# Eigenfrequency calculation

1. Jul 4, 2010

### dirk_mec1

1. The problem statement, all variables and given/known data
Determine the eigenfrequency of the disc with mass m = 20 kg of fig. 11-38. Assume that the disc doesn't slides.

http://img80.imageshack.us/img80/8408/86061698.jpg [Broken]

2. Relevant equations

3. The attempt at a solution
If the angle $$\theta$$ is measured in radians then the distance travelled downwards is $$R \cdot \theta$$, now Newton in the direction of the spring leads to:

$$m R \ddot{ \theta} + k R \theta = 0$$

Thus the eigenfrequency is:

$$\omega = \sqrt{ \frac{kR }{mR} } = \sqrt{60}$$

$$2\pi f= \omega \rightarrow f= \frac{\sqrt{60} }{2 \pi} = 1.23 Hz.$$

Last edited by a moderator: May 4, 2017
2. Jul 4, 2010

### vela

Staff Emeritus
There has to be a torque on the disk if it's going to rotate instead of just slide. That means there's another force on the disk that you need to account for.

3. Jul 4, 2010

### dirk_mec1

But in the picture there's no torque given, so the exercise is wrong?

4. Jul 4, 2010

### vela

Staff Emeritus
The picture doesn't show any of the forces and torques. As with every dynamics problem, you need to begin the analysis of the situation by identifying what forces are involved and go from there.

5. Jul 4, 2010

### dirk_mec1

Well I've got the gravity force and the spring force, but both go through in the center of the disc and

$$\sum M_{com} =I \ddot{\theta}$$

is then zero. What am I doing wrong?

6. Jul 4, 2010

### vela

Staff Emeritus
You're still missing a few forces. Think about the disk and the surface it's rolling on.

7. Jul 4, 2010

### dirk_mec1

But there isn't any information about friction forces, right? Then there's the normal force but it's perpendicular to the surface so it doesn't matter, right?

8. Jul 4, 2010

### vela

Staff Emeritus
There's information about friction. The problem states that the disc doesn't slide.

9. Jul 4, 2010

### dirk_mec1

So the friction force is mg*sin(30)?

10. Jul 4, 2010

### vela

Staff Emeritus
No. Just call it F for now. Write down the equations of motion for the disc.

11. Jul 7, 2010

### dirk_mec1

I got it thanks, Vela. I used sum of moments is zero around the contact point of the disc and newton along the direction of the disc, eliminated F and transformed the eigenfequency to Hertz this results in 1.006 Hz = 1.01 Hz.

12. Jul 7, 2010

### vela

Staff Emeritus
The sum of the moments about the contact point isn't 0 if the disc accelerates.