With this problem I have to get the answer: cosθ = r/L * cosα - √(3)/2 * (1 - r/L)
which in other words mean I need to find angle θ with arccos[r/L * cosα - √(3)/2 * (1 - r/L)].
Here's the picture:
Lcosθ is the vertical length of the string at its lowest point.
rcosα is a fraction of that same vertical string in terms of displacement "r" (which is from the start of the string to the peg)
∠β is the angle between the peg and the horizon.
(L-r) sinβ is the height from the end of the peg and the horizontal
(L-r) cosα is the horizontal length of that same peg.
(L-r)cosα is the vertical length of the string from the ball to the peg.
So, this is not really a physics issue but more like a math issue but since this is a physics problem I've decided to put it under here.
My problem is that I am unable to continue from this point as shown on the picture of my attempt. I don't know where to continue from here on out. I am trying to find "t" for the equation but I am unsure how. Where do I continue from now?
Newtonian Position Formula:
yf = yi +viyt + .5gt2
xf = xi +vixt + .5gt2
Work of hand - force of friction * displacement = delta Kinetic Energy + delta Potential Energy
Wh - fF*d = [.5*mvf2 - .5*mvi2] - [mghf - mghi]
The Attempt at a Solution
Picture of Attempt: