Two children (m = 34.0 kg each) stand opposite each other on the edge of a merry-go-round. The merry-go-round, which has a mass of 1.60 ✕ 10^2 kg and a radius of 1.6 m, is spinning at a constant rate of 0.34 rev/s. Treat the two children and the merry-go-round as a system. Both children walk half the distance toward the center of the merry-go-round. Calculate the final angular speed of the system.
I initial=Idisk+Ikida+Ikidb
and same for final If. just remember the Ikids is 1/2 mr^2, and the second r is 1/2 the first.
conservation of angular momentum
Iinitial*wi=Ifinal*wfinal
and you solve for Wfinal.
To calculate the final angular speed of the system, we can use the principle of conservation of angular momentum.
The total angular momentum of the system is given by the formula:
L = Iω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
Initially, the children and the merry-go-round are at rest. At this point, the total angular momentum is zero.
When the children walk half the distance toward the center of the merry-go-round, the moment of inertia changes. However, the total angular momentum remains constant.
The moment of inertia of the system is given by:
I = Imerry-go-round + 2Ichildren
The moment of inertia of the merry-go-round is given by the formula:
I_merry-go-round = 1/2 * m_merry-go-round * r^2
Substituting the given values:
I_merry-go-round = 1/2 * (160) * (1.6)^2 = 204.8 kg·m^2
The moment of inertia of the children can be approximated as two point masses rotating about the center at a distance equal to half the radius. So, we can use the formula for the moment of inertia of a point mass:
I_child = m_child * r^2
For each child, substituting the given values:
I_child = (34.0) * (0.8)^2 = 21.76 kg·m^2
Now, substituting these values into the equation for the moment of inertia of the system:
I = 204.8 + 2 * 21.76 = 248.32 kg·m^2
Since the total angular momentum remains constant, we can set the initial angular momentum of the system equal to the final angular momentum:
L_initial = L_final
Initially, the angular momentum of the system is zero. The final angular momentum can be calculated as:
L_final = I final * ω final
Since L = Iω, we have:
L_final = I * ω_final
Setting the initial and final angular momenta equal:
0 = 248.32 * ω_final
Solving for ω_final:
ω_final = 0
Therefore, the final angular speed of the system is 0.
To calculate the final angular speed of the system, we can use the principle of conservation of angular momentum. The angular momentum of the system is constant before and after the children walk towards the center.
The angular momentum of a rotating object can be calculated using the formula:
L = I * ω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia of a rotating object can be calculated using the formula:
I = m * r^2
where m is the mass and r is the radius.
Let's first calculate the moment of inertia of the merry-go-round. Given that the mass of the merry-go-round is 1.60 ✕ 10^2 kg and the radius is 1.6 m:
I_merry-go-round = (1.60 ✕ 10^2 kg) * (1.6 m)^2 = 409.6 kg*m^2
The initial angular momentum of the system is the sum of the angular momenta of the merry-go-round and the children. Since the children are standing opposite each other, their angular momenta cancel out, while the merry-go-round's angular momentum remains unchanged.
Therefore, the initial angular momentum of the system is:
L_initial = I_merry-go-round * ω_initial
Now, let's calculate the initial angular velocity of the merry-go-round. Given that it is spinning at a constant rate of 0.34 rev/s, we need to convert this to angular velocity in radians per second:
ω_initial = (0.34 rev/s) * (2π rad/rev) = 2.136 rad/s
Now, we can calculate the initial angular momentum of the system:
L_initial = (409.6 kg*m^2) * (2.136 rad/s) = 876.1856 kg*m^2/s
The final angular momentum of the system remains the same because of the conservation of angular momentum.
Finally, we can calculate the final angular speed of the system using the formula for angular momentum:
L_final = I_merry-go-round * ω_final
Solving for ω_final:
ω_final = L_final / I_merry-go-round
Since the angular momentum is conserved, L_final is equal to L_initial:
ω_final = L_initial / I_merry-go-round
Plugging in the values:
ω_final = 876.1856 / 409.6 ≈ 2.137 rad/s
Therefore, the final angular speed of the system is approximately 2.137 rad/s.