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/sec).
The angular momentum of the system (merry-go-round + Lindsey) is conserved.
Angular momentum before = Angular momentum after
The angular momentum before Lindsey jumps onto the merry-go-round is zero, since it is initially at rest.
Angular momentum after Lindsey jumps onto the merry-go-round:
L_after = I_1 * ω_f
where I_1 is the moment of inertia of the merry-go-round and ω_f is the angular velocity of the merry-go-round after Lindsey jumps on it.
The moment of inertia of a solid disk rotating about its axis of symmetry is given by:
I_1 = (1/2) * m_1 * r_1^2
where m_1 is the mass of the merry-go-round and r_1 is its radius.
Plugging in the given values:
I_1 = (1/2) * 78 kg * (2.20 m)^2
= 387.12 kg*m^2
The linear velocity of Lindsey before she jumps on the merry-go-round is equal to the linear velocity of a point on the outer edge of the merry-go-round. As she jumps onto the merry-go-round, her linear velocity becomes the same as the linear velocity of the merry-go-round.
Since the angular velocity ω is related to the linear velocity v by the equation:
v = r * ω
where r is the radius, we can rearrange the equation to solve for ω:
ω = v / r
Plugging in the given values:
ω = 9.0 m/s / 2.20 m
= 4.09 rad/s
Therefore, the angular velocity of the merry-go-round after Lindsey jumps on it is approximately 4.09 rad/s.