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:  \tau_{net}=I\alpha , where  I is the moment of inertia,  \tau is torque, and  \alpha 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.

Rotational Inertia and Moment of Inertia

Before we can consider the rotation of anything other than a point mass like the one in Figure 10.11, we must extend the idea of rotational inertia to all types of objects. To expand our concept of rotational inertia, we define the moment of inertia I of an object to be the sum of m r^{2} for all the point masses of which it is composed. That is, I=\sum m r^{2}. Here I is analogous to m in translational motion. Because of the distance r, the moment of inertia for any object depends on the chosen axis.

Actually, calculating I is beyond the scope of this text except for one simple case-that of a hoop, which has all its mass at the same distance from its axis. A hoop's moment of inertia around its axis is therefore M R^{2}, where M is its total mass and R its radius. (We use M and R for an entire object to distinguish them from m and r for point masses).

In all other cases, we must consult Figure 10.12 (note that the table is piece of artwork that has shapes as well as formulae) for formulas for I that have been derived from integration over the continuous body. Note that I has units of mass multiplied by distance squared \left(\mathrm{kg} \cdot \mathrm{m}^{2}\right), as we might expect from its definition.

The general relationship among torque, moment of inertia, and angular acceleration is

 \text{net} \, \tau=I \alpha

or

\alpha=\frac{\text { net } \tau}{I},

where net \tau is the total torque from all forces relative to a chosen axis. For simplicity, we will only consider torques exerted by forces in the plane of the rotation. Such torques are either positive or negative and add like ordinary numbers. The relationship in \tau=I \alpha, \alpha=\frac{\text { net } \tau}{I} is the rotational analog to Newton's second law and is very generally applicable. This equation is actually valid for any torque, applied to any object, relative to any axis.

As we might expect, the larger the torque is, the larger the angular acceleration is. For example, the harder a child pushes on a merry-go-round, the faster it accelerates. Furthermore, the more massive a merry-go-round, the slower it accelerates for the same torque. The basic relationship between moment of inertia and angular acceleration is that the larger the moment of inertia, the smaller is the angular acceleration. But there is an additional twist. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates. For example, it will be much easier to accelerate a merry-go-round full of children if they stand close to its axis than if they all stand at the outer edge. The mass is the same in both cases, but the moment of inertia is much larger when the children are at the edge.

Take-Home Experiment: Problem-Solving for Rotational Dynamics

Cut out a circle that has about a 10 cm radius from stiff cardboard. Near the edge of the circle, write numbers 1 to 12 like hours on a clock face. Position the circle so that it can rotate freely about a horizontal axis through its center, like a wheel. (You could loosely nail the circle to a wall). Hold the circle stationary and with the number 12 positioned at the top, attach a lump of blue putty (sticky material used for fixing posters to walls) at the number 3.

How large does the lump need to be to just rotate the circle? Describe how you can change the moment of inertia of the circle. How does this change affect the amount of blue putty needed at the number 3 to just rotate the circle? Change the circle's moment of inertia and then try rotating the circle by using different amounts of blue putty. Repeat this process several times.

Problem-Solving Strategy for Rotational Dynamics

  1. Examine the situation to determine that torque and mass are involved in the rotation. Draw a careful sketch of the situation.

  2. Determine the system of interest.

  3. Draw a free body diagram. That is, draw and label all external forces acting on the system of interest.

  4. Apply net \tau=I \alpha, \alpha=\frac{\text { net } \tau}{I}, the rotational equivalent of Newton's second law, to solve the problem. Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation.

  5. As always, check the solution to see if it is reasonable.

Making Connections

In statics, the net torque is zero, and there is no angular acceleration. In rotational motion, net torque is the cause of angular acceleration, exactly as in Newton's second law of motion for rotation.

Illustrations of ten different objects accompanied by their rotational inertias.

Figure 10.12 Some rotational inertias.