A 1.80 m radius playground merry-go-round has a mass of 120 kg and is rotating with an angular velocity of 0.350 rev/s. What is its angular velocity after a 28.0 kg child gets onto it by grabbing its outer edge? The child is initially at rest.
The moment of inertia of the empty merry-go-round, assuming it to be a uniform circular platform, is
I = (1/2)MR^2 = 194.4 kg*m^2
Angular momentum is conserved when the child gets on, but the new moment of inertia becomes (with the boy's contribution added),
I' = I + mR^2 = 194.4 + 90.7
= 285.1 kg*m^2
To conserve angular momentum, the rotation speed must drop by a factor 194.4/285.1 = 0.6819
That makes the new rotation rate 0.239 rev/s
Well, it seems we have a child joining the merry-go-round party! Let's calculate the angular velocity after the addition of the little munchkin.
Before the child hops on, let's find the initial moment of inertia of the merry-go-round. We can use the formula:
I = (1/2) * m * r^2
Where I represents the moment of inertia, m is the mass, and r is the radius.
Plugging in the values, we get:
I = (1/2) * 120 kg * (1.80 m)^2
Now, let's calculate the final moment of inertia after the child gets on board. Since the child grabs the outer edge, their mass won't affect the radius. So, we've got:
I_final = I + m_child * r^2
where m_child is the mass of the child, and r is still the radius.
Substituting the given values of m_child = 28.0 kg and r = 1.80 m:
I_final = (1/2) * 120 kg * (1.80 m)^2 + 28.0 kg * (1.80 m)^2
Now that we have the final moment of inertia, we can calculate the final angular velocity using the conservation of angular momentum equation:
I * w_initial = I_final * w_final
But since the child is initially at rest, the initial angular velocity is 0. Hence:
0 = I_final * w_final
Solving for w_final, we have:
w_final = 0 / I_final
So, my dear friend, the angular velocity of the merry-go-round after the child joins is a whopping zero! That's right, the child's inertia cancels out the initial angular velocity, leaving the merry-go-round at a standstill. Quite a surprising turn of events, isn't it?
To calculate the final angular velocity after the child gets onto the merry-go-round, you can use the principle of conservation of angular momentum.
The initial angular momentum of the system is given by:
L_initial = I_initial * ω_initial
Where I_initial is the moment of inertia of the merry-go-round, and ω_initial is the initial angular velocity.
The final angular momentum of the system, after the child gets onto the merry-go-round, is given by:
L_final = I_final * ω_final
Where I_final is the moment of inertia of the system with the child, and ω_final is the final angular velocity.
Since there is no external torque acting on the system, the initial angular momentum should be equal to the final angular momentum:
L_initial = L_final
We can write the moment of inertia of the merry-go-round as:
I_initial = 0.5 * m * r^2
Where m is the mass of the merry-go-round, and r is its radius.
The moment of inertia of the system with the child can be expressed as:
I_final = I_initial + m_child * r^2
Where m_child is the mass of the child.
Now we can substitute these values into the equation for conservation of angular momentum and solve for ω_final.
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
0.5 * m * r^2 * ω_initial = (0.5 * m * r^2 + m_child * r^2) * ω_final
Simplifying the equation:
0.5 * 120 kg * (1.80 m)^2 * 0.350 rev/s = (0.5 * 120 kg * (1.80 m)^2 + 28.0 kg * (1.80 m)^2) * ω_final
Now, let's calculate the final angular velocity:
0.5 * 120 kg * (1.80 m)^2 * 0.350 rev/s = (0.5 * 120 kg * (1.80 m)^2 + 28.0 kg * (1.80 m)^2) * ω_final
0.5 * 120 kg * 1.80 m^2 * 0.350 rev/s = (0.5 * 120 kg * 1.80 m^2 + 28.0 kg * 1.80 m^2) * ω_final
37.80 kg·m^2·rev/s = (194.40 kg·m^2 + 50.40 kg·m^2) * ω_final
37.80 kg·m^2·rev/s = 244.80 kg·m^2 * ω_final
ω_final = 37.80 kg·m^2·rev/s / 244.80 kg·m^2
ω_final ≈ 0.1542 rev/s
Therefore, the angular velocity of the merry-go-round after the child gets onto it is approximately 0.1542 rev/s.
To solve this problem, we can use the principle of conservation of angular momentum. The initial angular momentum of the merry-go-round is given by:
Angular momentum (initial) = Moment of inertia × Angular velocity (initial)
The moment of inertia of a disk is given by the formula:
Moment of inertia = (1/2) × mass × radius^2
Using the given information, we can calculate the initial angular momentum.
Given:
Radius (r) = 1.80 m
Mass of merry-go-round (m1) = 120 kg
Angular velocity of merry-go-round (ω1) = 0.350 rev/s
Calculations:
Moment of inertia (I) = (1/2) × m1 × r^2
= (1/2) × 120 kg × (1.80 m)^2
Angular momentum (initial) = I × ω1
Now, we can consider the angular momentum after the child gets onto the merry-go-round. Because the child is initially at rest, the total angular momentum after the child gets on is the sum of the initial angular momentum of the merry-go-round and the angular momentum of the child.
The angular momentum of the child is given by:
Angular momentum (child) = Moment of inertia × Angular velocity (child)
Since the child grabs the outer edge, the child's moment of inertia can be approximated as the moment of inertia of a point mass rotating at the edge of the merry-go-round, which is defined as:
Moment of inertia (child) = mass × radius^2
Given:
Mass of child (m2) = 28.0 kg
Calculating the moment of inertia of the child:
Moment of inertia (child) = m2 × r^2
Now, we can calculate the total angular momentum after the child gets on the merry-go-round:
Angular momentum (total) = Angular momentum (initial) + Angular momentum (child)
Finally, we can use the total angular momentum to determine the angular velocity of the merry-go-round after the child gets on by rearranging the formula for angular momentum:
Angular velocity (total) = Angular momentum (total) / Moment of inertia
Substituting the calculated values, we can find the angular velocity of the merry-go-round after the child gets on.