Recent content by achap01

is there an online or a video proof of this that i can get

harmonic motion, so angular frequency (rate of oscillation) would be g/R, which have a derivative of 0 with respect to theta?

so the ODE would be ẍ=a=-gx/R? Im a little confused

would ẍ just be the second derivative of x with respect to time, so ẍ=a=9.8x/R?

sorry i mean 9.8x/R=a

I go to 9.8/R=a but i dont know what to do now

is there a proof online bc i think im close now

cos(arccos(x/r)) = x/r If this is right im gonna feel so braindead

arccos(x/r)=theta cos(theta)=x/r

the only way i see to derive that theta(angles between x and r) is with the angle I mentioned previously, but by subtracting it from 90deg. So theta =90-(θx/2rsin(θ/2)). I don't understand how to get it without the central angle

cos(theta) of Fg. Is there any way I could see a reference for this proof bc I feel like this convo is going to take a while with the responses. I can ask you about the reference too, but I feel like I'd learn much better with it.

cos(θ) of Fg

I mistypes in the first post. it should be L = 2rsin(θ/2). I got the angle by finding the fraction of the total chord length x is, and then multiplying that with the central angle

force of gravity towards the center would be (r/R(radius of earth))*weight. Angle of Fg would be θx/2rsin(θ/2), so force parallel to chord would be w(r/R)*sin(θx/2rsin(θ/2)). Then plug in to f=ma

the forces would just be gravity(and normal force perpendicular to the chord) gravity would just be linear so (dist from center)/R * weight(at surface). To get an eq relating gravitational force towards the center with force parallel to the chord would require both time and dist which would have...