Irodov 1.241: A Hole Is Negative Mass – Amit & Deepali Sinha Foundation

Problem Irodov 1.241

A uniform disc of radius $R = 20$ cm has a circular hole cut clean through it. The hole is placed so that its own rim passes through the center of the disc and, on the far side, touches the rim of the disc from the inside. What is left of the disc has mass $m = 7.3$ kg. Find the moment of inertia of that remaining body about the axis that passes through its center of inertia perpendicular to the plane of the disc.

The disc with its circular cut. The cut's rim passes through the disc's center O and touches the disc's own rim on the far side, which fixes the cut's radius r and the distance d from O to the cut's center O'. — The disc with its circular cut. The cut’s rim passes through the disc’s center $O$ and touches the disc’s own rim on the far side, which fixes the cut’s radius $r$ and the distance $d$ from $O$ to the cut’s center $O'$ .

Given

$R = 20$ cm
$m = 7.3$ kg (the remaining body)
the cut’s rim passes through $O$ and touches the disc’s rim

Find

I

of the remaining body about the axis through its center of inertia, perpendicular to the disc

Solution

Step 1The cut’s size and position, from the two contacts

Nothing can be computed until the cut is pinned down, and the two contacts in the figure do that. Let the cut have radius $r$ and its center lie a distance $d$ from $O$ . Its rim passing through $O$ says one thing about $d$ and $r$ ; its rim touching the disc’s rim from inside, along the same line of centers, says another.

\begin{align*}d &= r \\ d + r &= R\end{align*}

The two together leave one possibility.

r = d = \frac{R}{2}

Step 2The three masses in play

The plate is uniform, so mass follows area. Write $\sigma$ for the mass per unit area, and call the complete uncut disc $m_{0}$ and the piece removed $m_{1}$ ; what is given is the difference, $m$ .

\begin{align*}m_{0} &= \sigma\pi R^{2} \\ m_{1} &= \sigma\pi\left(\frac{R}{2}\right)^{2} = \frac{1}{4}m_{0} \\ m &= m_{0} - m_{1} = \frac{3}{4}m_{0}\end{align*}

Reading the last line backwards expresses both unknown masses through the one that is given.

m_{0} = \frac{4}{3}m,\qquad m_{1} = \frac{1}{3}m

(1)

The same body read as a superposition: the complete disc of mass m_0 plus a ghost piece of mass -m_1 where the hole is. The center of inertia C lands on the line of centers, away from the hole. — The same body read as a superposition: the complete disc of mass $m_{0}$ plus a ghost piece of mass $-m_{1}$ where the hole is. The center of inertia $C$ lands on the line of centers, away from the hole.

Step 3Where the center of inertia sits

Treat the body as the complete disc with a piece of negative mass $-m_{1}$ laid over the hole: mass and first moment are both additive, so this reproduces the real body exactly. Measure $x$ from $O$ along the line of centers, positive toward the hole.

\begin{align*}m\,x_{C} &= m_{0}\cdot 0 + (-m_{1})\cdot\frac{R}{2} \\ &= -\frac{1}{3}m\cdot\frac{R}{2}\end{align*}

x_{C} = -\frac{R}{6}

(2)

Step 4The moment of inertia about the disc’s own center

Moments of inertia about one and the same axis add, so the body’s value about the axis through $O$ is the complete disc’s minus the removed piece’s. The disc is centered on that axis and contributes $\frac{1}{2}m_{0}R^{2}$ directly; the piece is not, so its own $\frac{1}{2}m_{1}r^{2}$ must be carried across the distance $R/2$ by the parallel-axis theorem before it can be subtracted.

\begin{align*}I_{O} &= \frac{1}{2}m_{0}R^{2} - \left[\frac{1}{2}m_{1}\left(\frac{R}{2}\right)^{2} + m_{1}\left(\frac{R}{2}\right)^{2}\right] \\ &= \frac{1}{2}\cdot\frac{4}{3}mR^{2} - \frac{1}{3}m\left(\frac{R^{2}}{8} + \frac{R^{2}}{4}\right) \\ &= \frac{2}{3}mR^{2} - \frac{1}{3}m\cdot\frac{3R^{2}}{8} \\ &= \frac{2}{3}mR^{2} - \frac{1}{8}mR^{2}\end{align*}

I_{O} = \frac{13}{24}mR^{2}

(3)

Step 5Moving the axis to the center of inertia

The axis the problem asks about passes through $C$ , not through $O$ , and those two are the $R/6$ apart found in (2). The parallel-axis theorem carries (3) across that gap, with the whole remaining mass $m$ riding on it, and with $R = 0.20$ m and $m = 7.3$ kg the number follows.

\begin{align*}I_{O} &= I_{C} + m\left(\frac{R}{6}\right)^{2} \\ I_{C} &= \frac{13}{24}mR^{2} - \frac{1}{36}mR^{2} \\ &= \left(\frac{39}{72} - \frac{2}{72}\right)mR^{2}\end{align*}

I_{C} = \frac{37}{72}mR^{2} = 0.15\ \mathrm{kg}\cdot\mathrm{m}^{2}

(4)

Check

Rebuild the answer by a route that never visits $O$ : apply the parallel-axis theorem to each piece separately, straight to the axis through $C$ . The complete disc’s center is $R/6$ from $C$ and the removed piece’s center is $R/2 + R/6 = 2R/3$ from it, so $I_{C} = [\frac{1}{2}m_{0}R^{2} + m_{0}(R/6)^{2}] - [\frac{1}{2}m_{1}(R/2)^{2} + m_{1}(2R/3)^{2}]$ , which with the masses of (1) is $(\frac{2}{3} + \frac{1}{27} - \frac{1}{24} - \frac{4}{27})mR^{2}$ . Over the common denominator 216 that is $(144 + 8 - 9 - 32)/216 = 111/216 = 37/72$ , the boxed coefficient. Numerically, $(37/72)(7.3)(0.20)^{2} = 0.150$ kg $\cdot$ m $^{2}$ .

Remark

The coefficient $37/72 = 0.514$ is larger than the $1/2$ of a complete uniform disc of the same mass and the same radius $R$ , which looks wrong until the two effects are separated. What was cut away was not average material: its own moment of inertia per unit mass about $O$ is $\frac{3}{8}R^{2}$ , well under the disc’s $\frac{1}{2}R^{2}$ , so the mass that survives is on average further from the axis than the mass that went. That raises the coefficient about $O$ to $13/24 = 0.542$ . Shifting the axis out to the true center of inertia then gives back $1/36 = 0.028$ , and no more, leaving the crescent stiffer to spin than a full disc of the same mass.