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?
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.
KE (original) = (1/2)Iù2
(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 don't have I and i don't 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.
That is not right at all guys -_-
I' w' = I'' w''
anyone know the answer?
To solve this problem, we can apply the principle of conservation of angular momentum. The angular momentum of an object is given by the product of its moment of inertia and its angular velocity.
Initial angular momentum (L1) = Moment of inertia (I) * Angular velocity (ω1)
Final angular momentum (L2) = Moment of inertia (I) * Angular velocity (ω2)
According to the problem, the initial angular velocity of the turntable (ω1) is such that it completes one revolution in 6.00 s. Since one revolution corresponds to 2π radians, we can calculate ω1 as follows:
ω1 = 2π / (6.00 s)
ω1 ≈ 1.05 rad/s
When the child moves towards the outer edge, the total moment of inertia (I) of the system (turntable + child) remains constant. Therefore, we can write the equation:
I * ω1 = I * ω2
Now, we need to calculate the final angular velocity (ω2) when the child is 2.00 m from the center. Since the child is treated as a particle, we can use the equation for the moment of inertia of a particle about an axis at a given distance r from its center of mass:
I = m * r^2
Here, m is the mass of the child and r is the distance of the child from the center of the turntable. Plugging in the values, we have:
I = 40.0 kg * (2.00 m)^2
I = 160 kg·m^2
Now, we can rearrange our earlier equation and solve for ω2:
ω2 = (I * ω1) / I
ω2 = (160 kg·m^2 * 1.05 rad/s) / 160 kg·m^2
ω2 ≈ 1.05 rad/s
Therefore, the angular speed of the turntable when the child is 2.00 m from the center is approximately 1.05 rad/s.