Rotational Inertia
As you read, pay attention to the diagram of an object rotating on a frictionless table in Figure 10.11. We can see the radius from the center of the table (the pivot point) and the mass at the end of the radius.
When solving dynamics problems, we first need to identify the system and draw a free-body diagram of all the forces acting upon the system. Once the forces acting upon the system are defined, we can use the torque equation and angular acceleration equations to solve the problem: , where
is the moment of inertia,
is torque, and
is the rotational acceleration due to the torque.
Example 10.7 shows how to use these equations to determine the angular acceleration of a person pushing a merry-go-round. Here, the first step is to calculate torque. The next step is to calculate the moment of inertia. Finally, torque and moment of inertia are used to calculate the angular acceleration on the merry-go-round.
Example 10.7 Calculating the Effect of Mass Distribution on a Merry-Go-Round
Example 10.7 Calculating the Effect of Mass Distribution on a Merry-Go-Round
Consider the father pushing a playground merry-go-round in Figure 10.13. He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius.
Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction.
Figure 10.13 A father pushes a playground merry-go-round at its edge and perpendicular to its radius to achieve maximum torque.
Strategy
Angular acceleration is given directly by the expression :
To solve for , we must first calculate the torque
(which is the same in both cases) and moment of inertia
(which is greater in the second case). To find the torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that
Solution for (a)
The moment of inertia of a solid disk about this axis is given in Figure to be
Now, after we substitute the known values, we find the angular acceleration to be
Solution for (b)
We expect the angular acceleration for the system to be less in this part, because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia , we first find the child's moment of inertia
by considering the child to be equivalent to a point mass at a distance of
from the axis. Then,
The total moment of inertia is the sum of moments of inertia of the merry-go-round and the child (about the same axis). To justify this sum to yourself, examine the definition of :
Substituting known values into the equation for gives
Discussion
The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it.
In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case. Summer Olympics, here he comes! Confirmation of these numbers is left as an exercise for the reader.