Karl's Calculus Tutor - Solution to Exercise 8.2-3

Solution to Exercise 8.2-3KCT logo

© 1999 by Karl Hahn

This one is barely even a related rate problem. It is more a problem of finding derivatives. But if you are going into engineering, the relationships that flow out of this problem will have to become second nature to you.

An object (could be a train or anything else) traveling around a circular path of radius r centered at the origin is at point:

   ( r cos(wt), r sin(wt) )
at time t seconds if its angular velocity is w radians/second. We need to take first and second derivatives of the expressions above to find the x and y components of the object's velocity and acceleration.

Taking the first time derivative of each of the components of the position described above gives:

   ( -rw sin(wt), rw cos(wt) )
So when the object is in the first quadrant, for example, its x velocity component is negative -- that is its net x motion there is toward the y axis. Its y velocity component is positive in the first quadrant, so its net y velocity there is away from the x axis.

But most important is that the velocity vector is at right angles to the position vector. To confirm this, pick any point, (x,y). Plot it on graph paper and draw a line connecting it to the origin. Now plot (-y,x) and draw a line connecting it to the origin. No matter what x and y you chose, the two lines you drew were at right angles to each other. Indeed, you could have scaled the second point by some value, k (which simply means multiply both the x and y components by k), and the lines would still have been at right angles to each other. Do you see how the velocity components relate to the position components in this problem in exactly the same way?

Now lets find the acceleration by taking the derivative of the velocity:

   ( -rw2 cos(wt), -rw2 sin(wt) )
Observe that the acceleration points in exactly the opposite direction as the position. That is the train is accelerated toward the center of the circle. You have surely noticed that when a car accelerates forward you feel a reaction pressing you back into your seat -- that is you feel a reaction opposite the acceleration. Here as the train goes around the circle it is accelerated inward, and somebody aboard the train would feel a reaction outward. The outward reaction is what is commonly referred to as centrifugal force.

Finally, w is the angular rate in radians per second that the train goes around the circle. If you wanted to convert it to revolutions per second you would divide w by 2p. If you double w, you should be able to see that you consequently double the magnitude of the velocity. It should be equally clear that you would quadruple the magnitude of the acceleration, since acceleration is shown to be proportional to the square of the angular rate, w. This also means that the centrifugal force increases with the square of w.

For your own amusement, see what happens if the circle is not centered at the origin. Suppose it is centered at (x0,y0). Then the position is given by

   ( r cos(wt) + x0, r sin(wt) + y0 )
Remember that x0 and y0 are both constants. What effect does this have on the velocity and acceleration of the train?

How about if the train does not start at the angular position of zero radians? Suppose it starts at f radians. Then its position is given by

   ( r cos(wt + f), r sin(wt + f) )
What effect does f have on the magnitude of the velocity and the acceleration? Remember that sin2(x) + cos2(x) = 1 for all x.


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