In a playground, there is a small merry-go-round of radius 1.20 m and mass 190 kg. Its radius of gyration (see Problem 85 of Chapter 10) is 91.0 cm. A child of mass 44.0 kg runs at a speed of 4.50 m/s along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round.
(b) Calculate the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round.
(c) Calculate the angular speed of the merry-go-round and child after the child has jumped on.
Answer:
(b) The magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round is 0.198 kg m2/s.
(c) The angular speed of the merry-go-round and child after the child has jumped on is 0.845 rad/s.
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.
To solve this problem, we can start by calculating the angular momentum of the child while running (part b) and then use the conservation of angular momentum to calculate the angular speed of the merry-go-round and child after the child has jumped on (part c).
(b) To calculate the magnitude of the angular momentum of the child while running, we can use the formula:
Angular momentum (L) = moment of inertia (I) * angular speed (ω)
The moment of inertia of a point mass rotating around a fixed axis is given by:
I = m * r^2
Where:
m = mass of the child = 44.0 kg
r = radius of the merry-go-round = 1.20 m
Now, we need to find the angular speed (ω). Since the child is running along a path tangent to the rim, the linear speed (v) of the child is equal to the distance traveled divided by the time taken:
v = 4.50 m/s
The distance traveled by the child along the rim of the merry-go-round is equal to the circumference of the merry-go-round, which is given by:
2 * π * radius = 2 * π * 1.20 m
Now, we can use the formula for angular speed:
ω = v / r
Substituting the given values:
ω = 4.50 m/s / 1.20 m
Calculate the angular speed (ω).
(c) To calculate the angular speed of the merry-go-round and child after the child has jumped on, we can use the conservation of angular momentum. Since there is no external torque acting on the system, the initial angular momentum of the child while running will be equal to the final angular momentum of the system (merry-go-round + child).
Initial angular momentum = Final angular momentum
The initial angular momentum of the child while running is given by:
L_initial = I_child * ω_child
Substituting the given values for the moment of inertia (I_child) and angular speed (ω_child) of the child while running, calculate the initial angular momentum (L_initial).
The final angular momentum of the system (merry-go-round + child) is given by:
L_final = I_merry-go-round * ω_final
We need to find the moment of inertia (I_merry-go-round) of the merry-go-round and the final angular speed (ω_final) of the system.
The moment of inertia of a solid disk rotating around its axis is given by:
I = (1/2) * m * r^2
Where:
m = mass of the merry-go-round = 190 kg
r = radius of gyration of the merry-go-round = 91.0 cm = 0.91 m
Since the child jumps onto the merry-go-round in a way that doesn't change the moment of inertia of the merry-go-round, I_merry-go-round remains the same before and after the jump.
Now, we can solve for the final angular speed (ω_final) using the equation:
L_initial = L_final
Substituting the known values for L_initial, I_child, I_merry-go-round, and solving for ω_final, calculate the final angular speed (ω_final).
Note: Make sure all units are consistent throughout the calculations.