is) and the point of rotation is at the edge of the ball (where the ball makes contact with the inclined plane.) If you sketch a diagram you can calculate the magnitude of the torque to be:

|(tau)| = rw*sin(theta) = mgr*sin(theta)

This is the net torque on the ball. So

Sum(tau) = I(alpha) = (2/5)mr^2 * (alpha) = mgr*sin(theta)

Thus

(alpha) = (5g)/(2r*sin(theta))

We know that for rolling without slipping

a = r*(alpha)

Thus:

g*[sin(theta) - (mu)*cos(theta)] = (5g)/(2*sin(theta))

is our condition on (theta).

Solving this for (theta) is NOT trivial. I get that:

sin(theta) - (mu)*cos(theta) = (5)/(2*sin(theta))

sin^2(theta) - (mu)*sin(theta)*cos(theta) = 5/2

(mu)*sin(theta)*cos(theta) = sin^2(theta) - 5/2 <-- Square both sides:

(mu)^2*sin^2(theta)*cos^2(theta) = [sin^2(theta) - 5/2]^2

Now note that cos^2(theta) = 1 - sin^2(theta):

(mu)^2*sin^2(theta)*[1 - sin^2(theta)] = sin^4(theta) -5*sin^2(theta) + 25/4

For the sake of writing this a bit easier I'm going to define:

y = sin^2(theta)

and

c = (mu)^2

cy(1 - y) = y^2 - 5y + 25/4

-cy^2 + cy = y^2 - 5y + 25/4

(c + 1)y^2 - (c + 5)y + 25/4 = 0

Using the quadratic equation:

y = [(c + 5) (+/-) sqrt{(c + 5)^2 - 4(c + 1)*25/4}]/[2(c + 1)]

y = [(c + 5) (+/-) sqrt{(c + 5)^2 - 25(c + 1)}]/[2(c + 1)]

Now, c = (mu)^2 is a positive number, so (c + 5)^2 - 25(c + 1) < (c + 5)^2 so we need to discard the "-" possibility. Thus:

y = [(c + 5) + sqrt{(c + 5)^2 - 25(c + 1)}]/[2(c + 1)]

But y = sin^2(theta) so:

sin^2(theta) = [(c + 5) + sqrt{(c + 5)^2 - 25(c + 1)}]/[2(c + 1)]

sin(theta) = (+/-)[(c + 5) + sqrt{(c + 5)^2 - 25(c + 1)}]/[2(c + 1)]

Since (theta) is physically a first quadrant angle, we discard the "-" solution here as well.

sin(theta) = [(c + 5) + sqrt{(c + 5)^2 - 25(c + 1)}]/[2(c + 1)]

(theta) = asn([(c + 5) + sqrt{(c + 5)^2 - 25(c + 1)}]/[2(c + 1)])

(theta) = asn([((mu)^2 + 5) + sqrt{((mu)^2 + 5)^2 - 25((mu)^2 + 1)}]/[2((mu)^2 + 1)])

(We can expand inside the square root, but it gives us nothing in particular, so I won't.)

-Dan