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Energy created by a falling object

by emandelli
Tags: energy, falling, object
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emandelli
#1
Jul14-06, 02:43 PM
P: 3
Hello All I hope someone can help.

I am dropping a weight on a bar to find out deflection of this bar. When I drop a 72lbs weight @ 1.0meter from the bar, the impact energy is (32.7KG x 9.8m/s2 x 1.0m) right? = ~311 N/m

The total deflection when I drop this weight on the bar is 80mm.

On a static deflection test, I am placing the bar simply supported with the same condition as above and then I slowly apply a load with a shop press and a load cell tells me what the load is. When the exactly same bar reaches the 80mm, the display shows 1500lbs, that means I need 1500lbs to deflect the bar the same distance a weight of 72lbs dropped at 1m needs.

How can I co-relate these two number,s 1500lbs and 311 N/m? Does it mean that the 311N/m is equal 1500lbs?

Once I found this relationship, I plan to perform only one of the test, make sense?

Thanks!
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Q_Goest
#2
Jul14-06, 02:49 PM
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The bar has a spring rate which can be calculated from the deflection versus force equations. The energy of the falling object is absorbed by the bar as if the bar were a spring. Equate the two and you can determine the deflection in the bar when the weight comes to rest at the most extreme point.
Janus
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Jul14-06, 06:37 PM
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Quote Quote by emandelli
Hello All I hope someone can help.

I am dropping a weight on a bar to find out deflection of this bar. When I drop a 72lbs weight @ 1.0meter from the bar, the impact energy is (32.7KG x 9.8m/s2 x 1.0m) right? = ~311 N/m
That should be ~311 N-m not N/m

emandelli
#4
Jul14-06, 09:29 PM
P: 3
Energy created by a falling object

Quote Quote by Q_Goest
The bar has a spring rate which can be calculated from the deflection versus force equations. The energy of the falling object is absorbed by the bar as if the bar were a spring. Equate the two and you can determine the deflection in the bar when the weight comes to rest at the most extreme point.

Should I use the 1/2kx2 for springs?
andrevdh
#5
Jul15-06, 12:37 PM
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A "normal" helical spring gives a linear relation ship between its extension and applied force. The area under the curve is the absorbed energy of the spring (this will also be true for your bar even if it is not linear). If your bar obeys Hook's law you can use the formula

[tex]U = 0.5 k x^2[/tex]

for the stored energy where k is the bar's force constant if the graph is linear. So I would suggest to plot a few points and see if it gives a linear graph (over the range you intend to predict its behaviour).
emandelli
#6
Jul16-06, 12:21 AM
P: 3
Yes, the graph is very much linear, force x deflection, actually the bar is linear until its breaking point which is sudden. The bar material has 65% glass fibers. In the case of using Hook's formula, what am I getting? The constant K that I can use to find the force to deflect at certain distances?
andrevdh
#7
Jul16-06, 05:11 AM
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Yes, the force constant of the bar will be the gradient of your load versus deflection graph.

The assumption you make in calculating the deflection caused by a dropped weight using

[tex]U_G = U_E[/tex]

, that is the gravitational potential energy is converted to stored elastic potential energy of the bar at maximum deflection assumes perfect energy conversion, which we all know is not true. The question is just how much is lost in the process to other forms (not converted to the stored elastic potential energy of the bar) of energy. I don't think anyone can answer this on a theoretical basis. You will have to investigate it experimentally.


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