Rotational Inertia and Moment of Inertia

Before we can consider the rotation of anything other than a point mass like the one in Figure 69.2, 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

[latex]I[/latex] of an object to be the sum of

[latex]{\text{mr}}^{2}[/latex] for all the point masses of which it is composed. That is,

[latex]I=\sum {\text{mr}}^{2}[/latex]. Here

[latex]I[/latex] is analogous to

[latex]m[/latex] in translational motion. Because of the distance

[latex]r[/latex], the moment of inertia for any object depends on the chosen axis. Actually, calculating

[latex]I[/latex] 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

[latex]{\text{MR}}^{2}[/latex], where

[latex]M[/latex] is its total mass and [latex]R[/latex] its radius. (We use [latex]M[/latex] and [latex]R[/latex] for an entire object to distinguish them from [latex]m[/latex] and [latex]r[/latex] for point masses.) In all other cases, we must consult Figure 69.3 (note that the table is piece of artwork that has shapes as well as formulae) for formulas for [latex]I[/latex] that have been derived from integration over the continuous body. Note that [latex]I[/latex] has units of mass multiplied by distance squared ([latex]\text{kg}\cdot {\text{m}}^{2}[/latex]), as we might expect from its definition.

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

or

where net [latex]\tau[/latex] 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 [latex]\tau =\mathrm{I\alpha }\mathrm{, }\alpha =\frac{\text{net τ}}{I}[/latex] 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.

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.

69.1 Calculating the Effect of Mass Distribution on a Merry-Go-Round

Consider the father pushing a playground merry-go-round in Figure 69.4. 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.

Strategy

Angular acceleration is given directly by the expression [latex]\alpha =\frac{\text{net τ}}{I}[/latex]
:

[latex]\alpha =\frac{\tau }{I}.[/latex]

To solve for [latex]\alpha[/latex], we must first calculate the torque [latex]\tau[/latex] (which is the same in both cases) and moment of inertia [latex]I[/latex] (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

[latex]\mathrm{\tau }=\mathrm{rF}\phantom{\rule{0.25em}{0ex}}\text{sin θ}=\left(\text{1.50 m}\right)\left(\text{250 N}\right)=\text{375 N}\cdot \text{m.}[/latex]

Solution for (a)

The moment of inertia of a solid disk about this axis is given in Figure 69.3 to be

[latex]\frac{1}{2}{\text{MR}}^{2}\text{,}[/latex]

where [latex]M=\text{50.0 kg}[/latex] and [latex]R=\text{1.50 m}[/latex], so that

[latex]I=(0\text{.}500)(\text{50.0 kg})(\text{1.50 m})^{2}=\text{56.25 kg}\cdot {\text{m}}^{2}\text{.}[/latex]

Now, after we substitute the known values, we find the angular acceleration to be

[latex]\alpha =\frac{\tau }{I}=\frac{\text{375 N}\phantom{\rule{0.25em}{0ex}}\text{⋅}\phantom{\rule{0.25em}{0ex}}\text{m}}{\text{56.25 kg}\cdot {\text{m}}^{2}}=\text{6.67}\frac{\text{rad}}{{\text{s}}^{2}}\text{.}[/latex]

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 [latex]I[/latex], we first find the child’s moment of inertia [latex]{I}_{\text{c}}[/latex] by considering the child to be equivalent to a point mass at a distance of 1.25 m from the axis. Then,

[latex]{I}_{\text{c}}={\mathrm{MR}}^{2}=\left(\text{18.0 kg}\right)\left(\text{1.25 m}\right)^{2}=\text{28.13 kg}\cdot {\text{m}}^{2}\text{.}[/latex]

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 [latex]I[/latex]:

[latex]I=\text{28.13 kg}\cdot {\text{m}}^{2}+\text{56.25 kg}\cdot {\text{m}}^{2}=\text{84.38 kg}\cdot {\text{m}}^{2}.[/latex]

Substituting known values into the equation for [latex]\alpha[/latex] gives

[latex]\alpha =\frac{\text{τ}}{I}=\frac{\text{375 N}\cdot \text{m}}{\text{84.38 kg}\cdot {\text{m}}^{2}}=\text{4.44}\frac{\text{rad}}{{\text{s}}^{2}}\text{.}[/latex]

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.