# 5.4: Moment of Inertia

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Suppose we have a mass m at the end of a massless stick of length \(r\), rotating around the other end of the stick. If we want to increase the rotation rate, we need to apply a tangential acceleration at

\[\boldsymbol{a}_{\mathrm{t}}=r \boldsymbol{\alpha} \nonumber\]

for which by Newton’s second law of motion we need a force

\[\boldsymbol{F}=m \boldsymbol{a}_{\mathrm{t}}=\operatorname{mr} \boldsymbol{\alpha}. \nonumber\]

This force in turn generates a torque of magnitude

\[\tau=r \cdot F=m r^{2} \alpha.\]

The last equality is reminiscent of Newton’s second law of motion, but with force replaced by torque, acceleration by angular acceleration, and mass by the quantity \(m r^2\). In analog with mass representing the inertia of a body undergoing linear acceleration, we’ll identify this quantity as the inertia of a body undergoing rotational acceleration, which we’ll call the moment of inertia and denote by \(I\):

\[\boldsymbol{\tau}=I \boldsymbol{\alpha} \label{torque}\]

Equation \ref{torque} is the rotational analog of Newton’s second law of motion. By extending our previous example, we can find the moment of inertia of an arbitrary collection of particles of masses \(m_\alpha\) and distances to the rotation axis \(r_\alpha\) (where \(\alpha\) runs over all particles), and write:

\[I=\sum_{\alpha} m_{\alpha} r_{\alpha}^{2}\]

which like the center of mass in Section 4.1 easily generalizes to continuous objects as^{2}

\[I=\int_{V}(\boldsymbol{r} \cdot \boldsymbol{r}) \rho \mathrm{d} V=\int_{V} \rho r^{2} \mathrm{d} V \label{integral}\]

Note that it matters where we choose the rotation axis. For example, the moment of inertia of a rod of length L and mass m around an axis through its center perpendicular to the rod is \(\frac{1}{12}mL^2\), whereas the moment of inertia around an axis perpendicular to the rod but located at one of its ends is \(\frac{1}{3}mL^2\). Also, moments of inertia are different for hollow and solid objects - a hollow sphere of mass m and radius R has \(\frac{2}{3}mR^2\) whereas a solid sphere has \(\frac{2}{5}mR^2\), and for hollow and solid cylinders (or hoops and disks) around the long axis through the center we find \(mR^2\) and \(\frac{1}{4}mR^2\) respectively. These and some other examples are listed in Table 5.1. Below we’ll relate the moment of inertia to the kinetic energy of a moving-and-rolling object, but first we present two handy theorems that will help in calculating them.

Object | Rotation Axis | Moment of Inertia |
---|---|---|

Stick | Center, perpendicular to stick | \(\frac{1}{12} ML^2\) |

Stick | End, perpendicular to stick | \(\frac{1}{3} ML^2\) |

Cylinder, hollow | Center, parallel to axis | \(MR^2\) |

Cylinder, solid | Center, parallel to axis | \(\frac{1}{2} MR^2\) |

Sphere, hollow | Any axis through center | \(\frac{2}{3} MR^2\) |

Sphere, solid | Any axis through center | \(\frac{2}{5} MR^2\) |

Planar object, size \(a \times b\) | Axis through center, in plane, parallel to side with length a | \(\frac{1}{12} Mb^2\) |

Planar object, size \(a \times b\) | Axis through center, perpendicular to plane | \(\frac{1}{12} M(a^2+b^2)\) |

^{2 }Like the one- and two-dimensional analogs of the center of mass of a continuous object (4.1.3), there are one- and two-dimensional analogs of \ref{integral}, which you get by replacing \(\rho\) with \(\lambda\) or \(\sigma\) and dV by dx or dA, respectively.