Question

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?

Answer 1

KE (original) = (1/2)Iω²

KE (new) = (1/2)I'w'²

where I'=1200 + mr²
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.

Answer 2

KE (original) = (1/2)Iù2
(1/2)(1360)(PI/3)^2 = 744.9475

do i do the same for new... the "w" looks different

Answer 3

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'.

Answer 4

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

Answer 5

I' is the sum of the turn-table <b>+</b> 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

Answer 6

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.

Answer 7

That is not right at all guys -_-

Answer 8

I' w' = I'' w''

Answer 9

anyone know the answer?