# Describing path of the object in an xy plane

• member 731016
In summary, the conversation discusses the equation of a circle centered at a given point and how it can be used to determine the motion of an object. The coordinates of the object satisfy the equation of the circle, which is how it is known that the object moves in a circular path with a radius of 5.00m centered at (0,4.00m). The equation of the circle is also simplified using trigonometric identities to better understand the motion of the object.
member 731016
Homework Statement
Relevant Equations
For part(d) of this problem,

The solution is,

However, how did they know that the object moves in a circle of radius 5.00m centered at (0,4.00m)?

Many thanks!

What is the equation of a circle centered at ##(x_0,y_0)##?

member 731016
The coordinates satisfy the equation of such a circle. This is how they know.

member 731016
Thank you for your replies @kuruman and @nasu!

The equation of a circle centered at ##(x_0, y_0)## is ##(x - x_0)^2 + (y - y_0)^2 = r^2##

Many thanks!

Can you bring what you have in that form? Remember that ##~\sin^2x+\cos^2x=1.##

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kuruman said:
Can you bring what you have in that form? Remember that ##~\sin^2x+\cos^2x=1.##

Here is what I got:

##x^2 + y^2 = r^2 ##
##(-5.00sin\omega t)^2 + (4.00 - 5cos\omega t)^2 = r^2 ##
##25sin^2\omega t + 16 - 40cos\omega t +25cos^2\omega t = r^2 ##
##25(sin^2\omega t + cos^2\omega t) +16 - 40cos\omega t = r^2 ##
## 41 - 40cos\omega t = r^2 ##

EDIT: I'm now thinking that I should put the individual position function in the x-components in the y-components in the circle formula I posted.

Many thanks!

Callumnc1 said:
Here is what I got:
Don't try to prove that way ...
Callumnc1 said:
EDIT: I'm now thinking that I should put the individual position function in the x-components in the y-components in the circle formula I posted.
Yes.
We know that ##x=-5sinωt## and ##y-4=-5cosωt## so we have:
##x^2+(y-4)^2=25sin^2ωt+25cos^2ωt=25## and we know that ##x^2+(y-4)^2=5^2## is equation of a circle with ##r=5## that is centered at ##(0,4)##.

member 731016
MatinSAR said:
Don't try to prove that way ...

Yes.
We know that ##x=-5sinωt## and ##y-4=-5cosωt## so we have:
##x^2+(y-4)^2=25sin^2ωt+25cos^2ωt=25## and we know that ##x^2+(y-4)^2=5^2## is equation of a circle with ##r=5## that is centered at ##(0,4)##.

MatinSAR

## 1. What is the equation of motion for an object moving in an xy plane?

The equation of motion for an object moving in an xy plane can be described using parametric equations: x(t) = x0 + vxt + 0.5axt2 and y(t) = y0 + vyt + 0.5ayt2, where x0 and y0 are the initial positions, vx and vy are the initial velocities, and ax and ay are the accelerations in the x and y directions, respectively.

## 2. How do you determine the trajectory of an object in an xy plane?

The trajectory of an object in an xy plane can be determined by eliminating the time variable (t) from the parametric equations of motion. By solving one of the equations for t and substituting it into the other, you can obtain a single equation relating x and y, which describes the path or trajectory of the object.

## 3. What is the significance of the velocity vector in describing the path of an object?

The velocity vector, represented as v = (vx, vy), is significant because it indicates both the speed and direction of the object's motion at any given point in time. The components vx and vy describe the rates of change of the object's position in the x and y directions, respectively. The direction of the velocity vector is tangent to the path of the object.

## 4. How do you calculate the position of an object at a specific time in an xy plane?

To calculate the position of an object at a specific time t in an xy plane, you use the parametric equations of motion: x(t) = x0 + vxt + 0.5axt2 and y(t) = y0 + vyt + 0.5ayt2. By substituting the given time t into these equations, you can find the coordinates (x, y) of the object's position at that time.