Ax and Ay As a linear combination of Atan and Arad

  • Thread starter spirof
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  • #1
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Homework Statement



The problem is about a rod that is set in a pendulum way but that has an angle high enough so the SHM doesn't apply to it. It starts at Pi/2 until it reachs 0. I was able to find the tangential and radial acceleration about the center of mass but now I need to know the x and y acceleration about the center of mass.


Homework Equations



It says that I should be using the angle between Atan and Ax as a reference

The Attempt at a Solution



Completly stuck!

Homework Statement





Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi spirof! welcome to pf! :smile:
I was able to find the tangential and radial acceleration about the center of mass but now I need to know the x and y acceleration about the center of mass.
the only difficulty is working out the angle between tan (or rad) and x (or y) …

the hint is simply telling you to write those angles as either ±θ or ±(90° - θ) …

go for it! :wink:
 
  • #3
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hi spirof! welcome to pf! :smile:


the only difficulty is working out the angle between tan (or rad) and x (or y) …

the hint is simply telling you to write those angles as either ±θ or ±(90° - θ) …

go for it! :wink:
Thanks u did put me on the track, I do possess all the θ already for a definite period of time. Actually, correct if I am wrong, if I have and angle of 1.40 rad, then if if i do pi/2 - 1.4, which will give me something around 0.17 rad.

From there, i know that the angle between arad and ax is 0.17 rad, the same angle betweenay and atan. I would then be tempted to simply do ax = ar cos 0.17. However, I cannot assume that they form a right angle triangle. Anymore inputs?
 
  • #4
tiny-tim
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hi spirof! :smile:
From there, i know that the angle between arad and ax is 0.17 rad, the same angle betweenay and atan. I would then be tempted to simply do ax = ar cos 0.17. However, I cannot assume that they form a right angle triangle. Anymore inputs?
i'm confused :redface: … there's right-angles everywhere …

which angle isn't a right- angle? :confused:
 
  • #5
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hi spirof! :smile:


i'm confused :redface: … there's right-angles everywhere …

which angle isn't a right- angle? :confused:
Well between Ar and Ax, If u look at my attached image, you will see that I cannot assume that they are parts of a right angle triangle. Ax seems to long to be the hypothenus of Ar, doesn't it?
 
  • #6
tiny-tim
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the length doesn't matter

all you need is the angle between ar and ax
 
  • #7
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the length doesn't matter

all you need is the angle between ar and ax
Thanks, I went over the internet and I have found the solution. I have to express as linear combination of the Ar and Atan.

It sums up to Ax = Ar cos (90-θ) + Atan sin (90-θ)
Ay = -Ar sin (90-θ) + Atan cos (90-θ)
 

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