Ground Impact Speed?

In another thread whose subject heading (pun) I dare not
mention, there seemed to be a claim that an object dropped
from--

Er, let us say, an object dropped from eye-level?

Anyway, the object will hit the pavement at the same speed
as the same object toppling sideways from the same height.

A rough analogy might be the middle of a rider's noggin
dropped in free fall versus the middle of a rider's noggin
toppling sideways like Benny Hill tipping over on a
tricycle.

But let's use my 62-inch metal sprinkler key instead to
avoid inflamed side-issues.

Some vague recollections about torque, rotation, toppling
chimneys breaking near the middle, things accelerated to
more than 1 G by the force of gravity, and balls dropping
into cups on short pivoted arms when released simultaneously
because the arm falls faster than the ball led me to look
into things and become confused.

Using Fogel Labs precision timer, I stood my 62" sprinkler
key upright on the lawn, warned the dog to stand out from
under, and tipped it . . .

One thousand one, one thousand two--

Thud! About two seconds to reach the ground in a graceful
arc.

Then I held the same sprinkler key horizontal at eye level
and dropped it . . .

One thousand one--

Thud! About one second to reach the ground in a bellyflop.

It kinda-sorta looked like the end of the toppling key
reached the ground more slowly than the same key in free
fall.

A generalized equation for the acceleration at any point on
a uniform beam rotating around a pivot is:

A = (3 * G * cos(angle) ) / ( 2 * L )

where A = acceleration in m/s^2
G = ~9.81 m/s^2
L = length of beam in meters
angle = 90 when beam is upright (0 acceleration)
0 when beam is flat (maximum acceleration)

See page 12 of this pdf, "Falling Chimneys":

http://hep.brown.edu/users/partridge/ph5/Lecture20.pdf

The cosine of 0 degrees is conveniently 1, so the
acceleration of the end of the beam at the instant of impact
should be:

(3G) / (2L)

beam-length acceleration
meters inches G's m/s^2
------ ----- ---- ----
0.50 19.7 3.00 29.42 /____cup and ball trick
1.00 39.4 1.50 14.71 \
1.50 59.1 1.00 9.81
1.575 62.0 0.952 9.34 --62" sprinkler key
2.00 78.7 0.75 7.35
3.00 118.1 0.50 4.90
6.00 236.2 0.25 0.15 --chimneys
50.00 1968.5 0.30 0.29

I like the inverted symmetry at the ends of the table, but
I'm not sure about the actual impact speed, which isn't the
same as the acceleration at the moment of impact and takes a
longer path--it takes roughly 2 seconds to cover an arc of
2.475 meters, but drops 1.5 meters in about 2 seconds.

For free fall, an object dropped from 62" should hit the
ground in 0.57 seconds (less than Fogel Labs measured it) at
about 5.6 m/s:

http://hyperphysics.phy-astr.gsu.edu...raj.html#ffall

Nor am I sure about whether the equation means the
acceleration at the end of the beam or in the middle.

I think that the end of a rigid, uniform 62"-long object
topples over and whomps into the ground significantly more
slowly than the same object held horizontal and dropped from
62", but I'm hoping that someone either has a simple
equation that I can follow, a more complicated equation that
I can't follow but will take on faith, or even an
explanation that the end of the toppling beam doesn't go
more slowly than it would in free fall.

. . - - - .
| / . - - - .
|/_ _ . - - - .

That is, how fast does the dot hit the ground in the two
figures above if it starts out 62" above the ground? I think
that it's going slower when it topples, but I want
reassurance.

Along similar lines, does anyone know of a sneaky
demonstration that would slow these two paths down enough to
make them clear, something like Galileo's trick of rolling
balls down inclines to slow the effect of gravity and show
that large and small balls accelerate at the same rate?

For anyone interested, here's a page with a miniature
falling chimney (two sections topple at different speeds)
and the short-armed cup and ball trick (a short beam
accelerates faster):

http://www.physics.umd.edu/lecdem/ou...arch3/a043.htm

Thanks,

Carl Fogel

writes:

The time it takes for the rod to topple is indeterminate. Imagine
that it were precisely balanced on its end---the duration would be
infinite. So timing the toppling is useless. The relevant measure
is the velocity at impact. In this case, the velocity at the tip
at impact. That is readily computed from conservation of energy.

Assume a uniform rod of length l, mass m. When standing on its end,
its potential energy is m*g*l/2. Assume that as it topples, its bottom
end does not slide (that is, acts as a pivot). When the tip strikes
the ground, all the potential energy has been converted to kinetic
energy. So

m*g*l/2 = 1/2*I*w^2

where

I = moment of inertia of rod, about one end = m*l^2/3
w = angular velocity of rod at impact

The velocity at the tip is

Vtip = w*l

so

Vtip = sqrt(3*g*l).

The same rod held horizontally and dropped from a height of l
has an impact velocity of

V = sqrt(2*g*l).

To achieve the same velocity as the toppling rod,
the horizontal rod must start at a height of 3/2*l.



Joe

wrote:
extensive snippage
I think that the end of a rigid, uniform 62"-long object
topples over and whomps into the ground significantly more
slowly than the same object held horizontal and dropped from
62", but I'm hoping that someone either has a simple
equation that I can follow, a more complicated equation that
I can't follow but will take on faith, or even an
explanation that the end of the toppling beam doesn't go
more slowly than it would in free fall.

. . - - - .
| / . - - - .
|/_ _ . - - - .

That is, how fast does the dot hit the ground in the two
figures above if it starts out 62" above the ground? I think
that it's going slower when it topples, but I want
reassurance.

Along similar lines, does anyone know of a sneaky
demonstration that would slow these two paths down enough to
make them clear, something like Galileo's trick of rolling
balls down inclines to slow the effect of gravity and show
that large and small balls accelerate at the same rate?

For anyone interested, here's a page with a miniature
falling chimney (two sections topple at different speeds)
and the short-armed cup and ball trick (a short beam
accelerates faster):

http://www.physics.umd.edu/lecdem/ou...arch3/a043.htm

Thanks,

Carl Fogel

There is no simple answer for the time it takes a toppling pole
to topple, since it depends on how big the initial imbalance
or disturbance was that caused it to fall. If the pole was very
well balanced, and undisturbed, it could take an arbitrarily
long time to fall, even though unstable.

If the disturbance that makes it fall is small enough not to
impart significant kinetic energy to the pole, it is fairly
straightforward to compute its speed of groundwhompage, at least
in the case of pivoting without its base slipping, and the usual
ignoring of air resistance. The potential energy of the center
of gravity of the pole falling to ground level is converted to
kinetic energy of the pole rotating about its pivot point.
Knowing the change in potential energy, and the moment of inertia
of the pole about its pivot, you can solve for its angular velocity,
and its speed of whompage of any point on the pole.

Dave Lehnen

wrote: (clip) Anyway, the object will hit the
pavement at the same speed as the same object toppling sideways from the
same height.(clip) (Stated as a conjecture, not as a fact.)
^^^^^^^^^^^^^^^^
Carl, there is a classic equality about falling objects which I think has
been distorted into this incorrect form.

A cannonball dropped from a given height will reach the ground at the same
time as a cannonball fired horizontally at the same time. Both will reach
the ground with the same vertical component. Both cannonballs will be
equidistant from the ground at all times.

This is equivalent to stating that the gravitational effect on the
cannonball is not related to any horizontal forces that are applied to it.
(With all the usual disclaimers about air friction.)

On Fri, 08 Jul 2005 23:54:02 GMT, "Leo Lichtman"
wrote:


wrote: (clip) Anyway, the object will hit the
pavement at the same speed as the same object toppling sideways from the
same height.(clip) (Stated as a conjecture, not as a fact.)
^^^^^^^^^^^^^^^^
Carl, there is a classic equality about falling objects which I think has
been distorted into this incorrect form.

A cannonball dropped from a given height will reach the ground at the same
time as a cannonball fired horizontally at the same time. Both will reach
the ground with the same vertical component. Both cannonballs will be
equidistant from the ground at all times.

This is equivalent to stating that the gravitational effect on the
cannonball is not related to any horizontal forces that are applied to it.
(With all the usual disclaimers about air friction.)


Dear Leo,

As Joe Riel's explanation makes clear, a beam rotating
around a pivot is not the same as a free-dropping
cannonball.

That is, a beam toppling sideways is not the same as
motionless beam released to drop straight down or a beam
fired horizontally out a cannon (which begins to drop at the
same rate).

Have a look at the cup and ball demonstration (the bottom
pictures) and its details:

http://www.physics.umd.edu/lecdem/ou...arch3/a043.htm

A beam of the right length at the right angle with a cup
near its end will pivot and fall faster than a ball when
both are released at the same time--the cup beats the ball
in the race to the table-top and catches it.

In contrast, a ball dropped from the edge of the table at
the same time as a ball shoots horizontally off the edge of
the table will reach the ground at the same time.

Carl Fogel

On Fri, 08 Jul 2005 22:59:44 GMT, Joe Riel
wrote:

writes:

The time it takes for the rod to topple is indeterminate. Imagine
that it were precisely balanced on its end---the duration would be
infinite. So timing the toppling is useless. The relevant measure
is the velocity at impact. In this case, the velocity at the tip
at impact. That is readily computed from conservation of energy.

Assume a uniform rod of length l, mass m. When standing on its end,
its potential energy is m*g*l/2. Assume that as it topples, its bottom
end does not slide (that is, acts as a pivot). When the tip strikes
the ground, all the potential energy has been converted to kinetic
energy. So

m*g*l/2 = 1/2*I*w^2

where

I = moment of inertia of rod, about one end = m*l^2/3
w = angular velocity of rod at impact

The velocity at the tip is

Vtip = w*l

so

Vtip = sqrt(3*g*l).

The same rod held horizontally and dropped from a height of l
has an impact velocity of

V = sqrt(2*g*l).

To achieve the same velocity as the toppling rod,
the horizontal rod must start at a height of 3/2*l.

Joe

Dear Joe,

So the speed-at-impact for the idealized toppling beam is
simpler than I expected? (As long as someone like you works
it out for me.)

And I had it backward? The toppling rod's impact speed when
it reaches the ground is always faster than than the impact
speed for free-fall?

That is, a horizontal rod one meter long reaches the ground
at 17.8 m/s when dropped from 1.5 meters:

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

So the tip of the same 1-meter rod just toppling over would
hit the ground at the same 17.8 m/s, even though it started
out only 1.0 meters off the ground, not 1.5 meters?

Sorry if I'm missing the obvious or confusing things.

Thanks,

Carl Fogel

Something extremely heavy at the anywhere on a falling ladder hits the
ground at the same speed with or without the ladder, whether or not
the ladder slips.

If it's not at the top, the top has to be going faster than normal to
achieve this.

Without the extremely heavy weight, the CG of the ladder serves to
impart the same tendency on the direction of change, answering the
question of faster or slower : the tip is going faster than it would
just falling straight down. The CG itself must therefore be going
slower. Somewhere between the CG and the tip is going the same
speed at impact with or without ladder.

--
Ron Hardin


On the internet, nobody knows you're a jerk.