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A falling pendulum question

  • Thread starter LiorE
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Homework Statement


Hi,

this is supposed to be an easy question, but for some reason I can't get it to work. The question is:

A weightless rod is hinged at O so that it can rotate without friction in a vertical plane. A mass m is attached to the end of the rod A, which is balanced vertically above O. At time t = 0, the rod moves away from the vertical with negligible initial angular velocity. Prove that the mass first reaches the position under O at t = √(OA/g) ln (1 + √2).



The Attempt at a Solution



First let OA = r, and let \theta be the angle from the upward vertical. Now, from conservation of energy:

[tex]\frac{1}{2}m r^2 \dot{\theta}^2 = mgr(1-\cos \theta)[/tex]

This is a seperable equation, and after rearranging we get:

[tex]dt = \sqrt{\frac{r}{2g(1-\cos\theta)}} d\theta[/tex]

This is where things start to go wrong. This integral diverges when I try to take it from theta=0. Why? Is it because I'm starting with zero velocity?

Also, the result of the indefinite integral is (using mathematica):
[tex]
t(\theta) = \sqrt{\frac{2r}{g}} \ln \left( \tan\frac{\theta}{4}\right).[/tex]

I'm trying to think what kind of angles I'm supposed to put in there to get the given result, but I can't work it out.

Thanks in advance,

Lior
 

Answers and Replies

  • #2
kuruman
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When the problem states that "At time t = 0, the rod moves away from the vertical with negligible initial angular velocity" this means something. What do you think it means? Have you used this assumption in your derivation?
 
  • #3
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Well, to be honest I'm not sure how I should use this assumption.

I thought about including it by adding a constant term to the energy, which changes the integral to having:

a + 2g(1-cos\theta)

instead of

2g(1-cos\theta)

in the denominator. But this evaluates to a hypergeometric function, and when I try to take the limit of this when a->0 I get nonsense.

Thanks,

Lior
 
  • #4
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That is - it gives an elliptic function.
 
  • #5
kuruman
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Well, to be honest I'm not sure how I should use this assumption.
If "at time t = 0, the rod moves away from the vertical with negligible initial angular velocity", how big do you think the pendulum's maximum angular displacement will be? Choose one from A to C.

A. Very big.
B. Medium big.
C. Not big at all.
 
  • #6
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Hi,

If I understand what you mean by your question (to take a small displacement approximation and linearize the problem), please note that the pendulum starts from *upward* position. This isn't a small amplitude oscillation problem.
 
  • #7
kuruman
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Hi,

If I understand what you mean by your question (to take a small displacement approximation and linearize the problem), please note that the pendulum starts from *upward* position. This isn't a small amplitude oscillation problem.
I am dreadfully sorry; I misread the problem :redface:

You might find what you need here.

http://en.wikipedia.org/wiki/Pendulum_(mathematics))

*** On edit ***
I am not convinced that the swing time is constant as the problem implies. Look at the animations in the above references, specifically the one where the pendulum "just barely" makes it around and imagine what would happen as the initial velocity becomes smaller and smaller.
 
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  • #8
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That's right - intuitively we can expect that the time will go to infinity as we take the velocity to zero, and it seems that this is what happens.

But since the question is from a Putnam final contest, I guess that the person who wrote it knew what he was doing.

What is also hard for me to understand is how exactly we managed to handle a second order ODE by specifying just one initial condition (displacement). Where does the initial velocity enter the equation? My guess is that it should enter through the total energy, but I am not sure, since a shift in the total energy can also simply correspond to a shift of the point of zero potential energy.

Maybe I should talk to someone who knows about dynamical systems about this problem.
 
  • #9
kuruman
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You can assume some (small) initial angular velocity ω0, conserve energy, solve the diff. eq. (if you can, because it doesn't appear to be separable) and then let the initial velocity to go to zero.

I will be very curious to see how this is done mathematically. I cannot help thinking physically. If you make any headway, please continue this thread. Thanks.
 
  • #10
Ray Vickson
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Homework Statement


Hi,

this is supposed to be an easy question, but for some reason I can't get it to work. The question is:

A weightless rod is hinged at O so that it can rotate without friction in a vertical plane. A mass m is attached to the end of the rod A, which is balanced vertically above O. At time t = 0, the rod moves away from the vertical with negligible initial angular velocity. Prove that the mass first reaches the position under O at t = √(OA/g) ln (1 + √2).



The Attempt at a Solution



First let OA = r, and let \theta be the angle from the upward vertical. Now, from conservation of energy:

[tex]\frac{1}{2}m r^2 \dot{\theta}^2 = mgr(1-\cos \theta)[/tex]

This is a seperable equation, and after rearranging we get:

[tex]dt = \sqrt{\frac{r}{2g(1-\cos\theta)}} d\theta[/tex]

This is where things start to go wrong. This integral diverges when I try to take it from theta=0. Why? Is it because I'm starting with zero velocity?

Also, the result of the indefinite integral is (using mathematica):
[tex]
t(\theta) = \sqrt{\frac{2r}{g}} \ln \left( \tan\frac{\theta}{4}\right).[/tex]

I'm trying to think what kind of angles I'm supposed to put in there to get the given result, but I can't work it out.

Thanks in advance,

Lior
It really does seem that the person who set the question is wrong. If you consult the Wikipedia article on pendulum mathematics, you will see an expression involving an Elliptic function for the period, T(theta_0), given a swing between -theta_0 and +theta_0 from the downward vertical. It can be shown that as theta_0 --> Pi from below, T(theta_0) --> infinity. The time you want (starting from rest at pi - theta_0 from the upward vertical) is T(theta_0)/2, so grows without bound as we take pi-theta_0 closer and closer to zero, and is +infinity in the limit. Basically, you were getting this from the divergence of the integral you used.

RGV
 
  • #11
kuruman
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That's right - intuitively we can expect that the time will go to infinity as we take the velocity to zero, and it seems that this is what happens.

But since the question is from a Putnam final contest, I guess that the person who wrote it knew what he was doing.
Not necessarily. With a little bit of googling, I was able to find the alleged solution at two sites
http://mks.mff.cuni.cz/kalva/putnam/psoln/psol395.html
http://www.math-olympiad.info/2nd-putnam-mathematical-competition-1939-problems.htm [Broken]

I will add some meat to the bones of this "solution". Start from the separated form as originally posted by LiorE and use the half-angle formula to get
[itex]dt=\sqrt{\frac{r}{4gsin^2(\theta/2)}}d\theta[/itex]
which, for a semicircular swing, leads to the integral
[itex]T=\sqrt{\frac{r}{g}}\int^{\pi}_{0}\frac{d(\frac{1}{2}\theta)}{sin(\theta/2)}[/itex]
Now define z = θ/2. The limits of integration (very important!) are from z = 0 to z = π/2. Then,
[itex]T=\sqrt{\frac{r}{g}}\int^{\pi/2}_{0}\frac{dz}{sinz}[/itex]
Now we believe (or we can verify by taking the derivative) that the indefinite integral is
[itex]\int\frac{dz}{sinz}=ln(sinz)-ln(1+cosz)=ln\left(\frac{sinz}{1+cosz}\right)[/itex]

Evaluating the definite integral at the limits gives
[itex]T=ln\left(\frac{sin(\pi/2)}{1+cos(\pi/2)}\right)-ln\left(\frac{sin0}{1+cos0}\right)[/itex]

The first term is clearly zero, but the second term is ???

I see no mathematically legitimate way that will lead to the proposed result. What bugs me is not that the original creator in the 1939 Putnam exam proposed this intractable problem. It is that on two occasions, like medieval monks copying manuscripts without understanding what was written, people posted the solution on the web without checking its correctness, not to mention the smug comment in the end. :yuck:
 
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  • #12
Ray Vickson
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Not necessarily. With a little bit of googling, I was able to find the alleged solution at two sites
http://mks.mff.cuni.cz/kalva/putnam/psoln/psol395.html
http://www.math-olympiad.info/2nd-putnam-mathematical-competition-1939-problems.htm [Broken]

I will add some meat to the bones of this "solution". Start from the separated form as originally posted by LiorE and use the half-angle formula to get
[itex]dt=\sqrt{\frac{r}{4gsin^2(\theta/2)}}d\theta[/itex]
which, for a semicircular swing, leads to the integral
[itex]T=\sqrt{\frac{r}{g}}\int^{\pi}_{0}\frac{d(\frac{1}{2}\theta)}{sin(\theta/2)}[/itex]
Now define z = θ/2. The limits of integration (very important!) are from z = 0 to z = π/2. Then,
[itex]T=\sqrt{\frac{r}{g}}\int^{\pi/2}_{0}\frac{dz}{sinz}[/itex]
Now we believe (or we can verify by taking the derivative) that the indefinite integral is
[itex]\int\frac{dz}{sinz}=ln(sinz)-ln(1+cosz)=ln\left(\frac{sinz}{1+cosz}\right)[/itex]

Evaluating the definite integral at the limits gives
[itex]T=ln\left(\frac{sin(\pi/2)}{1+cos(\pi/2)}\right)-ln\left(\frac{sin0}{1+cos0}\right)[/itex]

The first term is clearly zero, but the second term is ???

I see no mathematically legitimate way that will lead to the proposed result. What bugs me is not that the original creator in the 1939 Putnam exam proposed this intractable problem. It is that on two occasions, like medieval monks copying manuscripts without understanding what was written, people posted the solution on the web without checking its correctness, not to mention the smug comment in the end. :yuck:
I would not call the problem intractable; I would call it wrong: the test-taker was being asked to prove a genuinely incorrect result. By looking at the period T(theta_0) of a pendulum swinging between -theta_0 and +theta_0 (from the downward vertical), we can find a formula for 1/2 the period (which is what the question seeks), all in terms of complete Elliptic integrals. It is known that for theta_0 = Pi the time = + infinity; but more to the point, T(theta_0) --> infinity as theta_0 --> Pi from below, so one can find angles theta_0 < Pi that make the time sought 1,000,000 times larger than that given as a "solution" to the problem. Appropriate asymptotic expansions for the Ellilptic integral (as theta_0 --> Pi) can be found in Ambramowitz and Stegun, or in online articles.

RGV
 
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