# Conservation of Energy

by mathmannn
Tags: conservation, dynamics, rigid bodies
 P: 15 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? Attached Thumbnails
 P: 44 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.