Describe the path of the object in an xy plot.

  • Thread starter Thread starter naaa00
  • Start date Start date
  • Tags Tags
    Path Plot
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 8K views
naaa00
Messages
90
Reaction score
0

Homework Statement



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.

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.
 
Physics news on Phys.org
naaa00 said:

Homework Statement



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.

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.
This question is best attempted in plane polar coordinates. Do you think you could write down the position in polar coordinates?
 
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:
naaa00 said:
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))
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.
 
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.
 
By the way, thanks for the link! It was very useful! I learned many things and still doing it. And, well, thanks for the help!