A 33.0 kg child named Lindsey runs as fast as she can and jumps onto the outer edge of a merry-go-round. The merry-go-round is initially at rest and has a mass of 78 kg and a radius of 2.20m . Lindsey's linear velocity was 9.0m/s at the moment she jumped onto the merry-go-round. What is the angular velocity of the merry-go-round after she jumps on it (in rad/s )?
To find the angular velocity of the merry-go-round after Lindsey jumps on it, we can use the principle of conservation of angular momentum.
The initial angular momentum of the system (merry-go-round + Lindsey) is given by:
L_initial = I_initial * ω_initial
Where I_initial is the moment of inertia of the system, and ω_initial is the initial angular velocity of the system.
The final angular momentum of the system is given by:
L_final = I_final * ω_final
Where I_final is the moment of inertia of the system after Lindsey jumps on, and ω_final is the final angular velocity of the system.
According to the conservation of angular momentum principle, L_initial = L_final. Therefore:
I_initial * ω_initial = I_final * ω_final
The moment of inertia of the merry-go-round is given by:
I_merry-go-round = 1/2 * m_merry-go-round * r^2
Where m_merry-go-round is the mass of the merry-go-round, and r is the radius.
Substituting the given values:
I_merry-go-round = 1/2 * 78 kg * (2.20m)^2 = 422.64 kg*m^2
The moment of inertia of Lindsey after she jumps on can be approximated as the moment of inertia of a point mass rotating about an axis at a distance r from the axis, which is given by:
I_Lindsey = m_Lindsey * r^2
Where m_Lindsey is the mass of Lindsey.
Substituting the given values:
I_Lindsey = 33.0 kg * (2.20m)^2 = 1627.32 kg*m^2
Since there are no external torques acting on the system, the initial angular momentum of Lindsey is zero (L_initial_Lindsey = 0). Therefore:
L_initial = I_initial * ω_initial = I_merry-go-round * ω_initial
ω_initial = L_initial / I_merry-go-round = 0 / 422.64 kg*m^2 = 0 rad/s
For the final angular velocity, using the conservation of angular momentum equation:
I_initial * ω_initial = I_final * ω_final
(422.64 kg*m^2) * (0 rad/s) = (422.64 kg*m^2 + 1627.32 kg*m^2) * ω_final
0 = 2050.96 kg*m^2 * ω_final
Therefore, the final angular velocity of the merry-go-round after Lindsey jumps on it is:
ω_final = 0 rad/s
So the angular velocity after Lindsey jumps on the merry-go-round is 0 rad/s.