Falling Physical pendulum Q [Math]

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From: Existential Angst on 31 Mar 2010 21:27

Awl --

I've found plenty of standard stuff on physical pendulums (meter stick
pendulum), where I = 1/12 MR^2, period, etc, but what I can't find is this:

With a meter stick standing straight up and pivoted at the *bottom*, what is
the final angular velocity as it hits the table?

My dim recollection, and perhaps why I can't find the solution so quick, is
that an elliptic integral is involved??

The application is an interesting one:
I would like to approximate the force one would have to generate from a
pushup position sufficient to thrust one's self up to a standing position.

If I had the required Vo, I could probably figger stuff out from there.

Any links, hints -- or solutions -- appreciated.
--
EA

From: Existential Angst on 31 Mar 2010 21:37

From: Androcles on 31 Mar 2010 22:45

"Existential Angst" <UNfitcat(a)UNoptonline.net> wrote in message
news:4bb3f677$0$31267$607ed4bc(a)cv.net...
> Awl --
>
> I've found plenty of standard stuff on physical pendulums (meter stick
> pendulum), where I = 1/12 MR^2, period, etc, but what I can't find is
> this:
>
> With a meter stick standing straight up and pivoted at the *bottom*, what
> is the final angular velocity as it hits the table?
>
> My dim recollection, and perhaps why I can't find the solution so quick,
> is that an elliptic integral is involved??
>
> The application is an interesting one:
> I would like to approximate the force one would have to generate from a
> pushup position sufficient to thrust one's self up to a standing position.
>
> If I had the required Vo, I could probably figger stuff out from there.
>
> Any links, hints -- or solutions -- appreciated.
> --
> EA
Assume a stick experiences a force vertically (90 degrees) of -g
(acceleration is negative when falling to give a lower height)
which is opposed (Newton's third law) by an upward force g when
the stick is vertical and zero (0 degrees) when horizontal.
It should be immediately apparent that for any angle phi between
vertical and horizontal the opposing force is g.cos(phi).
Thus the vertical acceleration at angle phi is g(1-cos(phi).
So what you want for the linear vertical velocity is:
the integral of acceleration, which is the integral of
g(1-cos(phi)) for phi = 90 to zero, and then find the angular
velocity at phi = 0.
(Because all masses fall at the same rate the mass isn't relevant.)
HTH, I've had a couple of vodkas while watching "Time Team"
so it could be booze talking.

From: Greg Neill on 31 Mar 2010 23:42

Existential Angst wrote:
> Awl --
>
> I've found plenty of standard stuff on physical pendulums (meter stick
> pendulum), where I = 1/12 MR^2, period, etc, but what I can't find is
this:
>
> With a meter stick standing straight up and pivoted at the *bottom*, what
is
> the final angular velocity as it hits the table?
>
> My dim recollection, and perhaps why I can't find the solution so quick,
is
> that an elliptic integral is involved??
>
> The application is an interesting one:
> I would like to approximate the force one would have to generate from a
> pushup position sufficient to thrust one's self up to a standing position.
>
> If I had the required Vo, I could probably figger stuff out from there.
>
> Any links, hints -- or solutions -- appreciated.

Hmmm. Calculate the total potential energy (w.r.t. the table)
of the standing meter stick (should be a trivial integral).

When the meter stick reaches horizontal at the table, the
potential energy should be zero, it having all gone into the
kinetic energy of the falling stick.

With the kinetic energy, which should all be rotational, you
should be able to find the angular velocity. Remember to use
the parallel axis theorem to massage the moment of inertia.

From: Existential Angst on 1 Apr 2010 02:02

"Greg Neill" <gneillRE(a)MOVEsympatico.ca> wrote in message
news:XCUsn.22083$4m6.9571(a)unlimited.newshosting.com...
> Existential Angst wrote:
>> Awl --
>>
>> I've found plenty of standard stuff on physical pendulums (meter stick
>> pendulum), where I = 1/12 MR^2, period, etc, but what I can't find is
> this:
>>
>> With a meter stick standing straight up and pivoted at the *bottom*, what
> is
>> the final angular velocity as it hits the table?
>>
>> My dim recollection, and perhaps why I can't find the solution so quick,
> is
>> that an elliptic integral is involved??
>>
>> The application is an interesting one:
>> I would like to approximate the force one would have to generate from a
>> pushup position sufficient to thrust one's self up to a standing
>> position.
>>
>> If I had the required Vo, I could probably figger stuff out from there.
>>
>> Any links, hints -- or solutions -- appreciated.
>
> Hmmm. Calculate the total potential energy (w.r.t. the table)
> of the standing meter stick (should be a trivial integral).
>
> When the meter stick reaches horizontal at the table, the
> potential energy should be zero, it having all gone into the
> kinetic energy of the falling stick.
>
> With the kinetic energy, which should all be rotational, you
> should be able to find the angular velocity. Remember to use
> the parallel axis theorem to massage the moment of inertia.

Heh, so much for elliptic integrals, eh?? :)
Mebbe the integral is required for the general solution?
But it would appear not, because the conservation of energy method could be
used for any angular interval..... hmmmm.....

I used I = 1/12 ML^2, for a stick pivoted at the bottom.

I got an ang. vel of about 11 rad/sec, for a "stick" (body) of about 2 m.

Assuming I did it right.
--
EA

>
>

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