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From: rabid_fan on 3 Jul 2010 14:06
On Thu, 01 Jul 2010 19:49:46 -0700, Mapsread wrote:

>
> Suppose you have a metal cylinder that has a symmetrical hollow cavity
> in the middle. You don't know the material composing the cylinder and
> don't know it has a cavity. Supposing you had a fulcrum and scale...
>

Consider a cylinder 1 meter in length with diameter of 2 cm
situated so that the left end is zero on the coordinate system
and the right end is at the point marked 1 meter:

0__________________________________________1


Consider that the cylinder has a hollow cavity from 0.25m to
0.75m on the coordinate axis (total length of cavity is 0.5m)
with a cavity diameter of 1 cm.

Assume material is of uniform density throughout.

Place a movable pivot (fulcrum) at point 0m. Place a fixed
scale at point 1m.

With pivot at 0m, the scale will show a maximum reading.

Now move the pivot slowly to the right. The reading on the
scale will drop to zero once the pivot reaches 0.5m.

However, look at the *rate* of decline. There will be
a discontinuity at the point 0.25m, where the cavity begins.
At this point, the rate of decline will abruptly lessen.

The change in the rate of decline will depend on the density
of the material and the size of the cavity, but it will be
there.

Use a spreadsheet and some sample values for density to
compute the forces and plot the decline. The change will
be quite apparent.

From: jmorriss on 3 Jul 2010 20:23
On Jul 3, 2:06 pm, rabid_fan <r...(a)righthere.net> wrote:
> On Thu, 01 Jul 2010 19:49:46 -0700, Mapsread wrote:
>
> > Suppose you have a metal cylinder that has a symmetrical hollow cavity
> > in the middle. You don't know the material composing the cylinder and
> > don't know it has a cavity. Supposing you had a fulcrum and scale...
>
> Consider a cylinder 1 meter in length with diameter of 2 cm
> situated so that the left end is zero on the coordinate system
> and the right end is at the point marked 1 meter:
>
> 0__________________________________________1
>
> Consider that the cylinder has a hollow cavity from 0.25m to
> 0.75m on the coordinate axis (total length of cavity is 0.5m)
> with a cavity diameter of 1 cm.
>
> Assume material is of uniform density throughout.
>
> Place a movable pivot (fulcrum) at point 0m.  Place a fixed
> scale at point 1m.
>
> With pivot at 0m, the scale will show a maximum reading.
>
> Now move the pivot slowly to the right.  The reading on the
> scale will drop to zero once the pivot reaches 0.5m.
>
> However, look at the *rate* of decline.  There will be
> a discontinuity at the point 0.25m, where the cavity begins.
> At this point, the rate of decline will abruptly lessen.
>
> The change in the rate of decline will depend on the density
> of the material and the size of the cavity, but it will be
> there.
>
> Use a spreadsheet and some sample values for density to
> compute the forces and plot the decline. The change will
> be quite apparent.

I beg to differ.

In the situation you describe, the condition that no rotation take
place depends on the balancing of clockwise and counteclockwise
torques about any convenient point: in this case, the fulcrum. These
in turn depend only on the location of the fulcrum and the LOCATION OF
THE CENTRE OF MASS.. The mass distribution in the cylinder is
irrelevant, once it has been used to find the centre of mass.

If your position were correct, it would mean that the concept of C of
M was insufficient for solving problems in rotational statics.

Of course, if the moments do NOT balance, then angular acceleration
will take place, and the mass distribution, as represented by the
moment of inertia does play a part in determining the size of the
acceleration

Are you sure that your spreadsheet does not shift the C of M as it
changes the density distribution?

Or have I once again totally missed the point?
From: Mapsread on 4 Jul 2010 00:27
On Jul 3, 7:23 pm, "jmorr...(a)idirect.com" <jmorr...(a)idirect.com>
wrote:
> On Jul 3, 2:06 pm, rabid_fan <r...(a)righthere.net> wrote:
>
>
>
>
>
> > On Thu, 01 Jul 2010 19:49:46 -0700, Mapsread wrote:
>
> > > Suppose you have a metal cylinder that has a symmetrical hollow cavity
> > > in the middle. You don't know the material composing the cylinder and
> > > don't know it has a cavity. Supposing you had a fulcrum and scale...
>
> > Consider a cylinder 1 meter in length with diameter of 2 cm
> > situated so that the left end is zero on the coordinate system
> > and the right end is at the point marked 1 meter:
>
> > 0__________________________________________1
>
> > Consider that the cylinder has a hollow cavity from 0.25m to
> > 0.75m on the coordinate axis (total length of cavity is 0.5m)
> > with a cavity diameter of 1 cm.
>
> > Assume material is of uniform density throughout.
>
> > Place a movable pivot (fulcrum) at point 0m.  Place a fixed
> > scale at point 1m.
>
> > With pivot at 0m, the scale will show a maximum reading.
>
> > Now move the pivot slowly to the right.  The reading on the
> > scale will drop to zero once the pivot reaches 0.5m.
>
> > However, look at the *rate* of decline.  There will be
> > a discontinuity at the point 0.25m, where the cavity begins.
> > At this point, the rate of decline will abruptly lessen.
>
> > The change in the rate of decline will depend on the density
> > of the material and the size of the cavity, but it will be
> > there.
>
> > Use a spreadsheet and some sample values for density to
> > compute the forces and plot the decline. The change will
> > be quite apparent.
>
> I beg to differ.
>
> In the situation you describe, the condition that no rotation take
> place depends on the balancing of clockwise and counteclockwise
> torques about any convenient point:  in this case, the fulcrum.  These
> in turn depend only on the location of the fulcrum and the LOCATION OF
> THE CENTRE OF MASS.. The mass distribution in the cylinder is
> irrelevant, once it has been used to find the centre of mass.
>
> If your position were correct, it would mean that the concept of C of
> M was insufficient for solving problems in rotational statics.
>
> Of course, if the moments do NOT balance, then angular acceleration
> will take place, and the mass distribution, as represented by the
> moment of inertia does play a part in determining the size of the
> acceleration
>
> Are you sure that your spreadsheet does not shift the C of M as it
> changes the density distribution?
>
> Or have I once again totally missed the point?- Hide quoted text -
>
> - Show quoted text -

So I decided to actually plug in numbers, AND, ..., the scale doesn't
reveal the answer. :-( But please check my reasoning and math:

For simplicity, I made g = 1 so mass and weight are the same. Consider
a bar with length 100 cm and mass 100 kg (and weight of 100 N if g =
1). I decided to make mine have each end 10 cm long and weigh 40 N.
The part with the cavity is therefore 80 cm long and 20 N, like so:

==----------------==

10 80 10 length (cm)
40 20 40 weight (N)

Put the fulcrum at 10 cm from the left and the scale all the way to
the right:

==----------------==
^ --

Sum (torques) = 0 = Left torque + Right torgue + Normal torque (at the
scale) = L + R + N

L = -5 cm * 40 N = -200 N cm (Note the center of mass of the left
"block" is at -5 cm).
R = 70 cm * 60 N = 4200 N cm (Why 70 cm? Center of mass from fulcrum =
sum(mass*distance)/sum(mass) = (40 cm * 20 kg + 85 cm * 40 kg) / (20
kg + 40 kg) = 70 cm)

0 = -200 N cm + 4200 N cm + Normal
Normal torque = -4000 N cm

The scale is at 90 cm from the fulcrum: -4000 N cm / 90 cm = -44.44 N.
The scale reads 44.44 N with the hollow bar of mass 100 kg.

Now take another bar that doesn't have a hollow, of mass 100 kg and
length 100 cm. Put the fulcrum at 10 cm. By symmetry, the portion to
the left of the fulcrum weighs 10 N. The center of mass at the left is
still at -5 cm. The portion to the right weighs 90 N and the center of
mass is 45 cm.

Sum (torques) = 0 = Left torque + Right torque + Normal torque (at the
scale) = L + R + N

L = -5 cm * 10 N = -50 N cm
R = 45 cm * 90 N = 4050 N cm

0 = 4050 - 50 + Normal
Normal = -4000 N cm

The scale is still at 90 cm: -4000 N cm / 90 cm = -44.44 N. The scale
still reads 44.44 N.

We can't tell that the bar has a hollow with just a fulcrum and scale.
Bummer, my first guess was wrong. Oh well, still a fun puzzle. It's
not intuitive, at least not for me. I guess I'll have to spin the bar
to determine it's hollow.

:-)
From: jmorriss on 4 Jul 2010 00:44
On Jul 4, 12:27 am, Mapsread <cw_...(a)yahoo.com> wrote:
> On Jul 3, 7:23 pm, "jmorr...(a)idirect.com" <jmorr...(a)idirect.com>
> wrote:
>
>
>
>
>
> > On Jul 3, 2:06 pm, rabid_fan <r...(a)righthere.net> wrote:
>
> > > On Thu, 01 Jul 2010 19:49:46 -0700, Mapsread wrote:
>
> > > > Suppose you have a metal cylinder that has a symmetrical hollow cavity
> > > > in the middle. You don't know the material composing the cylinder and
> > > > don't know it has a cavity. Supposing you had a fulcrum and scale....
>
> > > Consider a cylinder 1 meter in length with diameter of 2 cm
> > > situated so that the left end is zero on the coordinate system
> > > and the right end is at the point marked 1 meter:
>
> > > 0__________________________________________1
>
> > > Consider that the cylinder has a hollow cavity from 0.25m to
> > > 0.75m on the coordinate axis (total length of cavity is 0.5m)
> > > with a cavity diameter of 1 cm.
>
> > > Assume material is of uniform density throughout.
>
> > > Place a movable pivot (fulcrum) at point 0m.  Place a fixed
> > > scale at point 1m.
>
> > > With pivot at 0m, the scale will show a maximum reading.
>
> > > Now move the pivot slowly to the right.  The reading on the
> > > scale will drop to zero once the pivot reaches 0.5m.
>
> > > However, look at the *rate* of decline.  There will be
> > > a discontinuity at the point 0.25m, where the cavity begins.
> > > At this point, the rate of decline will abruptly lessen.
>
> > > The change in the rate of decline will depend on the density
> > > of the material and the size of the cavity, but it will be
> > > there.
>
> > > Use a spreadsheet and some sample values for density to
> > > compute the forces and plot the decline. The change will
> > > be quite apparent.
>
> > I beg to differ.
>
> > In the situation you describe, the condition that no rotation take
> > place depends on the balancing of clockwise and counteclockwise
> > torques about any convenient point:  in this case, the fulcrum.  These
> > in turn depend only on the location of the fulcrum and the LOCATION OF
> > THE CENTRE OF MASS.. The mass distribution in the cylinder is
> > irrelevant, once it has been used to find the centre of mass.
>
> > If your position were correct, it would mean that the concept of C of
> > M was insufficient for solving problems in rotational statics.
>
> > Of course, if the moments do NOT balance, then angular acceleration
> > will take place, and the mass distribution, as represented by the
> > moment of inertia does play a part in determining the size of the
> > acceleration
>
> > Are you sure that your spreadsheet does not shift the C of M as it
> > changes the density distribution?
>
> > Or have I once again totally missed the point?- Hide quoted text -
>
> > - Show quoted text -
>
> So I decided to actually plug in numbers, AND, ..., the scale doesn't
> reveal the answer. :-( But please check my reasoning and math:
>
> For simplicity, I made g = 1 so mass and weight are the same. Consider
> a bar with length 100 cm and mass 100 kg (and weight of 100 N if g =
> 1). I decided to make mine have each end 10 cm long and weigh 40 N.
> The part with the cavity is therefore 80 cm long and 20 N, like so:
>
> ==----------------==
>
> 10       80         10     length (cm)
> 40       20         40     weight (N)
>
> Put the fulcrum at 10 cm from the left and the scale all the way to
> the right:
>
> ==----------------==
>    ^                      --
>
> Sum (torques) = 0 = Left torque + Right torgue + Normal torque (at the
> scale) = L + R + N
>
> L = -5 cm * 40 N = -200 N cm (Note the center of mass of the left
> "block" is at -5 cm).
> R = 70 cm * 60 N = 4200 N cm (Why 70 cm? Center of mass from fulcrum =
> sum(mass*distance)/sum(mass) = (40 cm * 20 kg + 85 cm * 40 kg) / (20
> kg + 40 kg) = 70 cm)
>
> 0 = -200 N cm + 4200 N cm + Normal
> Normal torque = -4000 N cm
>
> The scale is at 90 cm from the fulcrum: -4000 N cm / 90 cm = -44.44 N.
> The scale reads 44.44 N with the hollow bar of mass 100 kg.
>
> Now take another bar that doesn't have a hollow, of mass 100 kg and
> length 100 cm. Put the fulcrum at 10 cm. By symmetry, the portion to
> the left of the fulcrum weighs 10 N. The center of mass at the left is
> still at -5 cm. The portion to the right weighs 90 N and the center of
> mass is 45 cm.
>
> Sum (torques) = 0 = Left torque + Right torque + Normal torque (at the
> scale) = L + R + N
>
> L = -5 cm * 10 N = -50 N cm
> R = 45 cm * 90 N = 4050 N cm
>
> 0 = 4050 - 50 + Normal
> Normal = -4000 N cm
>
> The scale is still at 90 cm: -4000 N cm / 90 cm = -44.44 N. The scale
> still reads 44.44 N.
>
> We can't tell that the bar has a hollow with just a fulcrum and scale.
> Bummer, my first guess was wrong. Oh well, still a fun puzzle. It's
> not intuitive, at least not for me. I guess I'll have to spin the bar
> to determine it's hollow.
>
> :-)- Hide quoted text -
>
> - Show quoted text -

Even simpler... Calculate moments around the C of M. After all, any
point will do.

But this way, as soon as you calculate the moment contribution from
gravity acting on the the symetrical cylinder itself, you find it's
zero...by definition. End of story...
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Prev: New Thread was Re: Holograms Re: Benefits was Re: Hazards was Re:Conversation was Re: School was Re: The Stage
Next: * Hates US * admits his homosexuality while committing EPIC FAIL in futile attempt to support his LYING CLAIM about PNAC, which of course never said anything remotely suggesting it "wanted" the 9/11 attacks