Recent content by huybinhs

It's on a paper exam, so not sure about the answer yet :(

If so, I calculated I(total) = 8.00321 * 10^-4 => w = 14.32 rad/s Could u check whether it's correct or not, and let me know, k? Thanks ;)

.5mv2^2 + .5Iw^2 = .5mv1^2 I have some problem on this equation. It's about the m, is it? : .5(m+M)v2^2 + .5Iw^2 = .5mv1^2 , correct?

Hang on. Let me calculate it now ;)

I think the I on the right side will = 0, so the final equation is: .5mv2^2 + .5Iw^2 = .5mv1^2 Right?

So, the I on both side will be the same as I(total) , right?

Now I can see that it fits you perfectly ;)

LOL .5mv2^2 + .5Iw^2 = .5mv1^2 + .5Iw^2 where v1 is initial given, v2 is the part b answer, correct? ;)

Great! Then I just plug it in: .5mv2^2 + .5Iw^2 = .5mv1^2 + .5Iw^2 to find w, correct?

Ok, so: I before collision = (1/2 m r^2 + 1/2 MR^2) + [(m+M) * (R+r)] then I after collision = (1/2 m r^2 + 1/2 MR^2) + [(m+M) * 0.034925 m] which 0.034925 m is the center of mass of part a, am I right?

Sorry! What do u mean by that? Are those equation set up right?

so KEafter collision = .5mv2^2 + .5Iw^2 and KEbefore collison = .5mv1^2 + .5Iw^2 then set it to equal: .5mv2^2 + .5Iw^2 = .5mv1^2 + .5Iw^2 (why are they equal?) with I = 1/2 m r^2 + 1/2 MR^2 Am I on the right track? Thanks so much for your patience ;)

If using this, I have: Isystem = 1/2 m r^2 + 1/2 MR^2 then how can I get KE in order to find w (angular velocity)?

If the answer is 0.533 m/s, this is not right because this is the answer for part b already, right? I'm confused :(

So because this is inelastic collision, we have: m1v1+m2v2 = (m1+m2)v ,correct? then what should i do next?