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Re: [Phys-L] real-world physics

This topic is worthy of a little more conversation, I do believe. First defining terms:
I take acceleration to be a change in velocity. This means that the acceleration I feel while standing on the ground is a very moderate 0.01 m/s^2 due to the rotating reference frame. The force on my feet is a much more onerous 98 kg X 9.8 N/kg and I will assert that there is a force pair due to the gravitational effect, and I can easily see that one would treat the force due to acceleration and the force due to gravity as indistinguishable effects given certain conditions.
Using the terms as I defined hem here, I assert that when Gary Connery jumped out of the helicopter, it was flown in a hovering position with respect to the ground, so that the vertical acceleration started at zero. On jumping, he experienced a vertical acceleration of about 9.8m/s^2 which dropped to zero as his speed and rate of descent stabilized near the runway of cardboard boxes.
It was necessary for him to lose kinetic energy at touch down, and he arranged the materials so that his momentary acceleration in coming to a standstill was provided by a braking force within his physiological limits.

If he had dropped like a stone, then his descent at 9.8 m/s^2 could have been arrested (eventually), by applying some force even slightly greater than the Gravity force in the opposite direction. This could certainly be less than a lower limit of 2g equivalent acceleration, wouldn't you say?

But wait: if I fly an aerobatic plane, I expect to see a G meter: a meter that indicates 1 G when I am quite stationary. This is agreeing with the aero engineer's understanding of what is crucial to an airplane - forces caused by acceleration through the air and forces due to Gravity, both of which can and will break or bend the structure if excessive. And if he chooses to account for both types of force in Gs, who am I to complain? - they are after all equivalent. Finally, if I drop a lead weight a few inches onto a marble floor, then I can confidently presume that the retarding force (the one that may let it bounce up a little), is in the hundreds of G equivalent force.

So 'I summarize by asserting that a falling object may be reversed in direction by providing an opposing force calculated to prove an
upward acceleration of very very little, to very very large values.

Brian Whatcott


On 6/7/2012 11:56 AM, Paul Nord wrote:
Brian,

The deceleration starts from some speed. You are correct in the case where this is done over a long distance with an ideal spring F = -kx. Collapsing bubble wrap is probably not an ideal spring. The ability of the bubble wrap to launch you back into the air with some velocity does indeed indicate that there is more than 1g of acceleration from the bubble wrap at maximum compression. (Note: the floor is exerting 1g of acceleration on you right now.) If the bounce height is the same as the drop height then you can say that the speed at the start of the collision is equal and opposite to the speed at the end of the collision. In that case, delta-v is doubled and likewise the average acceleration. Simply bouncing doesn't double the acceleration. It does indicate that it is larger than simply crushing something and coming to a halt.

Paul


On Jun 7, 2012, at 11:35 AM, brian whatcott wrote:

I fear that I shall need someone to draw a picture: take a case where an object dropping near Earth is accelerating at -1g and via a very long buffer is decelerated thru 0 to +1g. Does that object not bounce up? Where is my misconception? Does -1g to +1 g constitute this doubling? I am think doubling -1 is -2 or doubling one is two. Would you straighten me out please?

Brian


On 6/7/2012 9:23 AM, Paul Nord wrote:
If the bubble wrap acts like an ideal spring, completely reversing the motion does imply that a lower limit on the maximum acceleration of 2g.

In this case, it's simpler to consider the duration of the collision. Starting at a speed of 120 mph (50 m/s) and changing that to zero implies an average speed during collision of 25 m/s. Traveling a distance of 4 inches (0.1 m) at this speed means that the collision takes 0.004 seconds. Knowing our delta-v and delta-t we can say that the average acceleration during that time is 12,500 m/s/s (1250 g).

That is the calculation that the Mythbusters should have done before attempting to build this.

Paul


On Jun 6, 2012, at 5:46 PM, brian whatcott wrote:

Hmmm.. with a compressing spring, an increasing force, providing an increasing retardation.
So certainly not constant acceleration. I think I was responding to the idea of doubling the acceleration by the reversed direction, which seems also mistaken

Brian W


On 6/6/2012 11:54 AM, brian whatcott wrote:
Comparable speed at reciprocal heading, with constant acceleration possibly??

Brian W

On 6/6/2012 10:57 AM, Bernard Cleyet wrote:
On 2012, Jun 06, , at 08:19, John Clement wrote:

I wonder if the boxes were filled with bubble wrap.
A possible worsening, as if not popped, bouncing, doubling the acceleration.




Since these are
professionals I suspect they first tried a dummy and that they know how to
calculate forces.

So testing an idea is not all that farfetched.
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