Thank you, but it still doesn't seem like T = 2π(Lsinβ)/v and T = 2π√(Lcosβ/g) would equate. It seems like they would give different values for the period
https://i.imgur.com/4nynNBg.png
I cannot understand for the life of me why this problem doesn't just use T = 2πr/v to solve for the period to get T = 2π(Lsinβ)/v. Instead, it rearranges the formula for a(rad) = (4π^2R)/T^2 to solve for T, and arrives at a totally different answer, T =...