At the Republic Day parade on Kartavya Path, the drum major of a band twirls a long baton as he marches. He always grips it at the middle and spins it about that point. He never spins it about one end, although that would look grander. Try it with a ruler: about the end it turns sluggishly and pulls at the wrist.
One distance for the whole mass
The table of the last lesson gives \(I\) as a number times \(MR^2\) or \(ML^2\). There is a neat way to read each entry. Imagine taking all the mass of the body and placing it on a thin ring round the same axis. At what radius must the ring be, to have the same moment of inertia as the body? That radius is the radius of gyration \(k\).
The radius of gyration \(k\) of a body of mass \(M\) about an axis is defined by \(I = Mk^2\). Here \(I\) is its moment of inertia about that axis. So \(k = \sqrt{I/M}\). Like \(I\), it depends on the axis.
Fig 5.1 turns the fractions of the table into distances. The ring needs no change: its mass is already at \(R\). The disc behaves as if all its mass were at \(0.71R\), and the solid sphere as if it were at \(0.63R\). The smaller \(k\), the more the body's mass crowds towards the axis. So \(k\) is a single, visible measure of how the mass is spread.
A uniform rod of mass \(2\,\text{kg}\) and length \(1.2\,\text{m}\) turns about an axis through one end, perpendicular to it. Find its radius of gyration.
\(I = \tfrac13 ML^2 = \tfrac13 \times 2 \times 1.44 = 0.96\,\text{kg m}^2\).
\(k = \sqrt{I/M} = \sqrt{0.96/2} = \sqrt{0.48} \approx 0.69\,\text{m}\).
In general \(k = L/\sqrt3 \approx 0.58L\) for a rod about its end. The rod acts as if all its mass sat a little beyond the middle, not at the middle itself, because the far half counts more.
What is the radius of gyration of a uniform disc of radius \(R\) about its central perpendicular axis?
A machine part of mass \(4\,\text{kg}\) has a moment of inertia of \(0.16\,\text{kg m}^2\) about its shaft. Find its radius of gyration about the shaft, in m.
Moving the axis: the parallel-axis theorem
The table gives most bodies about an axis through the centre of mass. Real problems often need another axis, parallel to it: a rod pivoted at its end, a disc turning about a pin on its rim. The parallel-axis theorem makes the step without a new integral.
Let \(I_{\text{cm}}\) be the moment of inertia of a body of mass \(M\) about an axis through its centre of mass. About any parallel axis at a distance \(d\) from it, the moment of inertia is \(I = I_{\text{cm}} + Md^2\).
\[ r^2 = (x - x_0)^2 + (y - y_0)^2 \]
Why. put the origin at the centre of mass, with both axes along z; the new axis passes through \((x_0, y_0)\), and \(d^2 = x_0^2 + y_0^2\).
\[ r^2 = (x^2 + y^2) - 2x_0x - 2y_0y + d^2 \]
Why. expand the squares and collect \(x_0^2 + y_0^2 = d^2\).
\[ I = \int (x^2 + y^2)\,dm - 2x_0\int x\,dm - 2y_0\int y\,dm + d^2\int dm \]
Why. \(I = \int r^2\,dm\); integrate each term over the body.
\[ \int x\,dm = \int y\,dm = 0 \]
Why. these are \(M\) times the coordinates of the centre of mass, which is at the origin.
\[ I = I_{\text{cm}} + Md^2 \]
Read it. the first integral is \(I_{\text{cm}}\), and \(\int dm = M\).
The proof shows where the condition comes from. The two cross terms vanish only because the origin is the centre of mass. About any other starting axis they would not vanish. Step 4 is the whole theorem, and it holds for the centre of mass alone.
Two results follow at once. First, the rod about its end: \(d = L/2\), so \(I = \tfrac{1}{12}ML^2 + M\left(\tfrac{L}{2}\right)^2 = \tfrac13 ML^2\), the value of the last lesson's integral. Second, a disc about an axis perpendicular to it through a point of its rim: \(d = R\), so \(I = \tfrac12 MR^2 + MR^2 = \tfrac32 MR^2\).
Since \(Md^2\) is never negative, the smallest \(I\) of all parallel axes is the one with \(d = 0\). The axis through the centre of mass is the easiest of them to spin about.
Slide the axis in the figure above and read the numbers. At C, \(0.3\,\text{m}\) from the heavy ball, \(I_{\text{cm}} = 3 \times 0.3^2 + 1 \times 0.9^2 = 1.08\,\text{kg m}^2\). At the heavy ball, \(d = 0.3\,\text{m}\) and \(I = 1.08 + 4 \times 0.09 = 1.44\,\text{kg m}^2\). At the middle of the rod, also \(0.3\,\text{m}\) from C, \(I\) is again \(1.44\). At the light ball, \(d = 0.9\,\text{m}\) and \(I = 1.08 + 4 \times 0.81 = 4.32\,\text{kg m}^2\). The least value sits at C, not at the middle of the rod.
Among all the axes parallel to a given direction, which has the least moment of inertia for a given rigid body?
A solid sphere of mass \(5\,\text{kg}\) and radius \(0.2\,\text{m}\) turns about an axis that touches its surface. Find its moment of inertia about this tangent, in kg m².
The classic trap
The theorem connects any axis to the axis through the centre of mass, and to nothing else. Suppose you know \(I_1\) about an axis at distance \(d_1\) from the centre of mass. You want \(I_2\) about a parallel axis at distance \(d_2\). You may not add \(M\) times the square of the distance between the two axes. Instead, step back to the centre of mass and then step out again.
A uniform rod of mass \(2\,\text{kg}\) and length \(1.2\,\text{m}\) has \(I = 0.96\,\text{kg m}^2\) about an axis through one end, perpendicular to it. Find \(I\) about the parallel axis \(0.2\,\text{m}\) from that end.
The centre of mass is at the middle, \(0.6\,\text{m}\) from the end. The end axis is at \(d_1 = 0.6\,\text{m}\) from it; the new axis is at \(d_2 = 0.6 - 0.2 = 0.4\,\text{m}\).
\(I_{\text{cm}} = I_1 - Md_1^2\), then \(I_2 = I_{\text{cm}} + Md_2^2\).
\(I_{\text{cm}} = 0.96 - 2 \times 0.36 = 0.24\,\text{kg m}^2\). \(I_2 = 0.24 + 2 \times 0.16 = 0.56\,\text{kg m}^2\).
\(I = 0.56\,\text{kg m}^2\). The tempting shortcut, \(0.96 + 2 \times 0.2^2 = 1.04\,\text{kg m}^2\), is wrong: the new axis is closer to the centre of mass, so \(I\) must fall, not rise.
A uniform disc of mass \(M\) and radius \(R\) has axis A perpendicular to it through a point of its rim. Axis B is parallel to A, through the point of the rim diametrically opposite. How is \(I_B\) related to \(I_A\)?
A student writes \(I_2 = I_1 + Ma^2\), where \(a\) is the distance between two parallel axes, neither of which passes through the centre of mass. Explain the error.
The parallel-axis theorem was proved with the origin at the centre of mass; there the cross terms \(\int x\,dm\) vanish. About any other axis they do not vanish, so the theorem cannot be used between two arbitrary axes. The correct way is to go through the centre of mass: \(I_2 = I_1 - Md_1^2 + Md_2^2\), with both distances measured from the centre of mass.
Huygens used the idea of the parallel-axis theorem in 1673, in his work on pendulum clocks. In many countries, though, it carries the name of Jakob Steiner, a Swiss geometer of the nineteenth century. Steiner was the son of a farmer and could not write until he was fourteen. He became one of the great geometers of Berlin, famous for proofs that used pictures instead of algebra. Students in Europe still learn the parallel-axis theorem as Steiner's theorem.
A door turns about its hinge line, which runs along one edge, not through its middle. By the parallel-axis theorem, a uniform door of width \(w\) has \(I = \tfrac13 Mw^2\) about its hinges. That is four times its value about a vertical line through its middle. A heavy door is therefore slow to start and slow to stop. That is why many offices fit a closer that brings the door to rest gently.
Now return to the parade. A baton of length \(L\) has \(I = \tfrac{1}{12}ML^2\) about its middle and \(\tfrac13 ML^2\) about its end, four times as much. The middle is the centre of mass, and no other axis across the baton does better. The drum major spins it there for a reason.
- Applying \(I = I_1 + Ma^2\) between two axes that both miss the centre of mass.Go through the centre of mass: \(I_2 = I_1 - Md_1^2 + Md_2^2\).
- Using the parallel-axis theorem for axes that are not parallel.The theorem needs parallel axes. For perpendicular axes in a flat body, use the perpendicular-axis theorem of the next lesson.
- Taking the geometric middle as the centre of mass of an uneven body.Find the centre of mass first. The least \(I\) is about it, not about the middle.
- Taking \(k\) as the average distance of the mass from the axis.\(k = \sqrt{I/M}\) comes from the squared distances, so it leans towards the outer mass.