Recent content by LZ27

So I got -mgh from solving the equation for v: (1/2)kΔl2 = (1/2)I(v2/r2) + (1/2)mv2 + mgh then pull all the terms with v to one side and get rid of the 1/2: I(v2/r2) + mv2 = kΔl2 - 2mgh get rid of the r2 fraction (not necessary, really, but I did it): v2(I + mr2) = r2(kΔl2 - 2mgh) Then...

Thanks for the help. I checked my calculations, and I got 1.89m/s again. My work is: v = sqrt[(r2(kΔl-2mgh))/(r2m + I)] v = sqrt[(.005)2(103*.00632 - 2*.005*9.8*.15))/(.0052*.005 + 5x10-8)] I left out units so as to simplify things, but they work out to be m/s. Should I be using a...

Alright, so I tried it again: a) Us = Kr + Kt + Ug (I think there is also gravitational potential energy once the marble is at the top of the loop) Giving me (1/2)kΔl2 = (1/2)I(v2/r2) + (1/2)mv2 + mgh And solved for l (h=30cm for the top of the loop). Which gave me 6.3 x 10-3m or .63 cm...

Hi all--Thanks for any help you might be able to provide, I've been lurking in the forum for a while and find everyone's comments to be extraordinarily helpful. Homework Statement The spring with a constant 103N/m launches a marble (m=5 g, r=.5cm) on a horizontal track with a loop (like a...