Recent content by Koolaid

okay i got it 16(1.26)=r6 r=3.36 thanks a lot phantomJay for all your help

well w=v/r so if L_(initial) =L_(final) and I=mr and L=Iw so mrw=mrw w=v/r mrw=mr(v/r) i canceled the m's out cause the mass is the same i also canceled the r's out on the right side r(1.26)=6 so from that we get r=4.76 but were not trying to get r were trying to get a length? mlw=ml(v/r) i...

well the conservation of angular momentum is L=Iw(w=omega) but i can't seem how to fit the givens in or come up with missing pieces like the moment of inertia

if i do that (.5*.3^2) times to for each ball... so .09 kg-m^2... doest hat make sence?

[SIZE="5"]i have two questions that i need help understanding 1a. Homework Statement A majorette takes tow batons and fastens them together in the middle at right angles to make an "x" shape. Each baton was 0.6m long and each ball on the end is 0.50kg. (Ignore the mass of the rods.) What is...

wow those were simple now that I think about it lol… well now i understand them.. thank you to everyone that helped

ok i got the first question all done. but for question 2. i am not sure how to incorporate the 1000 g into the problem. the radius also needs to be in meters os its .15 meters

i am still lost.. i am just not really making the "connection" for both

#1. still don't know g or m cause you only have G = 6.67[SIZE="1"]E-11 and r = 4000mi = 6.437[SIZE="1"]E6 m then multiply by 2 = 1.2874[SIZE="1"]E7 m . #2. a=rw [ centripetal acceleration = radius * angular velocity]

well for #1 i updated my post for #2 wouldn't be tangetal acceleration so [tangetal acceleration = radius * angular acceleration] (a=r*a) right? or would it be centripetal acceleration i am not sure [ centripetal acceleration = tangental velocity squared / radius] (a=v^2/r)

hello i am new to the forums. But I have been working on a packet of problems and I got them all done but i have been stuck on these two for a good hour. I have tried to search to internet for possible equations i could use but can't find any that will work or maybe i am just over looking them...