To solve this problem, we need to use the principle of conservation of angular momentum. This principle states that the total angular momentum before an event is equal to the total angular momentum after the event, as long as no external torques act on the system.
(b) To calculate the magnitude of the angular momentum of the child while running, we need to find the angular momentum of the child before she jumps onto the merry-go-round. The angular momentum of an object can be calculated by multiplying its moment of inertia by its angular velocity.
The moment of inertia of the child running along a path tangent to the rim of the merry-go-round can be calculated using the formula:
I_child = m_child * r_tangent^2
Where:
m_child is the mass of the child (44.0 kg)
r_tangent is the distance from the axis of rotation to the path the child is running along (which is equal to the radius of the merry-go-round, 1.20 m)
So, substituting the values into the formula:
I_child = 44.0 kg * (1.20 m)^2
I_child = 44.0 kg * 1.44 m^2
I_child = 63.36 kg.m^2
The angular velocity of the child while running is the speed of the child divided by the radius of the merry-go-round:
ω_child = v_child / r_merry-go-round
Where:
v_child is the speed of the child (4.50 m/s)
r_merry-go-round is the radius of the merry-go-round (1.20 m)
So, substituting the values into the formula:
ω_child = 4.50 m/s / 1.20 m
ω_child = 3.75 rad/s
Finally, we can calculate the angular momentum of the child while running by multiplying the moment of inertia by the angular velocity:
L_child = I_child * ω_child
L_child = 63.36 kg.m^2 * 3.75 rad/s
L_child = 237.6 kg.m^2/s
Therefore, the magnitude of the angular momentum of the child while running about the axis of rotation of the merry-go-round is 237.6 kg.m^2/s.
(c) After the child jumps onto the merry-go-round, the total angular momentum of the system (merry-go-round + child) remains constant.
To find the angular speed of the merry-go-round and child after the child has jumped on, we can use the conservation of angular momentum equation:
L_initial = L_final
The initial angular momentum is the angular momentum of the child while running, which we found to be 237.6 kg.m^2/s.
The final angular momentum can be calculated by summing the individual angular momenta of the merry-go-round and the child after the child has jumped on. Let's represent the angular momentum of the merry-go-round as L_merry-go-round and the angular momentum of the child after jumping on as L_child+jumped.
L_final = L_merry-go-round + L_child+jumped
Since the merry-go-round is initially at rest, its initial angular momentum is zero. Therefore, we have:
237.6 kg.m^2/s = 0 + L_child+jumped
To find L_child+jumped, we need to use the same formula as in part (b):
L_child+jumped = I_total * ω_final
The total moment of inertia, I_total, is the sum of the moment of inertia of the merry-go-round and the moment of inertia of the child after jumping on.
The moment of inertia of the merry-go-round, I_merry-go-round, can be calculated using the formula:
I_merry-go-round = m_merry-go-round * r_merry-go-round^2
Where:
m_merry-go-round is the mass of the merry-go-round (190 kg)
r_merry-go-round is the radius of the merry-go-round (1.20 m)
So, substituting the values into the formula:
I_merry-go-round = 190 kg * (1.20 m)^2
I_merry-go-round = 190 kg * 1.44 m^2
I_merry-go-round = 273.6 kg.m^2
The moment of inertia of the child after jumping on, I_child+jumped, can be calculated using the formula:
I_child+jumped = m_child * r_gyration^2
Where:
m_child is the mass of the child (44.0 kg)
r_gyration is the radius of gyration of the merry-go-round (91.0 cm = 0.91 m)
So, substituting the values into the formula:
I_child+jumped = 44.0 kg * (0.91 m)^2
I_child+jumped = 44.0 kg * 0.8281 m^2
I_child+jumped = 36.38 kg.m^2
Now we can substitute the moment of inertia values into the equation for L_child+jumped:
L_child+jumped = (I_merry-go-round + I_child+jumped) * ω_final
To solve for ω_final, rearrange the equation:
ω_final = L_child+jumped / (I_merry-go-round + I_child+jumped)
Substituting the values into the equation:
ω_final = 237.6 kg.m^2/s / (273.6 kg.m^2 + 36.38 kg.m^2)
ω_final = 237.6 kg.m^2/s / 309.98 kg.m^2
ω_final = 0.7677 rad/s
Therefore, the angular speed of the merry-go-round and child after the child has jumped on is approximately 0.7677 rad/s.