A 1.80 m radius playground merry-go-round has a mass of 120 kg and is rotating with an angular velocity of 0.300 rev/s. What is its angular velocity(rev/s)after a 22.0 kg child gets onto it by grabbing its outer edge? The child is initially at rest.
To find the angular velocity of the merry-go-round after the child gets onto it, we can use the principle of conservation of angular momentum.
The angular momentum of an object is given by the product of the moment of inertia and the angular velocity. In this case, the moment of inertia of the merry-go-round can be calculated using the formula:
I = 0.5 * m * r^2
where I is the moment of inertia, m is the mass of the merry-go-round, and r is the radius.
For the initial state, when only the merry-go-round is rotating, the moment of inertia (I_initial) can be calculated as:
I_initial = 0.5 * 120 kg * (1.80 m)^2
Now, let's consider the final state, when the child grabs the outer edge of the merry-go-round. The child's moment of inertia (I_child) can be calculated using the same formula:
I_child = 0.5 * 22.0 kg * (1.80 m)^2
According to the law of conservation of angular momentum, the initial angular momentum should be equal to the final angular momentum. Therefore,
I_initial * ω_initial = (I_initial + I_child) * ω_final
where ω_initial is the initial angular velocity and ω_final is the final angular velocity.
Rearranging the equation, we can solve for ω_final:
ω_final = (I_initial * ω_initial) / (I_initial + I_child)
Substituting the given values, we get:
ω_final = (I_initial * 0.300 rev/s) / (I_initial + I_child)
Now, we can calculate the angular velocity (ω_final).