A 24.5-kg child is standing on the outer edge of a merry-go-round which has moment of inertia 989 kg • m2 and radius 2.40 m. The entire system is initially rotating at 0.180 rev/s. Find the angular velocity if the child moves to a final position 1.10 m from the center of the merry-go-round.
Well, let's see how this child's movement affects the merry-go-round's angular velocity. First, we need to find the initial angular momentum of the system.
The initial moment of inertia of the merry-go-round is 989 kg • m^2. And since it's initially rotating at 0.180 rev/s, we can convert that to angular velocity (omega) by multiplying by 2π, which gives us 1.13 rad/s.
Now, when the child moves closer to the center, the moment of inertia of the system changes. The moment of inertia of the child can be approximated as mr^2, where m is the mass of the child (24.5 kg) and r is the distance from the center (1.10 m). So, the final moment of inertia of the system is 24.5 kg * (1.10 m)^2.
To find the final angular velocity (omega'), we can use the law of conservation of angular momentum, which states that the initial angular momentum of a system is equal to the final angular momentum of the system.
So,
Initial angular momentum = Final angular momentum
Initial angular momentum = Initial moment of inertia * Initial angular velocity
Final angular momentum = Final moment of inertia * Final angular velocity
Since the angular momentum is conserved, we have:
(989 kg • m^2) * (1.13 rad/s) = (24.5 kg * (1.10 m)^2) * omega'
Now we just need to solve for omega' (final angular velocity). Let me grab my calculator...
*Clown Bot reaches for a calculator but ends up juggling three rubber chickens instead*
Oops, sorry about that. Got a little carried away with the juggling act there. Give me a second, let me recalculate the value for you.
After crunching the numbers (and the rubber chickens), I found that the final angular velocity (omega') is approximately 10.68 rad/s.
So, when the child moves to the final position, the angular velocity of the merry-go-round changes to around 10.68 rad/s. That's quite a spin!
To find the final angular velocity, we can use the law of conservation of angular momentum, which states that the angular momentum of a system remains constant unless an external torque is applied.
The angular momentum (L) of the system can be calculated using the formula:
L = I * ω
Where:
L is the angular momentum
I is the moment of inertia
ω is the angular velocity
In the initial position, the child is standing on the outer edge of the merry-go-round. Therefore, the initial moment of inertia (I_initial) is the moment of inertia of the entire system (merry-go-round + child):
I_initial = 989 kg • m^2
The initial angular velocity (ω_initial) is given as 0.180 rev/s. To convert this to rad/s, we use the conversion factor:
1 revolution = 2π radians
Therefore, ω_initial = 0.180 rev/s * 2π rad/1 rev = 1.13 rad/s
After the child moves to the final position, the moment of inertia (I_final) changes since the child is now closer to the center. The final moment of inertia (I_final) can be calculated using the parallel axis theorem:
I_final = I_initial + m * r^2
Where:
m is the mass of the child
r is the distance of the child from the center of the merry-go-round
In this case, the mass of the child (m) is given as 24.5 kg, and the distance from the center (r) is given as 1.10 m. Plugging these values into the formula, we have:
I_final = 989 kg • m^2 + 24.5 kg * (1.10 m)^2 = 1018.43 kg • m^2
Now, using the law of conservation of angular momentum, we can equate the initial angular momentum to the final angular momentum:
I_initial * ω_initial = I_final * ω_final
Solving for ω_final, we get:
ω_final = (I_initial * ω_initial) / I_final
Plugging in the given values, we have:
ω_final = (989 kg • m^2 * 1.13 rad/s) / 1018.43 kg • m^2 ≈ 1.09 rad/s
Therefore, the final angular velocity is approximately 1.09 rad/s.