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Do you think it would be possible to get 17 joules out of a standard size mouse trap. By my math, it is a torsion coefficient of 3.45 or so out of the spring.

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Clearly 3.45 can't be the value of a torsion coefficient from dimensional analisys alone. –  Olin Lathrop Feb 22 at 17:20
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6 Answers 6

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I haven't used a mousetrap for several decades, but as I recall the moving arm is about 5cm long, so the tip moves 0.05$\pi$ or about 0.16m. To get 17J of work the force at the tip of the arm would need to be 100N. I'm fairly sure the force isn't anything like that great. I remember being able to pull the arm back with one finger. I would guess the force is nearer 10N, so you'd only get around 2J out.

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Yes, if you burn it. Neglecting the metal, a $25\ g$ wooden mousetrap at $15\ MJ/kg$ should yield about $375\ kJ$.

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I was going to make this a joke comment but then I realized that it's an answer :) –  Mark Beadles Feb 15 '12 at 18:21
    
+1: good answer :-) –  John Rennie Feb 15 '12 at 19:54
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Nice answer...I should have been a bit more specific. –  John Feb 16 '12 at 2:38
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And if you throw it into a black hole, you can get about a third of a megaton. I don't think this is in the spirit of the question, but it's cute. –  Ron Maimon Feb 16 '12 at 6:08
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Good guesstimating John Rennie. This fellow did some measurements (on a rat trap) and got slightly over 3 Joules:

http://www.instructables.com/id/Mouse-Trap-Speed/step4/Analysis-using-Basic-Physics/

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$$U=\frac{1}{2}k\theta^2$$. A standard mouse trap is usually set at $\approx 180^o=\pi\space$ radians. $$\therefore U=0.5*3.45*\pi^2\approx17.025\space J$$ Of course, the torsion law is only an approximation, and it may decrease after use of the moustrap, so the energy obtained would be slightly less.

EDIT: I seem to have misunderstood your question. I don't have any idea what the torsion constant of a spring should be, but 17 joules is pretty small.

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Even though you could not answer my question, thank you for the attempt. –  John Feb 15 '12 at 4:53
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Took the following data: measured force on 0.045 m trap arm and computed torque

(angle(rad), torque(n*m))

(0 , .1215) (1.57 , .2565) (3.14 , .378)

best fit line to data torque = .0816 * angle + 0.1238

area under this line is energy stored in spring

about 0.8 Joules

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Hi Steve, can you describe a bit more about this experimental setup? It should be possible to compute what a 17 J mouse trap arm would behave like (for a given size) which should give the answer at least a rough "sniff test" for whether it's possible. –  Brandon Enright Feb 18 at 3:39
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I clamped a small mousetrap to a heavy table and used a spring scale to measure the force when the hook was attached to the trap arm. Try to keep a 90 degree angle between the scale line of action and the arm of the mousetrap. The trap arm I measured was about 45 mm in length from the pivot point. I measured three forces, one to just lift the trap arm off the wooden base, a second when the arm has rotated 90 degrees (pi/2 rad) and a third when the trap are was almost in contact with the wood and having been rotated 180 degrees (pi rad). I had to trim some wood away for the last reading and clamp the trap overhanging the table. The force is multiplied times the lever arm of 45 mm to find the torque as a function of the angle of rotation. It is pretty much a straight line which is good since we would expect some sort of Hooke's law relationship. It can be shown that an increment of work is the torque times an increment of the angle of rotation. Thus integration of the torque function over the change in angle (area under torque-angle function) is the total work that can be done by the mousetrap when released to stop. I have repeated this experiment a few times and the work of the trap comes out to be between 0.7 and 0.8 joules.

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You could edit this in your old answer. –  jinawee Feb 22 at 16:00
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