How high can insect crawl in a bowl, if friction coefficient, radius of bowl given?

1. Homework Statement

If the coefficient of friction between an insect and bowl surface is u (mu) and the radius of bowl is r, what is the max height upto which the insect can crawl in the bowl?

a) r/root(1 + u2)
b) r [ 1 - 1/root(1 + u2)]
c) r [root(1 + u2)]
d)r [ root(1 + u2) -1 ]

2. Homework Equations

F=uR
tan theta = u

On a non-horizontal plane,

F=mg sin theta
R=mg cos theta

therefore, net f=mg(sin theta + u cos theta)

3. The Attempt at a Solution

Can't get the gist of how to approach this question.

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With problems like this, first thing you do is go about making a diagram. For all the purposes of this question, we can take the bowl to be a hemisphere. Let's assume that there is a single meridian line of the sphere that the insect travels across. Let's start by taking the cross section of such a line:

http://img267.imageshack.us/img267/5157/insecthelpqi1.jpg [Broken]

Here, i've assumed a right-handed co-ordinate system, with the circle:

$$x^2 + y^2 = r^2$$

Let the insect be at an arbitrary point on the circle (x, y). Now, a very small part of the circle can be taken to be as an inclined plane. The angle it makes can be found out using the angle the tangent on that point makes with the x-axis. It is given by:

$$\frac{dy}{dx} = \tan{(\theta)} = -\frac{x}{y}$$

Once, you've done that, you can easily find out the angles, $\sin(\theta)$ and $\cos(\theta)$, using Pythagoras theorem wherever necessary. And hence, you can find the angle of the inclination as a function of the co-ordinates (x, y). But, you need it in the terms of height. With a little algebra and geometry, you can find the co-ordinates as:

$$(x, y) \equiv (\sqrt{2hr - h^2}, r - h)$$

And hence, the inclination is given by:

$$\frac{dy}{dx} = \tan{(\theta)} = -\frac{x}{y} = \frac{\sqrt{2hr - h^2}}{h - r}$$

Now, you have the force components in terms of the height climbed by the insect. Now, you need to find 'h' such that, the frictional force (caused by the normal force mg cos(θ)) becomes equal to the restraining force [mg sin(θ)] and beyond which it is smaller.

I think you can do the problem now.

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