- #106
Milchstrabe
- 74
- 0
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...
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...
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