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

[srecko97]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I got a wrong result

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

[/B]

 

[TSny]

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

 

[srecko97]

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

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.
I still think that the answer is not correct, so everyone, please help me solve it.

 

[TSny]

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.

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 still think that the answer is not correct, so everyone, please help me solve it.

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.

 

[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 t_equation=1.82 s. I measured it t_measured=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...

 

[haruspex]

I do not know what sense would an oscillator with oscillating time 4,87 s have in relation to a clock.

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

 

[srecko97]

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

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

 

[haruspex]

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

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

 

[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?

 

[TSny]

I think nothing changes except the moment of inertia.

Are the moment arms of the forces affected by b?

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

OK (for small φ)

 

[srecko97]

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

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

 

[TSny]

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

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?

 

[haruspex]

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

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

 

[srecko97]

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

 

[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.

 

[srecko97]

.

 

[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.

 

[srecko97]

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

 

[srecko97]

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

 

[TSny]

 

[srecko97]

On the left the moment arm gets longer, but on the right it gets shorter

 

[TSny]

On the left the moment arm gets longer, but on the right it gets shorter

Yes, that's true.

 

[haruspex]

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.

Well, you won't get a numeric answer, but you should get a function of b that you can investigate.
Some labels for points:
A = Bottom left corner of the rod
A' = Top left corner
B = Point of contact.
B' = Point on top edge nearest B (so AB=A'B')
C = Mass centre of left hand portion (i.e. from AA' to BB')
D = Point on AB nearest to C
You already know the horizontal distance from B to D. You now want the horizontal distance from D to C to add to that. Can you see how to get that?

 

[srecko97]

the new moment arm is for b*tanφ longer?

 

[TSny]

the new moment arm is for b*tanφ longer?

Apparently, this came from a triangle. Can you describe the triangle?

 

[srecko97]

 

[TSny]

OK, We are probably using the term "moment arm" differently. I think of it as the perpendicular distance from the axis of rotation to the line of action of the force. You might be thinking of it as just the distance from the axis to the point of application of the force or something else?

 

[srecko97]

Ok, I know my previous attempt was wrong. You could think that I am just guessing the right equation, but no! Now I think that my answer is correct. The moment arm gets longer for b*sinφ, ok?

 

[TSny]

The moment arm gets longer for b*sinφ, ok?

Yes, good.

 

[srecko97]

On the other side it gets shorter for b*sinφ

 

[TSny]

On the other side it gets shorter for b*sinφ

Yes.

 

[srecko97]

But when talking about small φ that has no sense

 

[TSny]

But when talking about small φ that has no sense

Why is that?

 

[srecko97]

sin φ ≈ 0 for small φ

 

[TSny]

sin φ ≈ 0 for small φ

That's going too far in the approximation.

 

[srecko97]

I did the same approximation in the first part of my homework (calculating the oscillating time), when I say cos φ≈1 for small φ

 

[TSny]

Keep terms of first order in φ. cosφ ≈ 1 - φ²/2 ≈ 1 to first order. What about sinφ?

 

[srecko97]

sinφ=φ-(φ^3)/6

 

[TSny]

sinφ=φ-(φ^3)/6

So, to first order in φ?

 

[srecko97]

oh it is φ

 

[TSny]

oh it is φ

That's right.

 

[srecko97]

Well, I got that M_right - M_left = mg(rφ-bφ) ...M=torque

 

[srecko97]

The frequency gets smaller, as numerator gets smaller and denominator gets bigger. I hope it is correct! I would like to say you a big THANK YOU TSny! You helped me a lot... I learned a lot from solving this task with your help. I love this forum...

 

[TSny]

Your answer looks correct to me. (b ≥ r is interesting to think about.)

Glad I could help.

 

[srecko97]

ω² is then negative, so there is no oscillation in real?

 

[TSny]

ω² is then negative, so there is no oscillation in real?

Yes, there is no oscillation. Looking back at your post#43, I think you need to add an overall negative sign to the first equation so that it's

$J \ddot{\phi} = - mg(r-b)\phi$.

If $b > r$ then you can see that $\ddot{\phi} >0$ if $\phi$ is given an initial small, positive value. So, the rod rotates away from the equilibrium position when it is released and there is no oscillation. Of course, the differential equation for $\phi$ breaks down as $\phi$ continues to increase since the differential equation was derived under the assumption that $\phi$ is always small.

Another way to investigate the general behavior of the system for various values of b is to derive the potential energy of the system as a function of $\phi$ without making a small angle approximation. Then plots of the potential energy function for various values of b will reveal a lot about the behavior. But that's probably for a rainy day.