Asked by Sarah
A large turntable rotates about a fixed vertical axis, making one revolution in 6.00 s. The moment of inertia of the turntable about this axis is 1200 kg * mg^2. A childe of mass 40.0 kg, initually staning at the center of the turntable, runs out along a radius. What is the angular speed of the turntable when the child is 2.00 m from the center, assuming that you can treat the child as a paarticle?
Answers
Answered by
MathMate
KE (original) = (1/2)Iω<sup>2</sup>
KE (new) = (1/2)I'w'<sup>2</sup>
where I'=1200 + mr<sup>2</sup>
m=40 kg
r=2 m
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 (new) = (1/2)I'w'<sup>2</sup>
where I'=1200 + mr<sup>2</sup>
m=40 kg
r=2 m
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.
Answered by
Sarah
KE (original) = (1/2)Iù2
(1/2)(1360)(PI/3)^2 = 744.9475
do i do the same for new... the "w" looks different
(1/2)(1360)(PI/3)^2 = 744.9475
do i do the same for new... the "w" looks different
Answered by
MathMate
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'.
Note also that I changes to I'.
Answered by
Sarah
KE (original) = (1/2)I(PI/3)2
KE (new) = (1/2)(1360)w'2
right??? now im confused because i don't have I and i don't have w^2
KE (new) = (1/2)(1360)w'2
right??? now im confused because i don't have I and i don't have w^2
Answered by
MathMate
I' is the sum of the turn-table <b>+</b> the child. The contribution of the child is simply mr<sup>2</sup>, since it can be considered as a particle.
So I' = 1200 kg * mg^2 + 40 kg * 2<sup>2</sup> kg * mg^2
= 1320 kg * mg^2 (as already worked out above).
The only unknown is now omega<sup>2</sup> which can be obtained by equating KE before and after the child's participation, i.e.
KE original = KE new
So I' = 1200 kg * mg^2 + 40 kg * 2<sup>2</sup> kg * mg^2
= 1320 kg * mg^2 (as already worked out above).
The only unknown is now omega<sup>2</sup> which can be obtained by equating KE before and after the child's participation, i.e.
KE original = KE new
Answered by
MathMate
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.
I is given in the question, 1200 kg * mg^2, so the only unknown is omega (w) in the equation.
Answered by
John
That is not right at all guys -_-
Answered by
John
I' w' = I'' w''
Answered by
Ben
anyone know the answer?
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