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TAM 212: Introductory Dynamics

Moments of inertia #rem

The moment of inertia of a body, written IP,ˆa, is measured about a rotation axis through point P in direction ˆa. The moment of inertia expresses how hard it is to produce an angular acceleration of the body about this axis. That is, a body with high moment of inertia resists angular acceleration, so if it is not rotating then it is hard to start a rotation, while if it is already rotating then it is hard to stop.

The moment of inertia plays the same role for rotational motion as the mass does for translational motion (a high-mass body resists is hard to start moving and hard to stop again).

Moment of inertia about axis ˆa through point P.#rem‑ei

IP,ˆa=

The distance r is the perpendicular distance to dV from the axis through P in direction \hat{a}.

Warning: Mass moments of inertia are different to area moments of inertia.#rem‑wm

Observe that the moment of inertia is proportional to the mass, so that doubling the mass of an object will also double its moment of inertia. In addition, the moment of inertia is proportional to the square of the size of the object, so that doubling every dimension of an object (height, width, etc) will cause it to have four times the moment of inertia.

Moments of inertia about coordinate axes through point P.#rem‑ec

\begin{aligned} I_{P,x} &= I_{P,xx} = I_{P,\hat\imath} = \iiint_\mathcal{B} \rho (y^2 + z^2) \,dx\,dy\,dz \\ I_{P,y} &= I_{P,yy} = I_{P,\hat\jmath} = \iiint_\mathcal{B} \rho (z^2 + x^2) \,dx\,dy\,dz \\ I_{P,z} &= I_{P,zz} = I_{P,\hat{k}} = \iiint_\mathcal{B} \rho (x^2 + y^2) \,dx\,dy\,dz \end{aligned}

The coordinates (x,y,z) in the body are measured from point P.

Derivation

Example Problem: Moment of inertia of a square plate.#rem‑xs

Did you know?#rem‑i3

We are always considering the moment of inertia to be a scalar value I, which is valid for rotation about a fixed axis. For more complicated dynamics with tumbling motion about multiple axes simultaneously, it is necessary to consider the full 3 × 3 moment of inertia matrix: I = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{bmatrix} The three scalar moments of inertia from #rem-ec appear on the diagonal. The off-diagonal terms are called the products of inertia and are given by I_{xy} = -\iiint_{\mathcal{B}} \rho xy \,dx\,dy\,dz, and similarly for the other terms. The angular momentum of a rigid body is given by \vec{H} = I \vec\omega, which is the matrix product of the moment of inertia matrix with the angular velocity vector. This is important in advanced dynamics applications such as unbalanced rotating shafts and the precession of gyroscopes.

Transforming and combining moments of inertia #rem‑st

Parallel axis theorem.#rem‑el

I_{P,\hat{a}} = I_{C,\hat{a}} + m \, d^2

Here d is the perpendicular distance between the axes through P and C in direction \hat{a}, so d = \| \operatorname*{Comp}(\vec{r}_{CP}, \hat{a})\|.

Derivation

Warning: Parallel axis theorem must start from center of mass C.#rem‑wl

Warning: Parallel axis theorem d is perpendicular distance, not r_{CP}.#rem‑wp

Example Problem: Moment of inertia of a square plate about a corner.#rem‑xc

One consequence of the parallel axis theorem is that the moment of inertia can only increase as we move the rotation point P away from the center of mass C. This means that the point with the lowest moment of inertia is always the center of mass itself.

A second consequence of the parallel axis theorem is that moving the point P along the direction \hat{a} doesn't change the moment of inertia, because the axis of rotation is not changing as the point moves along the axis itself.

Adding moments of inertia.#rem‑ea

I^{\mathcal{B}}_{P,\hat{a}} = I^{\mathcal{B}_1}_{P,\hat{a}} + I^{\mathcal{B}_2}_{P,\hat{a}}

The whole body \mathcal{B} is composed of sub-bodies \mathcal{B}_1 and \mathcal{B}_2.

Derivation

Warning: To add moments of inertia they must be about the same axis.#rem‑wa

Example Problem: Moment of inertia of an L-shaped plate.#rem‑xl

Basic shapes #rem‑sb

The moments of inertia listed below are all computed directly from the integrals #rem-ec.

Point mass: moments of inertia#rem‑ep

\begin{aligned} I_{P,z} &= m r^2 \end{aligned}

Derivation

Rectangular prism: moments of inertia#rem‑er

\begin{aligned} I_{C,x} &= \frac{1}{12} m ({\ell_y}^2 + {\ell_z}^2) \\ I_{C,y} &= \frac{1}{12} m ({\ell_z}^2 + {\ell_x}^2) \\ I_{C,z} &= \frac{1}{12} m ({\ell_x}^2 + {\ell_y}^2) \end{aligned}

Derivation

Cylindrical thick shell: moments of inertia#rem‑ey

\begin{aligned} I_{C,x} &= I_{C,y} = \frac{1}{12} m (3({r_1}^2 + {r_2}^2) + \ell^2) \\ I_{C,z} &= \frac{1}{2} m ({r_1}^2 + {r_2}^2) \end{aligned}

Derivation

Spherical thick shell: moments of inertia#rem‑es

\begin{aligned} I_C &= \frac{2}{5} m \left(\frac{{r_2}^5 - {r_1}^5}{{r_2}^3 - {r_1}^3}\right) \end{aligned}

All axes through C have the same moment of inertia.

Derivation

Simplified shapes #rem‑ss

The moments of inertia listed below are all special cases of the basic shapes given in Section #rem-sb. Other special cases can be easily obtained by similar methods.

Rod: moments of inertia#rem‑eo

\begin{aligned} I_{C,z} &= \frac{1}{12} m \ell^2 \\ I_{P,z} &= \frac{1}{3} m \ell^2 \\ I_{C,x} &= I_{P,x} = 0 \end{aligned}

Derivation

Solid cylinder or disk: moments of inertia#rem‑ek

\begin{aligned} I_{C,x} &= I_{C,y} = \frac{1}{12} m (3r^2 + \ell^2) \\ I_{C,z} &= \frac{1}{2} m r^2 \end{aligned}

Derivation

Hollow cylinder or hoop: moments of inertia#rem‑eh

\begin{aligned} I_{C,x} &= I_{C,y} = \frac{1}{12} m (6 r^2 + \ell^2) \\ I_{C,z} &= m r^2 \end{aligned}

Derivation

Solid ball: moments of inertia#rem‑eb

\begin{aligned} I_C &= \frac{2}{5} m r^2 \end{aligned}

All axes through C have the same moment of inertia.

Derivation

Hollow sphere: moments of inertia#rem‑ew

\begin{aligned} I_C &= \frac{2}{3} m r^2 \end{aligned}

All axes through C have the same moment of inertia.

Derivation