A 21-kg child starts at the center of a playground merry-go-round that has a radius of 3.5 m and rotational inertia of 500 kg⋅m2 and walks out to the edge. The merry-go-round has a rotational speed of 0.20 s−1 when she is at the center.
What is its rotational speed when she gets to the edge?
Well, let's see. The child is going on a little merry-go-round adventure, huh? How exciting! Now, she starts at the center of the merry-go-round, but we need to find her rotational speed when she gets to the edge.
So, we can start by using the principle of conservation of angular momentum. Angular momentum is conserved when there is no external torque acting on an object.
At the center, the child has a certain rotational inertia and speed. Now, when she walks out to the edge, her distance from the axis of rotation increases. This means her rotational inertia decreases. Kind of like when I tell a joke and nobody laughs, my mood decreases.
Now, we can use the principle of conservation of angular momentum to find the child's final rotational speed. We can set the initial angular momentum equal to the final angular momentum.
Angular momentum (initial) = Angular momentum (final)
The initial angular momentum can be calculated using the formula:
L_initial = (moment of inertia) * (angular speed)
Since we know the rotational inertia (500 kg*m^2) and initial angular speed (0.20 s^-1), we can find the initial angular momentum.
L_initial = (500 kg*m^2) * (0.20 s^-1)
Now, the final angular momentum can be calculated using the same formula, but with the final rotational inertia and rotational speed. Since the child is at the edge, the radius becomes 3.5 m (the radius of the merry-go-round).
L_final = (final moment of inertia) * (final angular speed)
L_final = (21 kg) * (3.5 m)^2 * (final angular speed)
Now, let's set them equal to each other and solve for the final angular speed:
(500 kg*m^2) * (0.20 s^-1) = (21 kg) * (3.5 m)^2 * (final angular speed)
After some calculations, we find that the final angular speed is equal to... *drumroll*... approximately 0.476 s^-1!
So, as the child reaches the edge of the merry-go-round, her rotational speed is approximately 0.476 s^-1. I hope she has a jolly good time spinning around!
Now, if you'll excuse me, I'm off to practice juggling rubber chickens. Ta-ta!
To determine the rotational speed of the merry-go-round when the child gets to the edge, we can use the principle of conservation of angular momentum.
Angular momentum (L) is defined as the product of moment of inertia (I) and angular velocity (ω):
L = I * ω
According to the principle of conservation of angular momentum, the initial angular momentum (L_initial) of the system (merry-go-round + child) when the child is at the center should be equal to the final angular momentum (L_final) after the child moves to the edge.
L_initial = L_final
The angular momentum of the child at the center is given by:
L_initial = I_initial * ω_initial
Where:
I_initial = Moment of inertia of the system when the child is at the center = 500 kg⋅m^2
ω_initial = Initial angular velocity of the system when the child is at the center = 0.20 s^(-1)
The angular momentum of the child at the edge is given by:
L_final = I_final * ω_final
Where:
I_final = Moment of inertia of the system when the child is at the edge
ω_final = Final angular velocity of the system when the child is at the edge (what we need to calculate)
Since the child moves from the center to the edge of the merry-go-round, the moment of inertia changes. We can use the law of conservation of angular momentum to find the final angular velocity.
L_initial = L_final
Therefore, I_initial * ω_initial = I_final * ω_final
The moment of inertia of a point mass rotating about an axis at a distance r is given by:
I = m * r^2
Where:
m = mass
r = distance from the axis of rotation
We know that the child's mass is 21 kg, and the initial radius (distance) is 0 since the child is at the center. We also know the final radius is 3.5 m, which is the radius of the merry-go-round.
Using these values, we can calculate the moment of inertia at the edge (I_final):
I_final = m * r^2
= 21 kg * (3.5 m)^2
= 308.7 kg⋅m^2
Now we have all the required values to calculate the final angular velocity (ω_final):
I_initial * ω_initial = I_final * ω_final
Solving for ω_final:
ω_final = (I_initial * ω_initial) / I_final
= (500 kg⋅m^2 * 0.20 s^(-1)) / 308.7 kg⋅m^2
≈ 0.323 s^(-1)
Therefore, the rotational speed of the merry-go-round when the child gets to the edge is approximately 0.323 s^(-1).
To determine the rotational speed of the merry-go-round when the child gets to the edge, we can use the principle of conservation of angular momentum. Angular momentum is the product of rotational inertia and rotational speed, and it remains constant when no external torques act on a rotating object.
1. First, find the initial angular momentum of the system when the child is at the center of the merry-go-round.
Angular momentum (L1) = rotational inertia (I) * rotational speed (ω)
L1 = 500 kg⋅m^2 * 0.20 s^(-1)
2. Next, find the final angular momentum of the system when the child reaches the edge of the merry-go-round. Since the child walks out to the edge, the moment of inertia of the system changes. We need to consider the conservation of angular momentum as follows:
I1 * ω1 = I2 * ω2
I1: initial moment of inertia (at the center)
I2: final moment of inertia (at the edge)
ω1: initial angular speed (0.20 s^(-1))
ω2: final angular speed (what we want to find)
3. Determine the final moment of inertia (I2) when the child reaches the edge. The moment of inertia can be calculated using the following formula:
I = m * r^2
m: mass of the child (21 kg)
r: radius of the merry-go-round (3.5 m)
I2 = m * r^2
4. Substitute the known values into the conservation of angular momentum equation from step 2 and solve for ω2:
(I1 * ω1) = (I2 * ω2)
(500 kg⋅m^2 * 0.20 s^(-1)) = (m * r^2 * ω2)
Once you evaluate the expression, you'll find the rotational speed (ω2) when the child reaches the edge.
This is a conservation of angular momentum problem.
Now, the formula for angular momentum is:
L = Iw
So basically, L before = L after:
I1w1 = I2w2
The trick to solving these is to figure out what the change in moment of inertia is, and then apply the concept of conservation of angular momentum to it.
With a person on a merry-go-round, the moment of inertia would be the moment of inertia of the person plus the moment of inertia of the merry-go-round.
In this case the person starts at the center of the rotating merry-go-round, and so I think we are to say their moment of inertia is zero or negligible (it would be small), but then they move to the outside edge of the merry-go-round, and in the after picture, have a significant moment of inertia. The moment of inertia of the merry-go-round is the same before and after and given as 670 kgm2