Prev: "The Most Miserable People Are The Ones Who Constantly Put Others Down." --
Next: X-intercepts in circular functions
From: Greg Neill on 1 Apr 2010 07:42
Existential Angst wrote:

> 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.

Check into your expression for the moment of inertia. What you've
stated looks suspiciously like the moment of inertia for a thin
rod about its center, not its end. The parallel axis theorem comes
to your aid here.


From: Uncle Ben on 1 Apr 2010 09:37
On Mar 31, 10:45 pm, "Androcles" <Headmas...(a)Hogwarts.physics_x>
wrote:

(snip)

> 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.

Exercise for the better students:

Newton believed his laws to be in effect at all times, not just when
the stick is in certain positions.

The force on the stick (action) due to the earth's gravitational
attraction is constant. What is the (constant) reaction to this force,
boys and girls? In particular, on what object is it active?

Hint: It's not the stick.

Nit-picking Bonehead
From: Tater Gumfries on 1 Apr 2010 09:49
On Mar 31, 7:27 pm, "Existential Angst" <UNfit...(a)UNoptonline.net>
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.
> --
> EA

Same thing as a pendulum. It falls through pi/2 from infinitesimally
close to pi. Integrate your pendulum equation numerically usin Maple
or some such and see what the velocity is at Pi/2.

Tater

Tater
From: Uncle Al on 1 Apr 2010 11:08
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:

Already wrong. You left out the trig expansion tail for a non-zero
angle of swing.


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

mgh is the total energy at start. It is all kinetic energy at the
end. Integrate along the length, mgh at the start to (mv^2)/ at the
end, and see if you get consistent results at discrete points along
the length.

> 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.

Can you survive a rigid spine fall upon your hands re impulse? If
not, the reverse will not go well. You know the total energy needed,
mgh integrated along the length. Will relevant muscles and bone
leverage deliver that in the powered path and thrust interval
allotted?

You might go for two meters.

> If I had the required Vo, I could probably figger stuff out from there.
>
> Any links, hints -- or solutions -- appreciated.
> --
> EA


--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz4.htm
From: Existential Angst on 1 Apr 2010 13:33
"Greg Neill" <gneillRE(a)MOVEsympatico.ca> wrote in message
news:QE%sn.53664$V43.49305(a)unlimited.newshosting.com...
> Existential Angst wrote:
>
>> 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.
>
> Check into your expression for the moment of inertia. What you've
> stated looks suspiciously like the moment of inertia for a thin
> rod about its center, not its end. The parallel axis theorem comes
> to your aid here.

Heh, so does wiki! :) (1/3)mL^2
http://en.wikipedia.org/wiki/List_of_moments_of_inertia

Thanks.
--
EA
>
>


First  |  Prev  |  Next  |  Last
Pages: 1 2 3 4
Prev: "The Most Miserable People Are The Ones Who Constantly Put Others Down." --
Next: X-intercepts in circular functions