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 78.0 kg and a radius of 2.20 m. Lindsey's linear velocity was 9.00 m/s at the moment she jumped onto the merry-go-round. What is the initial angular momentum of the system (in kg m^2/s)? [Hint: Although she started with linear velocity, consider the moment just before she landed on the merry-go-round.]
And what is the angular velocity of the merry go round after she jumps on it in rad/sec?
To find the initial angular momentum of the system, we need to consider just before Lindsey lands on the merry-go-round.
The initial angular momentum of the system is given by the formula:
L_initial = I_initial * ω_initial
Where:
I_initial = moment of inertia of the system
ω_initial = angular velocity of the system
First, we need to calculate the angular momentum of Lindsey just before she lands on the merry-go-round.
Angular momentum of Lindsey just before landing = m * v * r
= 33.0 kg * 9.00 m/s * 2.20 m
= 646.8 kg m^2/s
Now, since the merry-go-round is initially at rest and Lindsey's angular momentum just before landing is equal to the initial angular momentum of the system, we have:
L_initial = 646.8 kg m^2/s
Next, we need to calculate the moment of inertia of the system. The moment of inertia of a merry-go-round is given by:
I = 1/2 * MR^2
= 1/2 * 78.0 kg * (2.20 m)^2
= 47.124 kg m^2
So, the initial angular momentum of the system is:
L_initial = I_initial * ω_initial
646.8 kg m^2/s = 47.124 kg m^2 * ω_initial
ω_initial = 13.71 rad/s
Therefore, the initial angular momentum of the system is 646.8 kg m^2/s and the angular velocity of the merry-go-round after Lindsey jumps on it is 13.71 rad/s.