Simple Harmonic Motion with Cylinder Attached to Spring
1. The problem statement, all variables and given/known data
Okay, so we've got a solid cylinder attached to a horizontal massless spring that rolls without slipping along a horizontal surface. The system is released from rest with the spring stretched past the equilibrium point. (The first two parts of this problem ask for translational and rotational kinetic energy values given the spring constant and the amount the spring is stretched. I already solved those and got those correct.) Show that under these conditions, the center of mass of the cylinder executes simple harmonic motion, and that the period T can be expressed in terms of the mass M of the cylinder as follows: T=2[itex]\pi[/itex]√(3M/2k) I'm given the hint: Find the time derivative of the total mechanical energy. 2. (Maybe?) Relevant equations T=2[itex]\pi[/itex]/ω ω[itex]^{2}[/itex]=k/m x'(t) for a spring = ωxsin(ωt) 3. The attempt at a solution My process seems a little iffy. Am I on the right track? Okay, so let's try to use the hint. I have the total mechanical energy equation: 1/2kx[itex]^{2}[/itex]=1/2mv[itex]^{2}[/itex]+1/2Iω[itex]^{2}[/itex]=3/4mv[itex]^{2}[/itex] I also know that v=X'(t)=ωxsin(ωt) 1/2kx[itex]^{2}[/itex]= 1/2m(ω[itex]^{2}[/itex]x[itex]^{2}[/itex]sin[itex]^{2}[/itex](ωt) x cancels... 2k/3m=ω[itex]^{2}[/itex]sin[itex]^{2}[/itex](ωt) This is starting to look like the equation given...but now I take the time derivative? Won't I get sines and 't's in that equation? I'm not really sure how to do this final step. How do I get rid of the ugly sine? I did a similar problem without rolling, but I got a sine squared and a cosine squared which went to one. How does this one work? 
Re: Simple Harmonic Motion with Cylinder Attached to Spring
Quote:

Re: Simple Harmonic Motion with Cylinder Attached to Spring
Quote:

All times are GMT 5. The time now is 10:14 PM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums