At a tyre shop in any Indian town, the mechanic fits a new tyre, mounts the wheel on a balancing machine and spins it. The machine shows a number, and the mechanic clips a small lead weight, perhaps fifty grams, to one point of the rim. Without it the car would shake the steering wheel at high speed. A wheel of many kilograms is corrected by a few grams.
A hole is a negative mass
A plate with a hole cut out is a composite body, but one of its pieces is missing rather than added. Think of it this way. The full plate, with no hole, is made of two parts: the plate that remains, and the disc that was cut out. We know the centre of mass of the full plate and of the cut-out disc. The unknown is the centre of mass of what remains.
Let the full plate have mass \(M\) with centre at \(x_{\text{full}}\), and let the cut-out piece have mass \(m\) with centre at \(x_{\text{hole}}\). The composite rule for the full plate says
\[ M x_{\text{full}} = (M - m)\,x_{\text{rem}} + m\,x_{\text{hole}} \]
so the remaining plate has
\[ x_{\text{rem}} = \dfrac{M x_{\text{full}} - m\,x_{\text{hole}}}{M - m} \]
This is the particle formula for two masses, \(M\) at \(x_{\text{full}}\) and \(-m\) at \(x_{\text{hole}}\). The hole acts as a negative mass placed at its own centre.
For a body with a piece removed, treat the full body as a positive mass at its own centre of mass. Treat the removed piece as a negative mass at its own centre of mass, and apply the particle formula.
The minus sign in front of \(m\) carries a clear meaning. Adding mass at a point pulls the centre of mass towards that point. Removing mass pushes it away. So the centre of mass of a plate with a hole always moves to the side opposite the hole.
A uniform disc of radius \(30\,\text{cm}\) has a circular hole of radius \(15\,\text{cm}\) cut from it. The hole touches the rim of the disc. Find the centre of mass of the remaining plate.
Put the origin at the centre O of the disc, and the x axis through the centre of the hole. The hole's centre is at \(x_{\text{hole}} = 30 - 15 = 15\,\text{cm}\).
The disc is uniform, so masses go as areas. The hole's radius is half the disc's, so its area, and its mass, is one quarter: \(m = M/4\).
\(x_{\text{rem}} = \dfrac{M \times 0 - (M/4) \times 15}{M - M/4} = \dfrac{-15/4}{3/4} = -5\,\text{cm}\).
The centre of mass is \(5\,\text{cm}\) from O on the side away from the hole. In general, for a hole of radius \(R/2\) touching the rim, the shift is \(R/6\).
For a hole of radius \(r\) touching the rim, the removed mass is \(M r^2/R^2\) at \(x = R - r\). The negative-mass formula gives a shift of \(r^2(R - r)/(R^2 - r^2) = r^2/(R + r)\) away from the hole. Move the slider and watch this grow. A small hole barely moves C: \(0.05R\) for \(r = R/4\). A hole of \(r = 0.8R\) removes nearly two thirds of the disc and moves C by \(0.36R\).
A uniform disc of radius \(10\,\text{cm}\) has a circular hole of radius \(5\,\text{cm}\) cut from it. The centre of the hole is \(4\,\text{cm}\) from the centre of the disc. How far from the centre of the disc is the centre of mass of the remaining plate, in cm?
A circular hole is punched in the right half of a uniform square plate. Which way does the centre of mass of the plate move?
A cavity inside a solid
The method works in three dimensions without change. A solid ball with a hollow space inside, a cavity, is the full ball plus a negative ball where the cavity is. The only difference is how the masses compare. For a uniform solid, mass goes as volume, and the volume of a sphere goes as the cube of its radius.
Take a uniform solid sphere of radius \(R\) with a spherical cavity of radius \(R/2\) that touches its surface. The cavity's volume is \((1/2)^3 = 1/8\) of the sphere's, so \(m = M/8\) at a distance \(R/2\) from the centre. The shift of the centre of mass is
\[ \dfrac{(M/8)(R/2)}{M - M/8} = \dfrac{R/16}{7/8} = \dfrac{R}{14} \]
away from the cavity. A flat disc with a hole of the same proportions gave \(R/6\). The cavity removes only one eighth of the sphere, while the hole removed one quarter of the disc, so the sphere's centre moves much less.
A uniform solid sphere of radius \(21\,\text{cm}\) has a spherical cavity of radius \(10.5\,\text{cm}\) inside it, touching its surface. How far from the centre of the sphere is the centre of mass of what remains, in cm?
A uniform disc has a hole of radius \(R/2\) touching its rim. A uniform sphere has a cavity of radius \(R/2\) touching its surface. Why does the sphere's centre of mass move less?
Cut corners and added weights
The removed piece need not be round, and it need not be inside the body. Cut a square of side \(a/2\) from one corner of a uniform square plate of side \(a\). Put the origin at the centre of the full plate. The removed square has one quarter of the area, and its centre is at \((a/4, a/4)\). Along each axis the remaining plate has
\[ x_{\text{rem}} = \dfrac{0 - \tfrac14 M \cdot \tfrac{a}{4}}{\tfrac34 M} = -\dfrac{a}{12} \]
and the same for \(y\). So C moves by \(a/12\) along each axis, along the diagonal away from the cut corner.
Adding a piece is the same formula with a plus sign. A wheel whose centre of mass lies a little off the axle shakes as it spins. A small weight clipped to the rim on the light side pulls the centre of mass back onto the axle. The weight \(m\) at radius \(r\) must cancel the product \(M d\) of the wheel's mass and its offset, so \(m r = M d\).
A uniform square plate of side \(12\,\text{cm}\) has a square of side \(6\,\text{cm}\) cut from one corner. Find the centre of mass of the remaining plate, measured from the centre of the full square.
Take the origin at the centre of the full square, with axes parallel to its sides and the cut corner in the first quadrant. The cut square has area \(36\,\text{cm}^2\), one quarter of \(144\,\text{cm}^2\), and its centre is at \((3, 3)\,\text{cm}\).
\(x_{\text{rem}} = \dfrac{144 \times 0 - 36 \times 3}{144 - 36}\), and the same for \(y_{\text{rem}}\).
\(x_{\text{rem}} = -108/108 = -1\,\text{cm}\) and \(y_{\text{rem}} = -1\,\text{cm}\).
C is at \((-1, -1)\,\text{cm}\), that is \(\sqrt2 \approx 1.41\,\text{cm}\) from the centre along the diagonal, away from the cut corner. This is \(a/12\) along each axis for \(a = 12\,\text{cm}\).
A student objects that there is no such thing as negative mass. Explain why the negative-mass method still gives the correct centre of mass of a plate with a hole.
No negative mass exists; it is a bookkeeping step. The full plate is truly made of two real parts: the plate that remains and the piece that was cut out. The composite rule therefore gives \(M x_{\text{full}} = m_{\text{rem}} x_{\text{rem}} + m\,x_{\text{hole}}\). Solving this real equation for \(x_{\text{rem}}\) gives exactly the particle formula with \(-m\) placed at the hole. The minus sign only records that the piece has been taken away.
A car wheel with its tyre has a mass of \(20\,\text{kg}\), and its centre of mass is \(0.5\,\text{mm}\) off the axle. The mechanic clips a weight to the rim, \(0.2\,\text{m}\) from the axle, on the opposite side. Estimate the mass of the weight, in grams.
Many mobile phones vibrate with a tiny electric motor that carries a half-disc of metal on its shaft. The half-disc has its centre of mass off the axis, at \(4R/3\pi\) for a flat half-disc. As the motor spins, the centre of mass swings in a circle, and the shaft must keep pulling it round. The phone feels that changing pull as a buzz. A wheel is balanced to stop exactly this shaking; the phone's motor is unbalanced on purpose.
This is the tyre shop of the opening. The balancing machine measures how far the wheel's centre of mass is off the axle and in which direction. The mechanic then clips a weight to the opposite side of the rim. A few tens of grams far from the axle cancel a small offset of the whole wheel. A missing mass pushes the centre of mass away; an added mass pulls it back.
- Adding the mass of the hole instead of subtracting it.A removed piece enters the formula with a minus sign, in the numerator and in the total mass.
- Using the area ratio for a solid with a cavity.For a uniform solid, masses go as volumes: halving the radius removes \(1/8\) of the mass, not \(1/4\).
- Measuring the hole's position from the rim instead of from the chosen origin.Place every centre, of the full body and of the hole, with the same origin, usually the centre of the full body.