Describe the path of the object in an xy plot.

  • Thread starter naaa00
  • Start date
  • Tags
    Path Plot
In summary, the problem statement is finding a way to express the position, velocity, and acceleration vectors in an xy plot, and describing the path of the object in that plane. The coordinates of the object vary with time according to the equations: x = - 5.00m sin(wt), y = 4.00m + 5.00m cos(wt). The Attempt at a Solution is to find expression for the position, velocity, and acceleration vectors in polar coordinates, and describe the path of the object in that coordinate system.
  • #1
naaa00
91
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
  • #2
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?
 
  • #3
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:
  • #4
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.
 
  • #5
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
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!
 

Related to Describe the path of the object in an xy plot.

1. What is an xy plot?

An xy plot is a type of graph that displays the relationship between two variables, typically an independent variable (x-axis) and a dependent variable (y-axis).

2. How is the path of an object represented on an xy plot?

The path of an object can be represented on an xy plot by plotting the position of the object at different time intervals. This creates a line or curve on the graph that shows the object's movement in the x-y plane.

3. What do the x and y axes represent in an xy plot?

The x-axis represents the independent variable, which is usually time in most cases. The y-axis represents the dependent variable, which can be any physical quantity such as distance, velocity, or acceleration.

4. How can an xy plot be used to analyze the path of an object?

An xy plot can be used to analyze the path of an object by examining the shape of the line or curve on the graph. The slope of the line can indicate the direction and speed of the object, while the curvature can show changes in speed or acceleration.

5. What factors can affect the path of an object on an xy plot?

The path of an object on an xy plot can be affected by several factors, including the initial velocity of the object, the presence of any external forces such as friction or gravity, and changes in the object's mass or shape.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
686
  • Introductory Physics Homework Help
Replies
7
Views
998
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
29
Views
1K
  • Introductory Physics Homework Help
Replies
27
Views
4K
Back
Top