Simple Harmonic Motion with Cylinder Attached to Spring

In summary, the conversation discusses a problem involving a solid cylinder attached to a horizontal massless spring that rolls without slipping on a horizontal surface. The first two parts of the problem ask for the translational and rotational kinetic energy values, which were solved correctly. The main focus is on showing that 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. The relevant equations and relevant attempts at a solution are also mentioned. The final step involves finding the total time derivative of the total energy and getting rid of the sine term to obtain the correct equation for the period T.
  • #1
DoTell
6
0

Homework Statement



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)

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?
 
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  • #2
DoTell said:
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?
 
  • #3
DoTell said:
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.
 

1. What is simple harmonic motion with a cylinder attached to a spring?

Simple harmonic motion occurs when an object moves back and forth in a regular pattern, such as a pendulum swinging or a mass on a spring bouncing up and down. When a cylinder is attached to a spring, it will also exhibit simple harmonic motion as it moves back and forth due to the restoring force of the spring.

2. How is the frequency of simple harmonic motion determined with a cylinder and spring?

The frequency of simple harmonic motion with a cylinder and spring is determined by the mass of the cylinder and the stiffness of the spring. The larger the mass or the stiffer the spring, the lower the frequency will be.

3. What factors affect the amplitude of simple harmonic motion with a cylinder and spring?

The amplitude of simple harmonic motion with a cylinder and spring is affected by the initial displacement of the cylinder, the mass of the cylinder, and the stiffness of the spring. A larger initial displacement, higher mass, or stiffer spring will result in a larger amplitude of motion.

4. Can the period of simple harmonic motion be changed with a cylinder and spring?

Yes, the period of simple harmonic motion with a cylinder and spring can be changed by altering the mass of the cylinder or the stiffness of the spring. Increasing the mass or stiffness will result in a longer period, while decreasing them will result in a shorter period.

5. What is the relationship between the frequency and period of simple harmonic motion with a cylinder and spring?

The frequency and period of simple harmonic motion with a cylinder and spring are inversely proportional. This means that as the frequency increases, the period decreases, and vice versa. This relationship is described by the equation f=1/T, where f is the frequency and T is the period.

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