Bouncing Half Pendulum Math (Arccos problems)

[brandonhil7]

So what I've got is a ball on a string hanging against a wall. When lifted and dropped, it forms a sort of half pendulum motion, but it looses energy very quickly as compared to a full pendulum. I need to calculate energy loss on each bounce, given only the time between each bounce. The solution to me seems to be relating height to half the time between each bounce, and then using height to calculate potential energy at the maximum point in each swing. Its the math that is giving me problems.

v = $\frac{s}{t}$
s = rθ
∴ v = $\frac{rθ}{t}$

E = 0.5mv² = mgh
0.5v² = gh
v² = 19.62h
∴ $\frac{r^{2}θ^{2}}{t^{2}}$ = 19.62h

to relate theta to height, a horizontal line can be imagined from the wall to the ball creating a right angled triangle with the hypotenuse being r, the adjacent wall being equal to r-h.

∴ cos(θ)r = r - h
θ = arccos($\frac{r-h}{r}$)

given this:
$\frac{r^{2}(arccos(\frac{r-h}{r}))^{2}}{t^{2}}$ = 19.62h

and if r is then assumed to be 1 m:
$\frac{(arccos(1-h))^{2}}{t^{2}}$ = 19.62h

And this is where I'm stuck. I can't figure out how to get rid of the arccos, and i can't find any identities to help.
$(arccos(1-h))^{2}= 19.62ht^{2}$
what i need to do is solve for H
Please note that i do indeed understand what arccos is, it is the squared that id like to get rid of.

Any help would be appreciated.

 

[jbsteele]

There is no simple way of getting rid of the arccos, but it can be solved numerically by iteration. Start with an initial guess for h, then calculate the corresponding t2 from the equation, and adjust h until the equation is satisfied. This can be done using a computer program or spreadsheet, or you can do it manually with a calculator.