Describing path of the object in an xy plane

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The discussion centers on understanding the motion of an object in a circular path of radius 5.00m centered at (0, 4.00m). Participants clarify that the equation of a circle, given by (x - x0)² + (y - y0)² = r², confirms this motion. The position functions x = -5sin(ωt) and y - 4 = -5cos(ωt) are derived, leading to the conclusion that the object satisfies the circle's equation. The final consensus emphasizes the correctness of using these functions to describe the circular path. The mathematical derivation effectively demonstrates the object's trajectory in the xy-plane.
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Homework Statement
Please see below
Relevant Equations
Please see below
For part(d) of this problem,
1676517142176.png

The solution is,
1676517195998.png

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

Many thanks!
 
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What is the equation of a circle centered at ##(x_0,y_0)##?
 
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The coordinates satisfy the equation of such a circle. This is how they know.
 
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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.##
Thank you for your reply @kuruman !

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)##.
 
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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)##.
Thank you for your reply @MarinSAR! I see now :)
 
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