Register to reply

Conservation of Energy

by mathmannn
Tags: conservation, dynamics, rigid bodies
Share this thread:
mathmannn
#1
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\



EDIT:
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?
Attached Thumbnails
Screen Shot 2012-04-13 at 12.17.28 AM.png  
Phys.Org News Partner Science news on Phys.org
Fungus deadly to AIDS patients found to grow on trees
Canola genome sequence reveals evolutionary 'love triangle'
Scientists uncover clues to role of magnetism in iron-based superconductors
cng99
#2
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.
mathmannn
#3
Apr13-12, 05:07 PM
P: 15
Ugh, I'm an idiot haha. Thank you! I got it now!


Register to reply

Related Discussions
Conservation of momentum vs conservation of kinetic energy Introductory Physics Homework 18
Capacitors - Conservation of Charge vs Conservation of Energy Introductory Physics Homework 3
WHat is more fundamental - conservation of momentum or conservation of energy? Classical Physics 12
Linear momentum conservation vs mecanical energy conservation General Physics 1
Simulatenous Conservation of Momentum and Conservation of Energy Engineering Systems & Design 9