# Finding frequency of a specific mechanical oscillator -- horizontal rod on pivot

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1. Jan 10, 2017

### srecko97

1. The problem statement, all variables and given/known data

There is a cyllinder with radius 0.5 m fixed on the wall. We put a 6 metres long thin rod with mass 0.3 kg on it, which does not slip. I would like to calculate the oscillating time. It is a part of a clock, so the oscillating time is probably 1 or 2 seconds, but I got around 5 seconds. Please help me find a mistake.

2. Relevant equations

3. The attempt at a solution
I got a wrong result

I am used to write M (torque) and J (moment of inertia)

Last edited: Jan 10, 2017
2. Jan 10, 2017

### TSny

Your work looks correct. As a check, I used a different approach and still got the same answer.

3. Jan 10, 2017

### srecko97

Thanks for your reply ... I will wait for my schoolmates to solve it in order to compare our results. I am quite sure that everything is correct except the calculation of moment of inertia. Should I treat each part of the rod separately (the longest and the shortest part using a formula for a rod rotating around one end .. (mx^2)/3) and sum moments of inertia of both parts as it does not oscillate accurately around its centre of mass.

4. Jan 10, 2017

### TSny

You can show that the contribution of the center-of-mass motion to the inertia of the system is negligible for small oscillations. So, you are OK with just taking into account the rotational inertia about the center of mass, $\frac{1}{12}ML^2$.

I did a rough experiment and the result for the period was reasonably in agreement with the formula you derived. I used a yard stick on a cylinder of approximate diameter 2.75''. Measured period was about 2.6 s. Compare to the formula prediction of 2.8 s.

Last edited: Jan 10, 2017
5. Jan 10, 2017

### srecko97

Thanks for all your help! I really appreciate it! You inspired me to do the experiment. I put a thin tube (L=1.02 metres) on a ball (R=0.107 m). Formula prediction is tequation=1.82 s. I measured it tmeasured=1.92 s ... This is almost 5% accurate.
As older students write homework for us it is quite possible that the given data are not correct :D . Well, it is everything OK, but I do not know what sense would an oscillator with oscillating time 4,87 s have in relation to a clock. There is still an option that I have made a mistake... Correct me please, if so...

6. Jan 10, 2017

### haruspex

I see no difficulty. Pendulum clocks are not expected to be accurate to the second. Within 15 seconds would be considered excellent.

7. Jan 11, 2017

### srecko97

Do you think my solution is correct or do you just answer to my doubt about the relation to clock?

8. Jan 11, 2017

### haruspex

I was just responding to your doubt about suitability as a clock. If TSny says your work is correct, it is.

9. Jan 15, 2017

### srecko97

The next question is: What happens if we calculate the oscillating time considering that the rod is not thin, but it has width 2b (it is a cyllinder). So b=radius of this cylinder. I think nothing changes except the moment of inertia.
I am caluculating it in this way:

I found an equation for a rotating cyllinder around the perpendiculat axis (on the picture marked as xC or yC) in the middle of a cylinder.

J(moment of inertia) = 1/4 m*b^2 + 1/12 m*L^2
In our case the cylinder is not rotating around the axis in the middle of it, but around the axis which is a tangent on surface of this cylinder, so I think that the desired moment of inertia for our case considering Steiner's rule is:
J = m*b^2 + 1/4 m*b^2 + 1/12 m*L^2

Is my thinking correct?

10. Jan 15, 2017

### TSny

Are the moment arms of the forces affected by b?
OK (for small φ)

Last edited: Jan 15, 2017
11. Jan 15, 2017

### srecko97

Yep I forgot to consider that. The moment arms gets longer considering Pitagora's Theorem. arm^2= former arm^2 + b^2, right?

12. Jan 15, 2017

### TSny

Does this give the correct moment arm for each side when φ = 0? Does the moment arm on the left side and the moment arm on the right side both get longer due to b?

13. Jan 15, 2017

### haruspex

No, that's not a good guess. Draw a diagram.

14. Jan 15, 2017

### srecko97

I think it gets very very very very complicated if I consider the change of length of the momentum arm

15. Jan 15, 2017

### srecko97

I do not have any data about b, so I cannot caluclate numbers for alpha or R, so it would get more difficult if I wanted to calculate the omega without numbers.

16. Jan 15, 2017

### srecko97

.

17. Jan 15, 2017

### TSny

You are interested in how much the moment arm changes as you add the effect of b. Your R does not represent the change in the moment arm. Think about where the force on the left side acts before taking into account b and then after taking into account b.

18. Jan 15, 2017

### srecko97

The force acts in the center of mass of the left part

19. Jan 15, 2017

### srecko97

Honestly I do not understand clearly what you want to tell me. Can you help me please with some equations or sketches?

20. Jan 15, 2017