Posted by Sarah on Thursday, June 25, 2009 at 10:44pm.
KE (original) = (1/2)Iω2
KE (new) = (1/2)I'w'2
where I'=1200 + mr2
I' = 1200 + 160 = 1360
Setting KE (original) = KE (new)
the only unkown is ω' for which you can solve.
Note: ω=2π/6 rad/s = π/3 rad/s.
KE (original) = (1/2)Ił2
(1/2)(1360)(PI/3)^2 = 744.9475
do i do the same for new... the "w" looks different
I put w' instead of ω', but they are meant to be omega's. ω' is different from ω.
Note also that I changes to I'.
KE (original) = (1/2)I(PI/3)2
KE (new) = (1/2)(1360)w'2
right??? now im confused because i dont have I and i dont have w^2
I' is the sum of the turn-table + the child. The contribution of the child is simply mr2, since it can be considered as a particle.
So I' = 1200 kg * mg^2 + 40 kg * 22 kg * mg^2
= 1320 kg * mg^2 (as already worked out above).
The only unknown is now omega2 which can be obtained by equating KE before and after the child's participation, i.e.
KE original = KE new
Oops, I' should be 1360 as you had it.
I is given in the question, 1200 kg * mg^2, so the only unknown is omega (w) in the equation.
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