A physics problem

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January 6th 2006, 11:17 AM
ThePerfectHacker

A physics problem

This physics problem appeared in my Calculus book:
Given a rod with length $a$ it is hindged thus it can freely rotate. An object is attached to the end of the rod and it is brough all the way vertically up. Then a slight disturbance and it flys down (from equilibrium). How long does it take for it to come down to the minimum point? Just remember it travels in a circular motion.
My book gave the answer:
$\sqrt{\frac{a}{g}}\ln{(1+\sqrt{2})}$
Where $g$ I am guessing it the gravitational acceleration.
January 10th 2006, 01:48 PM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

This physics problem appeared in my Calculus book:
Given a rod with length $a$ it is hindged thus it can freely rotate. An object is attached to the end of the rod and it is brough all the way vertically up. Then a slight disturbance and it flys down (from equilibrium). How long does it take for it to come down to the minimum point? Just remember it travels in a circular motion.
My book gave the answer:
$\sqrt{\frac{a}{g}}\ln{(1+\sqrt{2})}$
Where $g$ I am guessing it the gravitational acceleration.

It seems to me that there is something wrong here. It looks to me that
as the disturbance goes to zero the time to fall goes to $\infty$ .

RonL
January 10th 2006, 02:11 PM
ThePerfectHacker

This is just like a free-falling problem from physics. Except this is one is in a circular path? Did you try to do the problem and got some unusual result?
January 10th 2006, 08:25 PM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

This is just like a free-falling problem from physics. Except this is one is in a circular path? Did you try to do the problem and got some unusual result?

What I did is derived the relevant DE for this problem (I was overcome
with curiosity because physical intuition was that it should take a arbitrarily
long time to fall from an unstable equilibrium as the initial disturbance becomes
arbitrarily small).

The DE can be solved but gives an answer in terms of for the (quarter)
period in terms of an elliptic integral (not necessarily a problem in itself as
we want a special case which could have in principle reduced to something
more easily manageable).

Examining the integral showed that there was a singularity at one end of the
range of integration for the given initial condition, and that near the
singularity the integral behaved like $1/x$ , and so the integral
should be divergent.

So I did some research, and as far as I can tell (the research did not turn
up this exact problem but something similar) it indicates that my initial
feelings about this may be correct.

RonL

The nearest reference to this problem is in Morris Klein's

"Calculus : An Intuitive and Physical Approach"
http://www.amazon.com/gp/product/048...lance&n=283155
January 11th 2006, 02:19 PM
ThePerfectHacker

But this problem cannot have some kind of extremely complicated solution (elliptic integral) it appeared in my Calculus book for univeristies.

Perhaps, I am thinking, this is somehow connected with the swinging pendulum?
January 11th 2006, 02:41 PM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

But this problem cannot have some kind of extremely complicated solution (elliptic integral) it appeared in my Calculus book for univeristies.

Perhaps, I am thinking, this is somehow connected with the swinging pendulum?

An elliptic integral is not complicated in principal, its just that in
general it does not have a closed form representation in terms
of elementary functions.

Morris Klein's book I referred to is an undergraduate calculus
text.

And yes it is a rigid pendulum (being rigid allows the thing to be
started with the weight above the pivot without it just falling
straight down initially).

RonL
January 11th 2006, 02:46 PM
ThePerfectHacker

Do you know how to solve it?
Because all my attempts failed.
January 12th 2006, 05:51 AM
CaptainBlack

1 Attachment(s)

There are two methods of obtaining the Differential Equation governing
the motion of the bob of a (rigid)pendulum.

(I will assume that the rod is "light" that is its mass is negligible compared
to that of the bob)

One is to just consider the energy of the bob. The total energy
is:

$TE=PE+KE$ ,

where $TE$ is the total energy, $PE$ is the potential energy and $KE$
is the kinetic energy.

(another is to resolve the force due to gravity on the bob into components
along the rod and perpendicularly to the rod, the perp component gives the
torque which gives rise to angular acceleration, and the other component
is cancelled by the tension/compression forces in the rod and the
reaction at the pivot)

Now we will describe the position of the bob by a single angle $\theta$ (see the diagram).

Take the reference height for the zero of potential energy to be the height
of the bob when $\theta=0$ , then:

$PE=m.g.(a-a\cos(\theta))$ ,

and:

$KE=\frac{1}{2}I\dot \theta ^2$ ,

where $I=a^2m$ is the moment of inertia (assuming a light rod)

So:

$m.g.a(1-\cos(\theta_0))=m.g.a(1-\cos(\theta))+\frac{1}{2}m.a^2.\dot \theta^2$ ,

where $\theta_0$ is the maximum angular amplitude. Rearranging:

$\dot \theta ^2 = \frac{2g}{a} (\cos (\theta)-\cos (\theta_0))$ .

Which is a DE of separable type which may be integrated to give:

$\pm \int_{\theta_0}^{\phi} \frac{1}{(\cos (\theta)-\cos (\theta_0))^{\frac{1}{2}}}d\theta=t_{\phi}\sqrt{ \frac{2g}{a} }$ ,

where $t_{\phi}$ is the time that the bob is a angle $\phi$
(after having been released from $\theta=\theta_0$ at $t=0$ ).

Now consider just the first swing the we want the signs in front
of the integral to be -ve, and the bob reaches the bottom of its swing when $\phi=0$ ,
so the time we are looking for is:

$\tau=\sqrt{ \frac{a}{2g} } \int_{0}^{\theta_0} \frac{1}{(\cos (\theta)-\cos (\theta_0))^{\frac{1}{2}}}d\theta$ .

Now I think that the integral on the RHS diverges as $\theta_0 \rightarrow \pi$ .

RonL

(apologies in advance for any errors that may have crept into my
LaTeX).
January 12th 2006, 04:47 PM
ThePerfectHacker

Thank you