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Friday, January 29, 2010

Edition 9: IIT Entrance Guidance

In this article we will see some tricky applications of vector algebra and Gravitation.
Problem 1 : How far would the drunkard go?
A drunkard, under the influence of alcohol, steps out of the pub. Having very little control on himself, he starts walking by taking steps of constant length L. However, he walks in random direction and hence he doesn't proceed towards any particular destination. The question is how far would he go from the pub if he takes N number of steps in random direction, where N is very large.
Such problems typically need the method of induction. Assume that the position vector of the drunkard from the origin is R(n), where n denotes the
number of steps taken.

Then he takes a step of length L in some direction and reaches a point whose position vector is R(n+1)
Therefore we can right
R(n+1) = R(n) + L (The quantities are all vectors)
Now, we want to find out the magnitude of R(n+1), which can be obtained by taking a dot product with itself. Remember, self dot product of any vector gives the square of the magnitude of the vector.
R(n+1) ^ 2 = R(n+1) . R(n+1)
= (R(n) + L) . (R(n) + L)
= R(n) . R(n) + L . L + R(n).L + L.R(n)
= R(n) ^2 + L^2 + 2 R(n)*L* Cosine(α)
Where α is the angle between R(n) and L.
Notice that for a large number of steps taken (n -> infinity) the following properties hold
1. 'α' would vary from 0 to 360 degrees for different values of 'n'
2. Therefore cosine(α) would vary from -1 to +1
3. For every positive value of R(n)*L* Cosine(α) , there is a equi-lue.
probable negative value.
Therefore when this induction is carried on for n = 1,2,3,4,……. Infinity, the terms R(n)*L* Cosine(α) would sum up to zero !
So for n -> infinity,
R(n+1) ^ 2 = R(n) ^2 + L ^2
Clearly, R(1) = L, and by induction we can say
R(n+1)^2 = (n+1)L^2
So R(n) = L * sqrt(n)
This behavior is observed with Brownian motion of particles.
Problem 2 : escape velocity from the center of the earth
The problem description is quite simple. Assume there is a tunnel dug through the center of the earth and a particle is resting at the center. How much velocity would you need to impart to the particle at the center of the earth so that it not only travels out of the tunnel but also escapes to infinity. The problem may look very innocuous at the outset but can be quite confusing to solve unless you stick to the basics.
Why I say that it could be confusing ? It is because of the fact that
1. In the theory of gravitation, you arnt that the escape velocity from the surface of earth is
learnt that the escape velocity from the surface of earth is
V(escape) = sqrt (2GM/R)
2. In the theory of Simple harmonic motion, you learnt that a particle left through an hypothetical tunnel through the center of the Earth perform a simple harmonic motion. And you also learnt that the maximum velocity it achieves at the center of the earth is
V(max) = sqrt (GM/R)
For a moment you may be enticed to add these two velocities to arrive at the desired result. But as you know that it is not the velocities that adds up, it is the energies that dictates whether the particle can escape or not.

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