Three children are riding on the edge of a merry-go-round that has a mass 100.0 kilograms, a 1.60-meter radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kilograms. If the child who has a mass of 28.0 kilograms moves to the center of the merry-go-round, what is the new angular velocity in revolutions per minute?
Let’s start by highlighting some of the vital information given to us in this problem statement. Were told that the merry-go-round itself has a mass of 100.0 kilograms and that its radius is 1.60 meters and that it spins at 20 revolutions per minute. The masses of the children on the merry-go-round are 22.0, 28.0, and 33.0 kilograms.
We’re asked to solve for angular velocity after one of the children has moved on the merry-go-round and we’ll call that angular velocity 𝜔 sub 𝑓. Let’s draw a diagram of this scenario. In this diagram, we see an aerial view of the merry-go-round, and the three dots on the outer edge of the merry-go-round represent the three children riding. We named the children 𝑐 one for child one, 𝑐 two for child two, and 𝑐 three for child three.
The mass and radius of the merry-go-round and the masses of the three children are all given. The merry-go-round rotates with an initial angular speed, given as 20.0 rpm. We’re told that child number then moves from the outer edge to the center of the merry-go-round.
As a consequence of that movement the rotational rate of the merry-go-round changes to 𝜔𝑓, and it’s that variable that we’re going to solve for. To begin, let’s recall the equation that tells us the angular momentum of this merry-go-round as it spins; the magnitude of an object’s angular momentum is equal to its moment of inertia 𝐼 times its rotational speed 𝜔.
We refer to angular momentum because we know that this quantity is conserved. So if we write the angular momentum of our merry-go-round under its initial conditions, 𝐿 sub 𝑖 equals 𝐼 sub 𝑖 times 𝜔 sub 𝑖, then by the conservation of angular momentum we know that this quality is equal to 𝐿 sub 𝑓, which equals 𝐼 sub 𝑓 times 𝜔 sub 𝑓. In each of these equations the 𝑖 subscript represents the initial state and the 𝑓 subscript represents the final state.
We see that 𝜔 sub 𝑓, the variable we’re working to solve for, is in our last equation. So let’s rearrange that equation so that it reads 𝜔 sub 𝑓 equals something else. To do that, let’s divide both sides by the final moment of inertia of the system, 𝐼 sub 𝑓.
When we do that division, 𝐼 sub 𝑓 cancels out of the right side of our equation. What remains is a relationship that reads 𝜔 sub 𝑓 equals 𝜔 sub 𝑖 times 𝐼 sub 𝑖 divided by 𝐼 sub 𝑓 In other words, the final angular speed equals the initial angular speed, 𝜔 sub 𝑖, times the ratio of the initial moment of inertia to the final moment of inertia of our system.
As our next step let’s, calculate the initial and final moments of inertia of our system. We know that our system consists of a disk rotating about a line through its center — that’s the merry-go-round — and three masses, each representing the children a distance 𝑟 away from the rotational center. Using a table for the moments of inertia of common shapes, we see that the moment of inertia of our disk 𝐼 sub disk equals one-half the mass of the desk multiplied by its radius squared. And when we look up the moment of inertia of a point mass rotating a distance 𝑟 away from a center of rotation, the moment of inertia of that type of object is equal to the mass of the object times its distance from the center of rotation squared; we’ll call that 𝐼 sub 𝑝 the moment of inertia of a point mass.
Using these two relationships, we can write out the initial and final moment of inertia for our system. That includes the merry-go-round and the three children. Our system’s initial moment of inertia equals the moment of inertia of the merry-go-round one-half 𝑚𝑟 squared plus the mass of each child multiplied by 𝑟 squared.
Now let’s consider 𝐼 sub 𝑓. The final moment of inertia of our system. What has changed? Child number two has moved from the outer edge of the merry-go-round to the center, so 𝐼 sub 𝑓 is written the same way as 𝐼 sub 𝑖 except that because 𝑚𝑐 two now has a radius value of because the child stands in the centre of the merry-go-round, we cancel out or drop that term from the equation. Not because the mass of the child has decreased, but because the child’s radius has gone to zero.
Let’s now enter in these values of 𝐼 sub 𝑖 and 𝐼 sub 𝑓 into our equation for the conservation of angular momentum. So the final angular speed of the merry-go-round equals the initial angular speed multiplied by one-half the mass of the merry-go-round plus the sum of the masses of the three children times 𝑟 squared divided by half the mass of the merry-go-round plus the sum of the mass of child one and child three again multiplied by 𝑟 squared.
With the equation in this form, we see that we can cancel out the 𝑟 squared term, which means that our solution is independent of the value of the radius of the merry-go-round. Now let’s plug in for the values of the masses and the initial angular speed we’ve been given: 𝜔 sub 𝑖 is 20.0 rpm; 𝑚 is 100.0 kilograms; 𝑚𝑐 one is 22.0 kilograms; 𝑚𝑐 two is 28.0 kilograms; and 𝑚𝑐 three is 33.0 kilograms.
Now that we’ve plugged in for our values of 𝜔 sub 𝑖 and the masses in our system, we’re ready to calculate out 𝜔 sub 𝑓, the final angular speed of our system, in units of rpm. When we enter these values into our calculator, we find that that final angular speed is 25.3 rpm or revolutions per minute. So as child number two moves to the centre of the merry-go-round, the angular speed of the merry-go-round increases.