Conservation of Energy of a rolling sphere

[mathmannn]

Homework Statement

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?

Homework Equations

T_1+V_1=T_2+V_2

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?

 

[cng99]

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 = \omegar, for rolling without slipping)

Solving this gives \frac{7}{10}mv^{2}.

Try putting that and you should get an answer.

 

[mathmannn]

Ugh, I'm an idiot haha. Thank you! I got it now!