.Solving the Physics of a Seesaw Catapult

[Milchstrabe]

I say the torque of gravity is positive, but then when i add it to the other clockwise torques, i make it negative you see?


Anyways, height i'd have to drop the mass:

[tex]\Delta x = \frac {1}{2}a(t^2) [/tex]

[tex] V = at + Vo [/tex]

[tex] t = \frac {.108356m/s}{9.8m/s} [/tex]

[tex] t = .011056 sec [/tex]

[tex] \Delta x = \frac {1}{2}(9.8 m/s)(.011056^2) [/tex]

[tex] \Delta x = .000599 m [/tex] THhat is .05 cm high that i would need to drop the 1 kg... eh? something might be wrong.


IT doesn't make sense because if I re-calculate it with a mass falling of only 50 grams (same mass as the hacky sack) then it only needs a 2.3502 velocity to launch a 50 gram hacky at 7.85 m/s, lift an arm 4 times heavier than it, 20 degrees, i don't know. I think our clockwise torques are just fine based on basic definitions of torque etc... Unless our launch torque is wrong. I think something is more likely wrong with our counter clockwise torque forumulas...

 

[whozum]

[tex] \tau_{counter} = (1kg)(9.8)(.1m) + \frac {(1kg)(V)}{.01}[/tex]

Your impulse torque doesn't have units of torque. You forgot to add the .1m

[tex] 11.8157985(clockwise torque) - .98(torque of obj) = \frac {V}{.01} [/tex]

I get V = 1083 when I do this, but factoring in the .1 from above, Its still really high, 108.3. Are you sure you did your torque-work correct?


Two things I also noticed. The torque isn't constant while the angle is swept, since gravity isn't always perpendicular to the radius vector, do you want to take this into account? Also, when the KG hits the seesaw, its impulse torque won't be perpendicular to the seesaw, so you are losing some torque right there. Just a few thoughts.

 

[Milchstrabe]

Are you sure about those numbers? watch the decimal places maybe? Because (11.8 - .98 )(.01) = V doesn't get you that big of a number. I won't take those things into account.

 

[whozum]

Oh right my bad, I can't multiply.

 

[Milchstrabe]

lol okay so if we both agree that hte number is too small, then what the heck is going on ::bangs head on desk::

 

[whozum]

[tex] \tau_{clock} - \tau_{obj} = \tau_{imp} [/tex]

[tex] 10.8 = \frac{mv}{t} x [/tex]

[tex] \frac{10.8t}{xm} = v [/tex]

[tex] v = 1.08m/s [/tex]

Try that. Sounds more reasonable.

I get 6cm.

 

[Milchstrabe]

That sounds more reasonable, 6 cm drop height.. approx: 2.5 inches. Maybe...

 

[whozum]

The biggest source of error is assuming the torque angles are perpendicular. You're probably going to need a larger impact velocity to get it to work perfect.

 

[Milchstrabe]

I'll cut the wood tomorrow and set it up and run some experiments to see what happens, and if it's not a whole lot off (meaning i don't have to adjust the drop height more then 3 cm or so, i'll do an error analysis, if not, i may factor it it in. However if i have to adjust by 10 cm + then we probably have a problem.

 

[whozum]

Sounds good, have fun with it. Dont break the wood :)

 

[Milchstrabe]

I tried setting something similar to this last night with the wood at the exact length and the ratio's were pretty close. The height of the fulcrum was a little less than it should be but I was just using my hand impacting the wood as the force. I found it EXTREMELY difficult to get much height out of the hacky sack, not even a meter. I was hitting it pretty hard too. Is there anyone you can verify that we did this right with?

 

[Milchstrabe]

Yeah, I know it won't work with a kg of mass from that small of a height, using my hand, with correct rations and everything, and a lip to keep the hacky sack from releasing early, I still can't get it to go 2.5 m horizontally. The best I can get hitting it very hard is about 1 m.

Should I start a new thread in the advance physics forum and see what we can do?

 

[whozum]

Wouldnt hurt, but try to make some error analysis to pinpoint the problem..

 

[whozum]

I wouldnt' be surprised if I missed something, but if we had a way to quantify the discrepancy it would be really helpful.

 

[Milchstrabe]

I posted on the college forum just to see if we can see what others have to say... if anything lol I have AP test next week, all kinds of things to do for this physics project, this is the most important part at least. THe rest the calculations i'll have to do are all basic, so I'm good. Refresh me tho, I just hada brain fart, how do you solve for the acceleration of a ball rolling down a ramp knowing the angle of course.

 

[whozum]

is it rolling or sliding..
if its sliding then friction is neglible and your acceleration is just g*sin(t)

 

[Milchstrabe]

can't we make a similar assumption if it's a marble (I know that it really isn't the same since the object is rolling, and gravity has an effect on how the ball spins and therefore changes the acceleration). However we never did it this way since it is high school physics after all. So i don't know if he wants me to go too deep with the whole rolling thing. Unless you can explain it fairly simply?