JUST

JUST

Is that supposed to be difficult?

for kids like OP, yeah

F = kx
Fc = (mv^2)/r

Wow 2 equations op so hard

well you know the radius of a circle is pir^2. so you have the circumference now. the squiggly w looks like roughly 45 degrees so thats just an eigth of the area. now to find the area of the sin wave you just take the integral from m to the radius. whats so hard?

>the squiggly w looks like 45 degrees

r''(t) = -k/m*(r(t)-R) + w^2 / r(t)

>radius of a circle is pir^2

>squiggly w

You bring shame to this board

>> Falling for that bait

lol yo shutup

>le epin bate
Take it to /r9k/

Just post the circumference

>squiggly w

[eqn]L=\frac{M}{2}(r\dot{\theta})^2 - kr^2/2 [/eqn]
solv'd

assuming its vertically oriented too
[eqn]L=\frac{M}{2}(r\dot{\theta})^2 - kr^2/2 - Mgr\sin\theta[/eqn]

Given M,k,w find R

>what is the equilibrium length

Question
When a term has a dot over it, what does that signify? I kept seeing it in my differential geometry class, but the professor never told us what it was.

for real

mRw^2 = 0.5kR^2

thus, R=(2m*w^2/k)

en.wikipedia.org/wiki/Time_derivative

Thank you friend.

Check dimensions: MRw^2 is a force while kR^2/2 is an energy.

You need a minimum radius r (unstretched spring or mass offset from center)
Mw^2R = k(R-r) -> R(k-Mw^2) = kr -> R = r/(1-Mw^2/k)

get headache from this stupidy