## Conservation of Energy

1. The problem statement, all variables and given/known data

A sphere rolling with an initial velocity of 30 ft/s starts up a plane inclined at an angle of 30o with the horizontal as shown. How far will it roll up the plane before it rolls back down?

2. Relevant equations

$T_1+V_1=T_2+V_2$

3. The attempt at a solution
We are doing rigid bodies so I started with

$T_1+V_1=T_2+V_2$ Where $V_1=T_2=0$ So I have

$T_1=V_2$

$.5 m v^2 = m g h$

$h=x\sin(30)$

Which gives me $.5(30)^2 = (32.2)(x \sin(30))$

$x=27.95$

And that is not one of the answers, I assume inertia is supposed to be used somewhere but I have no idea where to plug it in because no radius of the circle is give.. Any help would be very much appreciated\

EDIT:
I tried using Inertia like this:

$T_1 = V_2$

$T_1 = .5 I \omega^2 + .5mv^2 \quad, \qquad \omega = v/r$

$I=.5 m r^2$

$T_1 = .5((.5 m r^2)(\frac{v}{r})^2) + .5 m v^2$

$T_1 = \frac{1}{4} m v^2 + \frac{1}{2} m v^2 = \frac{3}{4}mv^2$

$\frac{3}{4}mv^2 = m g x \sin(30)$

$x = \frac{3v^2}{4 g \sin(30)} \qquad, x=41.9255$

Still not an answer but closer than I was.. Any suggestions?
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 It's a sphere! Assuming this to be a case of rolling without slipping, the the total energy in the sphere is $\frac{7}{10}$mv$^{2}$ Here's how: Total energy = Translational Kinetic energy + Rotational kinetic energy = $\frac{1}{2}$mv$^{2}$ + $\frac{1}{2}$I$\omega$$^{2}$ (m = mass v = velocity I= moment of iniertia about the centre $\omega$=angular velocity about the centre r=radius of the sphere ) =$\frac{1}{2}$mv$^{2}$ + $\frac{1}{2}$($\frac{2}{5}$mr$^{2}$)$\omega$$^{2}$ (Because I = $\frac{2}{5}$mr$^{2}$, for a sphere) =$\frac{1}{2}$mv$^{2}$ + $\frac{1}{2}$($\frac{2}{5}$mr$^{2}$)($\frac{v}{r}$)$^{2}$ (Because v = $\omega$r, for rolling without slipping) Solving this gives $\frac{7}{10}$mv$^{2}$. Try putting that and you should get an answer.
 Ugh, I'm an idiot haha. Thank you! I got it now!

 Tags conservation, dynamics, rigid bodies