# Describe the path of the object in an xy plot.

1. Sep 20, 2011

### naaa00

1. The problem statement, all variables and given/known data

The problem statement is a) to find expression for the position, velocity, and acceleration vectors, and to describe the path of the object in an xy plot. The coordinates of object moving in the xy plane vary with time according to the equations:

x = - 5.00m sin(wt), w is a constant.

y = 4.00m + 5.00m cos(wt) , w is a constant.

3. The attempt at a solution

For a)

r = xi + yj = (4.00 m)j + (5.00 m)[ -sin(wt)i - cos(wt)j ]

v = (5.00 m)w [ -cos(wt)i + sin(wt)j ]

a = (5.00 m)w^2 [ sin(wt)i + sin(wt)j ]

3. My problem

I have been told that the object moves in a circle of radius 5.00m and its centered at (0, 4.00m).

I really don't get it. I don't understand why.

2. Sep 20, 2011

### Hootenanny

Staff Emeritus
This question is best attempted in plane polar coordinates. Do you think you could write down the position in polar coordinates?

3. Sep 20, 2011

### naaa00

Hello! thank you for your answer, Hootenanny.

Well, I must say that I am learning polar coordinates for my first time. Probably what I am going to say is totally wrong, but anyways:

r = (4.00 m)j + (5.00 m)[ -sin(wt)i - cos(wt)j ]

or xi + yj = r

x = - 5.00m sin(wt),

so

r = -5 and theta = sin^-1(wt) or (-5, sin^-1(wt))

-------

y = 4.00m - 5.00m cos(wt) or y - 4.00 = - 5.00m cos(wt)

so r = -5, theta = cos^-1(wt) or (-5, cos^-1(wt))

So if I rotate a directed distance r from the origin through all the plane, I get a circle (correct?). In this case r happens to be -5, but 5, since raidius is always positive...

So the equation of a circle is of the form x^2 + y^2 = r^2

Then: x^2 + (y - 4)^2 = 5^2

Is this correct?

-------

May I asked another question? And if r is not -5??

I tried the following:

r = x + y or r = [-5sin(wt)] + [4 -5cos(wt)], let (wt) = O ,

so r = 4 - 5 [ sinO + cosO ] , (common factor)

r^2 = 4r - 5r[ sinO + cosO ] , (multyplied by r)

x^2 + y^2 = 4r - (x + y), (substituting: x^2 + y^2 = r^2, -5r sinO = x, -5r cosO = y)

x^2 + y^2 + x + y = 4r or [ x^2 + y^2 + x + y ]/4 = r ???

I have the feeling that this is redundant and a tautology (?)

Last edited: Sep 20, 2011
4. Sep 21, 2011

### Hootenanny

Staff Emeritus
Not quite. I'm not entirely sure what you're doing here. Take a look at this page on how to convert to polar coordinates: http://tutorial.math.lamar.edu/Classes/CalcII/PolarCoordinates.aspx and then try again.

5. Sep 21, 2011

### naaa00

So, the position vector is:

R = 4j + 5 [ -Sin(tw)i + Cos(wt)j ].

Components are:

(x) = -5sin(wt),

(y) = 4 - 5cos(wt) or (y - 4) = -5cos(wt).

So, converting to polar coordinates:

x^2 + y^2 = r^2,

(x)^2 + (y - 4)^2 = r^2,

[-5sin(wt)]^2 + [-5cos(wt)]^2 = r^2, (plugging components);

25sin^2(wt) + 25cos^2(wt) = r^2,

25[sin^2(wt) + cos^2(wt)] = r^2, [CF, and substituting: sin^2(wt) + cos^2(wt)= 1];

25(1) = r^2 or 5 = r.

And

(x)^2 + (y - 4)^2 = 5^2 => a circle with radius 5 and centered at (0,4).

I suppose that's the answer. I cannot believe it was that simple. This is frustrating.

6. Sep 22, 2011

### naaa00

By the way, thanks for the link! It was very useful! I learnt many things and still doing it. And, well, thanks for the help!

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