Ax and Ay As a linear combination of Atan and Arad

[spirof]

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!

 

[tiny-tim]

welcome to pf!

hi spirof! welcome to pf!

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! ​

 

[spirof]

hi spirof! welcome to pf!


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! ​

Thanks u did put me on the track, I do possesses 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?

 

[tiny-tim]

hi spirof!

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 … there's right-angles everywhere …

which angle isn't a right- angle? ​

 

[spirof]

hi spirof!


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

which angle isn't a right- angle? ​

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?

 

[tiny-tim]

the length doesn't matter

all you need is the angle between a_r and a_x

 

[spirof]

the length doesn't matter

all you need is the angle between a_r and a_x

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)