Recent content by TxRationalist

you need the equation of motion for rotation. the flywheel is undergoing uniform acceleration; you need to take that into consideration.

theta = 30 degrees.

I broke the problem down into the x and y components: x = (V0cosθ)t y = (V0sinθ)t - 0.5gt^2 I used the first equation to isolate V0, substituted this into the second equation and solved for t. I then used the first equation again to get my answer for V0. Hope this helps, I have to get...

well, i didn't cancel out the "t" the first time, so that my quadratic equation was in the general form. I corrected that, which led to the form of at^2+c=0. I solved this using my 0.97 m and got a result of 25.7 m/s. Awfully close. I'm going to run it with the same method using your numbers...

I'm not saying I'm right, just that that is what jumped out at me as the path to the solution when I looked at it. I ran the numbers and the quadratic equation gave me a negative for "b squared minus 4ac" also. Unless I did something wrong. I'm going to take another look at it.

you're given an angle of 30 degrees and told that it must not drop more than 3 cm...sounds like 0.97 meters since the question you quoted stated "drop", not rise...but it's your karma...good luck

Shouldn't that Max Height be 1-0.03 meters = 0.97 meters?

no, his math is correct. convert e to the i pi to cosine and sine (euler's equation). then take then treat the sum as doing exactly as he says. a sum that goes from 1 to infinity. it converges at a value. this is an oldy but a goodie.

I have to agree with you. seems they left some info out...maybe I'll print this out and look at it at home tonight...good luck

Since they give you the spring constant and the amount of compression, couldn't you figure out what the gain in kinetic energy was from going down the hill? Just a thought, I'm at work and can't spend time on it. Let me know how this works out!

You're very welcome, Nikieme

write the equation of motion for each partile, i.e. x0+v1t, etc. and use the distance equation sqrt(x^2 + y^2). then...

where are you taking the frictional force and V0 into consideration in your equations?

seems to me that you have to set up two equations, one for the v0 = 0 at time T and one for v0 = 72.52 m/s at time = (T - 3.7). that's my take...I haven't tried plugging these in, though.

I came up with the same answer. I didn't major in economics, though...I just assumed (because it sounds right) that you are right, that equilibrium occurs when the supply = demand.