A 3.7 m diameter merry-go-round is rotating freely with an angular velocity of 0.63 rad/s. Its total moment of inertia is 1600 kg\cdot m^2. Four people standing on the ground, each of mass 70 kg, suddenly step onto the edge of the merry-go-round.
You have not said what the question is. Presumably, they want to know the angular velocity after the four people step on.
You can solve for that by using the fact that total angular momentum is conserved. Let the initial and final angular velocity be w1 and w2, respectively.
I*w1 = (I + 4*M*R^2)w2
M is the mass of one person.
I = 1600 kg*m^2 is the merry-go-round moment of inertia (without people on it).
Solve for w2
To find the new angular velocity of the merry-go-round after the people step onto it, we can use the principle of conservation of angular momentum.
The initial angular momentum of the merry-go-round can be calculated using the formula:
L_initial = I_initial * ω_initial
Where:
L_initial is the initial angular momentum
I_initial is the initial moment of inertia of the merry-go-round
ω_initial is the initial angular velocity of the merry-go-round
Plugging in the given values, we can calculate the initial angular momentum:
L_initial = 1600 kg·m^2 * 0.63 rad/s
Next, let's consider the angular momentum after the people step onto the merry-go-round. The total mass of the people is 4 * 70 kg = 280 kg.
The final moment of inertia of the system (merry-go-round + people) can be calculated based on the principle of conservation of linear momentum. Since no external torques are acting on the system, angular momentum is conserved. Therefore, we can use the formula:
L_final = I_final * ω_final
Where:
L_final is the final angular momentum
I_final is the final moment of inertia of the system (merry-go-round + people)
ω_final is the final angular velocity of the system
Since the people are initially at rest, their initial angular velocity is zero:
ω_people_initial = 0 rad/s
When they step onto the spinning merry-go-round, they start rotating with the same angular velocity as the merry-go-round. Therefore, their final angular velocity is also 0.63 rad/s:
ω_people_final = 0.63 rad/s
We can calculate the final angular momentum of the system after the people step onto the merry-go-round:
L_final = (I_merry_go_round + I_people) * ω_final
Where:
I_merry_go_round is the moment of inertia of the merry-go-round
I_people is the moment of inertia of the people (treated as point masses)
ω_final is the final angular velocity of the system
The moment of inertia of the people can be calculated using the formula:
I_people = m_people * r^2
Where:
m_people is the mass of each person
r is the radius of the merry-go-round
Plugging in the given values, we can calculate the final moment of inertia of the system:
I_final = (1600 kg·m^2 + (280 kg * (3.7 m / 2)^2)) * 0.63 rad/s
Once we have the initial and final angular momenta, we can set them equal to each other because angular momentum is conserved:
L_initial = L_final
1600 kg·m^2 * 0.63 rad/s = (1600 kg·m^2 + (280 kg * (3.7 m / 2)^2)) * 0.63 rad/s
Simplifying and solving for the final angular velocity, ω_final:
ω_final = (1600 kg·m^2 * 0.63 rad/s) / (1600 kg·m^2 + (280 kg * (3.7 m / 2)^2))
Plug the numbers into the equation and simplify to find the final angular velocity of the merry-go-round after the people step onto it.