please help a particle moves along a circular path over a horizontal xy ........

Question

please help a particle moves along a circular path over a horizontal xy ...................

A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 3.90 s, it is at point (5.00 m, 6.10 m) with velocity (3.60 m/s) jhat and acceleration in the positive x direction. At time t2 = 12.6 s, it has velocity (-3.60 m/s) ihat and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit.

Explanation / Answer

There are two ways to do this question - both give the same answer.

Although the question asks for the center co-ordinates of the circle, what you're really looking for here is the radius of the circle, as the co-ordinates can be calculated from this.

First things first. Draw a diagram of what's happpening.

So, you have a particle at position 1, (5.8,6.4) It is being accelerated in the positive x direction and has a velocity in the positive y direction of 2.2 m/s. This means that the motion will circle upwards and to the right (up and right being positive in my system). 10.3s later, it is moving with velocity in the negative x direction and acceleration in the positive y direction.

What this means is that at t1 it is at the leftmost point of the circle - go on, draw it out and you'll see what I mean. At t2 it is at the bottom of it's curve. Follow the logic and you'll see that it has travelled 0.75 of a revolution in this time.

Right - here are the two methods.

1. Work it out on the basis of time.

This was how I originally approached the problem. I worked out that if it took 10.3s to do 3/4 of a revolution, then one revolution would take 13.73s. I then converted this into a frequency, by taking the reciprocal - giving 0.0728 /s, and then to an angular frequency by multiplying by 2*pi, giving 0.457 radians/s. The angular frequency w (should be omega) can then be used in v = w*r, the expression for linear velocity andangular velocity. Knowing v = 2.2m/s and now w = 0.457 rads/s, we can see that r = 4.81m. Now, the last trick is to note that the velocity in circular motion is always at right-angles to the radius (alternatively, the radial direction is parallel to the acceleration) So, at t1 if the velocity is up, the radius must extend to the right only. So the y-co-ordinate is the same as the position at t1, and the x-co-ordinate is 5.8+4.81m

2. Do it using distances.

You could also have done this on the basis of distances. This is probably the easier route in hindsight. I used this to check my result.

Go back to the argument that the movement between t1 and t2 is 3/4 of a revolution. Given that the speed of the particle remains constant (i.e. the vector without direction) at 2.2m/s, and you know the time taken is 10.3s, and finally by realising that thecircumference of a circle is 2*pi*r;

(3/4) * (2*pi*r) = 10.3 * 2.2

That is, using the simple d = v*t, 3/4 of a circumference equals speed times time. Solve this for r and you find that r = 4.81m. Then apply the same arguments for the direction from position 1 that the center of the orbit must be in as before.

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