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Conservation of Energy

by mathmannn
Tags: conservation, dynamics, rigid bodies
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Apr13-12, 12:26 AM
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

[itex] T_1+V_1=T_2+V_2 [/itex]

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

[itex] T_1+V_1=T_2+V_2 [/itex] Where [itex]V_1=T_2=0[/itex] So I have

[itex] T_1=V_2 [/itex]

[itex].5 m v^2 = m g h [/itex]

[itex] h=x\sin(30) [/itex]

Which gives me [itex] .5(30)^2 = (32.2)(x \sin(30)) [/itex]

[itex] x=27.95 [/itex]

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\

I tried using Inertia like this:

[itex] T_1 = V_2 [/itex]

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

[itex] I=.5 m r^2 [/itex]

[itex] T_1 = .5((.5 m r^2)(\frac{v}{r})^2) + .5 m v^2 [/itex]

[itex] T_1 = \frac{1}{4} m v^2 + \frac{1}{2} m v^2 = \frac{3}{4}mv^2 [/itex]

[itex] \frac{3}{4}mv^2 = m g x \sin(30) [/itex]

[itex] x = \frac{3v^2}{4 g \sin(30)} \qquad, x=41.9255 [/itex]

Still not an answer but closer than I was.. Any suggestions?
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Apr13-12, 01:01 AM
P: 44
It's a sphere! Assuming this to be a case of rolling without slipping, the the total energy in the sphere is [itex]\frac{7}{10}[/itex]mv[itex]^{2}[/itex]

Here's how:

Total energy
= Translational Kinetic energy + Rotational kinetic energy
= [itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex] + [itex]\frac{1}{2}[/itex]I[itex]\omega[/itex][itex]^{2}[/itex]

(m = mass
v = velocity
I= moment of iniertia about the centre
[itex]\omega[/itex]=angular velocity about the centre
r=radius of the sphere

=[itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex] + [itex]\frac{1}{2}[/itex]([itex]\frac{2}{5}[/itex]mr[itex]^{2}[/itex])[itex]\omega[/itex][itex]^{2}[/itex]

(Because I = [itex]\frac{2}{5}[/itex]mr[itex]^{2}[/itex], for a sphere)

=[itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex] + [itex]\frac{1}{2}[/itex]([itex]\frac{2}{5}[/itex]mr[itex]^{2}[/itex])([itex]\frac{v}{r}[/itex])[itex]^{2}[/itex]

(Because v = [itex]\omega[/itex]r, for rolling without slipping)

Solving this gives [itex]\frac{7}{10}[/itex]mv[itex]^{2}[/itex].

Try putting that and you should get an answer.
Apr13-12, 05:07 PM
P: 15
Ugh, I'm an idiot haha. Thank you! I got it now!

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