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. [hint: similiar to the spinning ice-skater problem]
I * omega = the same
I gets smaller so omega gets bigger
To solve this problem, we can use the conservation of angular momentum principle. The angular momentum of an object can be calculated by multiplying its moment of inertia and angular velocity.
The initial angular momentum of the system is given by:
L_initial = I * ω_initial
where I is the moment of inertia of the system and ω_initial is the initial angular velocity.
Since the system is initially rotating at 0.180 rev/s, we need to convert this to radians per second:
ω_initial = 0.180 rev/s * 2π rad/rev = 0.180 * 2π rad/s
The angular momentum of the child can be calculated by:
L_child = I_child * ω_child
where I_child is the moment of inertia of the child and ω_child is the angular velocity of the child.
After the child moves to a final position, the system still conserves angular momentum. Therefore, the total angular momentum of the system will be equal to the angular momentum of the child at this final position.
We can calculate the final angular momentum of the system with the child at a distance of 1.10 m from the center of the merry-go-round as follows:
L_final = I_final * ω_final
where I_final is the moment of inertia of the system at this final position and ω_final is the final angular velocity of the system.
Since the child is standing on the outer edge of the merry-go-round, the moment of inertia of the system changes. The moment of inertia can be calculated as:
I_final = I_merry-go-round + I_child
where I_merry-go-round is the moment of inertia of the merry-go-round and I_child is the moment of inertia of the child.
The moment of inertia of the merry-go-round is given in the problem as 989 kg • m^2.
The moment of inertia of a point mass rotating about an axis at a distance r from its center of mass is given by:
I_child = m_child * r^2
where m_child is the mass of the child and r is the distance of the child from the axis of rotation.
In this problem, the mass of the child is given as 24.5 kg and the child moves to a final position 1.10 m from the center of the merry-go-round.
Now, equating the initial and final angular momenta of the system, we have:
L_initial = L_final
Using the equations for angular momentum and the moment of inertia, we can write the equation as:
I * ω_initial = (I_merry-go-round + I_child) * ω_final
We can substitute the given values into this equation and solve for ω_final to find the final angular velocity of the system.