A 3.6 m diameter merry-go-round is rotating freely with an angular velocity of 0.78 rad/s. Its total moment of inertia is 1700 kg\m^2. Four people standing on the ground, each of mass 62 kg, suddenly step onto the edge of the merry-go-round.
1. What is the angular velocity of the merry-go-round now?
2. What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?
What do you think the answer is?
I keep getting 0.68 ?????
I get something close to that.
Total kinetic energy
= (1/2)Iω²
ω1=0.78 rad/s
I1=1700 kg-m²
I2=1700+4*(62*1.8²)=2503.5
(1/2)I1ω1²=(1/2)I2ω2²
Solve for
ω2
=√(I1/I2)ω1
=0.64 rad/s.
Check my calculations.
To answer these questions, we need to understand the concept of conservation of angular momentum. Angular momentum is the rotational equivalent of linear momentum and is conserved in the absence of external torques.
1. To find the new angular velocity of the merry-go-round when the people step onto its edge, we can set up an equation based on the conservation of angular momentum. The initial angular momentum is given by 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
Given:
I_initial = 1700 kg·m^2
ω_initial = 0.78 rad/s
We can substitute these values into the equation to calculate the initial angular momentum.
L_initial = (1700 kg·m^2) * (0.78 rad/s) = 1326 kg·m^2/s
Now, let's consider the final angular momentum when the people step onto the merry-go-round. The final angular momentum can be calculated using 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 merry-go-round (including the people)
ω_final is the final angular velocity of the merry-go-round
Since the people are initially at rest relative to the ground, they do not contribute to the initial angular momentum. Therefore, the final angular momentum is only due to the merry-go-round.
L_final = 1326 kg·m^2/s
We can rearrange the equation to solve for the final angular velocity:
ω_final = L_final / I_final
We know L_final and I_final is the sum of the initial moment of inertia and the moment of inertia due to the people.
I_final = I_initial + (m1 * r1^2 + m2 * r2^2 + m3 * r3^2 + m4 * r4^2)
Where:
m1, m2, m3, m4 are the respective masses of the four people (62 kg each)
r1, r2, r3, r4 are the respective distances of the four people from the axis of rotation (radius of the merry-go-round)
Given the diameter of the merry-go-round is 3.6 m, the radius is 1.8 m. Assuming the four people stand at the edge of the merry-go-round, their distances from the axis of rotation are also 1.8 m.
Substituting the values into the equation, we get:
I_final = 1700 kg·m^2 + (62 kg * (1.8 m)^2 + 62 kg * (1.8 m)^2 + 62 kg * (1.8 m)^2 + 62 kg * (1.8 m)^2)
By calculating this expression, we can find the value of I_final. Then, substituting this value and L_final into the equation, we can calculate ω_final, which will give us the answer to the first question.
2. If the people on the merry-go-round jump off in a radial direction relative to the merry-go-round, they will have no effect on its angular velocity. This is because their movement is opposite in direction to the rotation, and their moment of inertia does not change. Therefore, the angular velocity of the merry-go-round will remain the same as its initial angular velocity, which is 0.78 rad/s.
To summarize:
1. To find the new angular velocity when the people step onto the merry-go-round, calculate I_final and ω_final using the equations mentioned above.
2. If the people jump off in a radial direction, the angular velocity of the merry-go-round remains the same.