Lesson 4 · 35 min
Composite Bodies
A pendulum is a rod and a bob; a flywheel is a disk with a bore and a ring of holes. Split a part into standard shapes, move each one to the common axis with the parallel-axis theorem, and add, subtracting the holes.
Learning objectives
- Split a body into standard shapes and find the moment of inertia of each about a common axis.
- Organize the calculation in a table: \(m_i\), \(d_i\), \(\bar I_i\), \(m_i d_i^2\).
- Treat holes and cut-outs as parts with negative mass.
- Find the composite center of mass and the composite moment of inertia \(I_G\).
Add the parts
The moment of inertia is an integral over the body, and an integral over a whole is the sum of the integrals over its parts. So, about any one axis,
Composite body about a common axis
\[ I = \sum_i \left(\bar I_i + m_i d_i^2\right), \qquad m = \sum_i m_i \]\(\bar I_i\): moment of inertia of part \(i\) about its own centroidal axis parallel to the common axis (Lesson 2 table). \(d_i\): distance from that centroidal axis to the common axis. A hole is a part with negative \(m_i\), and its whole \(\bar I_i + m_i d_i^2\) is subtracted.
Every term must be about the same axis. The parallel-axis theorem moves each part's table value there. When the parts are not uniform, or the densities differ, compute each mass from its own density and volume.
Example 4.1 — A rod-and-disk pendulum
A pendulum is a uniform \(2\ \text{kg}\) slender rod, \(0.8\ \text{m}\) long, pinned at its top end \(O\), with a thin \(5\ \text{kg}\) disk of radius \(0.15\ \text{m}\) welded to its lower end so that the disk's center is \(0.95\ \text{m}\) below \(O\). The pendulum swings in the plane of the disk. Find \(I_O\), the center of mass, \(I_G\) and the period of small oscillations.
Show solution
The axis at \(O\) is perpendicular to the plane of the swing. Measure distances down from \(O\):
| Part | \(m_i\) | \(d_i\) | \(\bar I_i\) | \(m_i d_i^2\) | \(I_{O,i}\) |
|---|---|---|---|---|---|
| Rod | 2 | 0.40 | \(\tfrac1{12}(2)(0.8)^2 = 0.1067\) | 0.3200 | 0.4267 |
| Disk | 5 | 0.95 | \(\tfrac12(5)(0.15)^2 = 0.0563\) | 4.5125 | 4.5688 |
| Pendulum | 7 | 4.9954 |
So \(I_O = 4.995\ \text{kg·m}^2\). The center of mass is \(\bar y = (2 \cdot 0.40 + 5 \cdot 0.95)/7 = 0.7929\ \text{m}\) below \(O\), and
\[ I_G = I_O - m\bar y^2 = 4.995 - 7(0.7929)^2 = 0.5951\ \text{kg·m}^2 \]For small swings of a compound pendulum, \(\tau = 2\pi\sqrt{I_O/(m g \bar y)} = 2\pi\sqrt{4.995/(7 \cdot 9.81 \cdot 0.7929)} = 1.903\ \text{s}\).
Notice how the disk's own \(\bar I\) (0.056) is tiny next to its transfer term \(m d^2\) (4.51): far from the axis, a part behaves almost like a particle.
Holes are negative mass
A plate with a hole is the full plate minus a disk of the same material. The same goes for bores, lightening holes, slots and pockets: compute the missing piece as if it were there, then subtract its mass and its moment of inertia about the common axis, transfer term included.
Example 4.2 — A plate with two holes
A steel plate (\(\rho = 7850\ \text{kg/m}^3\)) is \(400 \times 300\ \text{mm}\) and \(10\ \text{mm}\) thick. Two holes of radius \(50\ \text{mm}\) are centered \(100\ \text{mm}\) either side of the plate's center, on its long centerline. Find the mass and the moment of inertia about the axis through the plate's center, perpendicular to the plate.
Show solution
Masses: full plate \(7850(0.4)(0.3)(0.01) = 9.420\ \text{kg}\); each hole \(7850\pi(0.05)^2(0.01) = 0.6165\ \text{kg}\).
| Part | \(m_i\) | \(d_i\) | \(\bar I_i\) | \(m_i d_i^2\) | \(I_i\) |
|---|---|---|---|---|---|
| Plate | 9.420 | 0 | \(\tfrac1{12}m(a^2 + b^2) = 0.19625\) | 0 | 0.19625 |
| Hole 1 | −0.6165 | 0.10 | \(-\tfrac12 m r^2 = -0.00077\) | −0.00617 | −0.00694 |
| Hole 2 | −0.6165 | 0.10 | \(-0.00077\) | −0.00617 | −0.00694 |
| Plate with holes | 8.187 | 0.18238 |
\(m = 8.187\ \text{kg}\) and \(I = 0.1824\ \text{kg·m}^2\). The holes remove 13% of the mass but only 7% of the moment of inertia, because they sit fairly close to the axis.
The composite center of mass
Often the axis you need is through the composite center of mass \(G\), which is not the center of any single part. Find it first, as in statics,
\[ \bar x = \frac{\sum m_i \bar x_i}{\sum m_i}, \qquad \bar y = \frac{\sum m_i \bar y_i}{\sum m_i}, \]then either transfer every part to \(G\) directly, or compute \(I\) about a convenient point \(O\) and step back once: \(I_G = I_O - m\,d_{OG}^2\).
Example 4.3 — A T-shaped bracket
A T-shaped bracket is made of two uniform slender rods: \(AB\) is \(0.6\ \text{m}\) long with mass \(1.2\ \text{kg}\), and \(CD\) is \(0.4\ \text{m}\) long with mass \(0.8\ \text{kg}\), welded at \(C\) to the midpoint of \(AB\) and perpendicular to it. Find the center of mass and the moment of inertia about the axis through \(G\) perpendicular to the plane of the T.
Show solution
Put the origin at \(C\), with \(x\) along \(AB\) and \(y\) along \(DC\) (so \(D\) is at \(y = -0.4\ \text{m}\)). By symmetry \(\bar x = 0\), and
\[ \bar y = \frac{1.2(0) + 0.8(-0.2)}{2.0} = -0.0800\ \text{m} \]About \(C\) (axis perpendicular to the plane): \(AB\) turns about its middle and \(CD\) about its end:
\[ I_C = \tfrac1{12}(1.2)(0.6)^2 + \tfrac13(0.8)(0.4)^2 = 0.0360 + 0.0427 = 0.0787\ \text{kg·m}^2 \] \[ I_G = I_C - m\,\bar y^2 = 0.07867 - 2.0(0.08)^2 = 0.06587\ \text{kg·m}^2 \]Check your understanding
Key takeaways
- \(I = \sum (\bar I_i + m_i d_i^2)\) about one common axis, with each part moved there by the parallel-axis theorem.
- Holes are parts with negative mass: subtract \(\bar I_i + m_i d_i^2\) in full.
- Find the composite \(G\) from \(\sum m_i \bar x_i / \sum m_i\), then \(I_G = I_O - m\,d_{OG}^2\).
- Far from the axis, a part's transfer term \(m_i d_i^2\) dominates its own \(\bar I_i\).
- Next: Lesson 5 adds products of inertia, the first step into three dimensions.