A 4.2 m diameter merry go round is rotating freely with an angular velocity of 0.8 rad/s. If its total moment of inertia is 1760 kg*m2, find the angular velocity of the merry go round when four people of 72 kg jump onto the edge of the merry go round. Treat the persons as point particles.
To solve this problem, we need to analyze the principle of conservation of angular momentum. The moment of inertia of the system is given by the sum of the moment of inertia of the merry go round and the moment of inertia of the people.
The moment of inertia of the merry go round can be calculated using the formula:
I_merry go round = 0.5 * m * r^2
Where:
m = mass of the merry go round
r = radius of the merry go round
Given that the diameter is 4.2m, the radius (r) is half of that, which is 2.1m.
So, the mass of the merry go round can be calculated using the formula:
m = 2 * π * r * ρ
Where:
ρ = density of the merry go round (assuming it to be uniform)
To find the angular velocity of the merry go round when the people jump onto the edge, we need to apply the conservation of angular momentum. According to this principle, the initial angular momentum is equal to the final angular momentum of the system.
Initially, when the merry go round is rotating freely, its angular momentum (L_initial) is given by:
L_initial = I_merry go round * ω_initial
Where:
ω_initial = initial angular velocity of the merry go round = 0.8 rad/s (given)
Once the four people of mass 72kg each jump onto the edge, the angular momentum (L_final) is given by:
L_final = (I_merry go round + I_people) * ω_final
Where:
I_people = 4 * m * r^2 (since there are four people)
As the problem assumes that the people are point particles, the moment of inertia of each person is m * r^2.
Substituting the values and rearranging the equation:
I_merry go round * ω_initial = (I_merry go round + (4 * m * r^2)) * ω_final
Dividing both sides by I_merry go round and simplifying:
ω_initial = (I_merry go round / (I_merry go round + (4 * m * r^2))) * ω_final
Solving for ω_final:
ω_final = (ω_initial * I_merry go round) / (I_merry go round + (4 * m * r^2))
Now, let's calculate the values and find the angular velocity of the merry go round when the people jump onto the edge.
Given:
ω_initial = 0.8 rad/s
I_merry go round = 1760 kg * m^2
r = 2.1m
m = 2 * π * r * ρ (to be calculated)
First, we need to calculate the mass of the merry go round using the equation:
m = 2 * π * r * ρ
Assuming a constant density of the merry go round, we'll use this equation to find ρ:
2 * π * r * ρ = m / V
Since the merry go round is a cylinder, we can calculate its volume (V):
V = π * r^2 * h
Since the diameter is given as 4.2m, we can assume the height (h) to be negligible compared to the radius (r). Therefore, we can approximate the volume (V) as:
V ≈ π * r^2
Substituting this value in the equation for mass:
2 * π * r * ρ = m / (π * r^2)
Cancelling out the common terms:
2 * ρ = 1 / r
Substituting the value of r = 2.1m:
2 * ρ = 1 / 2.1
Simplifying:
ρ = 1 / (2 * 2.1) = 1 / 4.2
ρ ≈ 0.238
Now, we have the density (ρ) of the merry go round. Let's substitute all the values into the equation for ω_final:
ω_final = (0.8 rad/s * 1760 kg * m^2) / (1760 kg * m^2 + (4 * (2 * π * 2.1 * 0.238) * (2.1)^2))
Simplifying further:
ω_final = (0.8 rad/s * 1760) / (1 + ( 4 * (2 * π * 2.1 * 0.238) * (2.1)^2))
Calculating the expression gives us the angular velocity of the merry go round when the four people jump onto its edge.