Physics Forums

Physics Forums (http://www.physicsforums.com/index.php)
-   Introductory Physics Homework (http://www.physicsforums.com/forumdisplay.php?f=153)
-   -   Simple Harmonic Motion with Cylinder Attached to Spring (http://www.physicsforums.com/showthread.php?t=555299)

DoTell Nov29-11 10:06 PM

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 iff-y. 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?

BruceW Nov30-11 01:57 PM

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

Quote by DoTell (Post 3641614)
2. (Maybe?) Relevant equations

T=2[itex]\pi[/itex]/ω
ω[itex]^{2}[/itex]=k/m
x'(t) for a spring = -ωxsin(ωt)

The first equation is correct. The second equation is definitely not right, as you can see , it would not give the correct answer for the period T. (The reason the second equation doesn't apply here is because the cylinder also has rotational motion). The third equation is almost correct. There should be -omega^2, not -omega. Also, I'm assuming lower-case x means the displacement at t=0?

BruceW Nov30-11 02:01 PM

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

Quote by DoTell (Post 3641614)
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]

You've essentially written KE=PE, which isn't correct. The total energy is KE+PE. And the hint says to find the total time derivative of the total energy, so that's what your next step should be.


All times are GMT -5. The time now is 10:48 PM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums