A 1.80 m radius playground merry-go-round has a mass of 120 kg and is rotating with an angular velocity of 0.600 rev/s. What is its angular velocity after a 27.0 kg child gets onto it by grabbing its outer edge? The child is initially at rest.
Well, well, well, we've got ourselves a merry-go-round situation here! So, let's see what happens when this 27.0 kg child decides to hop on for a wild ride.
First off, we need to understand the conservation of angular momentum, my friend. Before the child jumps on, the merry-go-round has its own angular momentum, given by the following equation:
L1 = I1 * ω1
Where L1 is the initial angular momentum, I1 is the moment of inertia of the merry-go-round, and ω1 is its initial angular velocity.
Now, when the child jumps on, we still have the conservation of angular momentum at play. So, we can use the equation:
L1 = (I1 + I2) * ω2
Where I2 is the moment of inertia of the child (considering the child as a point mass grabbing the outer edge of the merry-go-round), and ω2 is the angular velocity we want to find.
Now, let's substitute the values we have:
L1 = (I1 + I2) * ω2
I1 * ω1 = (I1 + I2) * ω2
Since the child is initially at rest, its initial angular velocity is zero (ω2 = 0). Therefore, the equation becomes:
I1 * ω1 = I1 * ω2
Okay, okay, pay attention, because here comes the fun part! The moment of inertia of the merry-go-round is given by:
I1 = 0.5 * m1 * r1^2
Where m1 is the mass of the merry-go-round and r1 is its radius. Similarly, the moment of inertia of the child (considering them as a point mass) is given by:
I2 = m2 * r2^2
Where m2 is the mass of the child and r2 is the distance of the child from the axis of rotation (which is the radius of the merry-go-round, given that they're grabbing its outer edge).
Now, let's plug in the values we have and solve for ω2:
0.5 * m1 * r1^2 * ω1 = 0.5 * m1 * r1^2 * ω2 + m2 * r2^2 * ω2
My humorous friend, I hope you're appreciating the beauty of equations as much as I am right now.
Anyway, now we can simplify this equation:
0.5 * m1 * r1^2 * ω1 = (0.5 * m1 * r1^2 + m2 * r2^2) * ω2
Divide both sides of the equation by (0.5 * m1 * r1^2 + m2 * r2^2):
ω2 = (0.5 * m1 * r1^2 * ω1) / (0.5 * m1 * r1^2 + m2 * r2^2)
Now, just plug in the given values and do some calculations, and voila! You'll find the angular velocity after the child gets on the merry-go-round.
But remember, my friend, always approach merry-go-rounds with caution. They might spin you around, but they can't tickle your funny bone like I can.
To find the angular velocity of the merry-go-round after the child gets onto it, we can use the principle of conservation of angular momentum.
The formula for angular momentum is given as:
L = Iω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Initially, the merry-go-round is rotating. The moment of inertia (I1) of the merry-go-round itself can be calculated using the formula:
I1 = (1/2)mr^2
Where m is the mass of the merry-go-round (120 kg) and r is the radius (1.80 m). Substituting the values into the formula, we get:
I1 = (1/2)(120 kg)(1.80 m)^2
I1 = 194.4 kg·m²
The moment of inertia (I2) of the merry-go-round after the child gets on can be calculated as follows:
I2 = I1 + m2r^2
Where m2 is the mass of the child (27.0 kg). Substituting the values into the formula, we get:
I2 = 194.4 kg·m² + (27.0 kg)(1.80 m)^2
I2 = 194.4 kg·m² + (27.0 kg)(3.24 m²)
I2 = 194.4 kg·m² + 87.48 kg·m²
I2 = 281.88 kg·m²
Now, we can use the conservation of angular momentum principle to find the final angular velocity (ω2). The initial angular momentum (L1) is equal to the final angular momentum (L2):
L1 = L2
I1ω1 = I2ω2 (where ω1 is the initial angular velocity)
Substituting the values, we get:
(194.4 kg·m²)(0.600 rev/s) = (281.88 kg·m²)ω2
Simplifying the equation:
116.64 kg·m² rev/s = 281.88 kg·m² ω2
Dividing both sides by 281.88 kg·m², we get:
ω2 = (116.64 kg·m² rev/s) / (281.88 kg·m²)
ω2 ≈ 0.414 rev/s
Therefore, after the child gets onto the merry-go-round, the angular velocity of the system will be approximately 0.414 rev/s.
To find the angular velocity after the child gets onto the merry-go-round, we need to apply the law of conservation of angular momentum. Angular momentum is given by the equation:
L = I * ω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
To calculate the angular velocity of the merry-go-round after the child gets onto it, we need to consider the initial and final states.
Initial state:
The moment of inertia (I) of the merry-go-round without the child can be calculated using the formula for the moment of inertia of a disk:
I_initial = (1/2) * m * r^2
where m is the mass of the merry-go-round and r is the radius.
I_initial = (1/2) * 120 kg * (1.80 m)^2
I_initial = 194.4 kg·m^2
The initial angular momentum (L_initial) can be calculated by multiplying the moment of inertia (I_initial) by the initial angular velocity (ω_initial):
L_initial = I_initial * ω_initial
Given that the initial angular velocity (ω_initial) is 0.600 rev/s, we need to convert it to rad/s:
ω_initial = 0.600 rev/s * (2π rad/rev)
ω_initial = 3.768 rad/s
L_initial = 194.4 kg·m^2 * 3.768 rad/s
L_initial = 731.52 kg·m^2/s
Final state:
When the child grabs the outer edge, he will start rotating together with the merry-go-round. The moment of inertia of the system (merry-go-round + child) can be calculated by adding the moment of inertia of the merry-go-round (I_initial) and the moment of inertia of the child (I_child).
The moment of inertia of a point mass rotating about a perpendicular axis passing through its mass is given by:
I_child = m_child * r_child^2
where m_child is the mass of the child and r_child is the distance from his center of mass to the axis of rotation.
For the child, we assume that his center of mass is at the outer edge of the merry-go-round, so r_child equals the radius of the merry-go-round.
I_child = 27.0 kg * (1.80 m)^2
I_child = 87.48 kg·m^2
The angular momentum of the final state (L_final) can thus be calculated using the moment of inertia of the system (I_system) and the final angular velocity (ω_final):
L_final = I_system * ω_final
Since the child starts from rest, we assume there is no initial angular momentum for the child.
So, the angular momentum of the system (L_system) is equal to the initial angular momentum (L_initial) plus the angular momentum of the child (L_child), which is zero:
L_system = L_initial + L_child
L_system = L_initial
Therefore:
L_initial = I_system * ω_final
Now we can solve for ω_final:
ω_final = L_initial / I_system
Substituting the values we have:
ω_final = 731.52 kg·m^2/s / (194.4 kg·m^2 + 87.48 kg·m^2)
ω_final = 731.52 kg·m^2/s / 281.88 kg·m^2
ω_final ≈ 2.60 rad/s
Therefore, the angular velocity of the merry-go-round after the child gets onto it is approximately 2.60 radians per second.