The Workbook

\(x_{\text{cm}} = \dfrac{2 \times 0 + 3 \times 5}{5} = 3\) m, closer to the heavier mass.

1. \(x_{\text{cm}} = \dfrac{1 \times 0 + 2 \times 4 + 3 \times 0}{6} = \dfrac{4}{3}\) m.

2. \(y_{\text{cm}} = \dfrac{0 + 0 + 3 \times 3}{6} = 1.5\) m.

3. Yes: \(\left(\dfrac{4}{3}, 1.5\right)\) lies below the line from \((4, 0)\) to \((0, 3)\), whose equation is \(3x + 4y = 12\) (\(4 + 6 = 10 < 12\)).

Corners at \((0, 0)\), \((2, 0)\) and \((1, \sqrt{3})\). Equal masses give the average of the coordinates: \(x = \dfrac{0 + 2 + 1}{3} = 1\), \(y = \dfrac{\sqrt{3}}{3} = \dfrac{1}{\sqrt{3}} \approx 0.58\) m. This is the centroid.

From the \(1\) kg mass: \(x = \dfrac{3 \times 4}{4} = 3\) m. So it is \(1\) m from the \(3\) kg mass. The centre of mass divides the distance in the inverse ratio of the masses.

\(\dfrac{1 \times 0 + 3 \times 4 + 2x}{6} = 2\) gives \(12 + 2x = 12\), so \(x = 0\): at the origin, with the \(1\) kg mass.

\(M = \lambda L\). \(x_{\text{cm}} = \dfrac{1}{\lambda L}\int_0^L x\lambda\,dx = \dfrac{1}{L}\left[\dfrac{x^2}{2}\right]_0^L = \dfrac{L}{2}\).

\(M = \int_0^L kx\,dx = \dfrac{kL^2}{2}\). \(x_{\text{cm}} = \dfrac{1}{M}\int_0^L x \cdot kx\,dx = \dfrac{kL^3/3}{kL^2/2} = \dfrac{2L}{3}\). For \(L = 3\) m: \(2\) m, nearer the heavy end.

1. The ring is symmetric about the \(y\)-axis: for every element at \(+x\) there is an equal one at \(-x\), so the contributions cancel.

2. \(y_{\text{cm}} = \dfrac{2 \times 7}{22/7} = \dfrac{14 \times 7}{22} = 4.45\) cm.

3. No. The ring is at distance \(7\) cm from \(O\); the centre of mass at \(4.45\) cm is in the empty space inside. A centre of mass need not lie on the body.

\(\dfrac{4 \times 7}{3 \times 22/7} = \dfrac{28 \times 7}{66} \approx 2.97\) cm. The disc has mass all the way in to the centre, which pulls the centre of mass down. The ring has all its mass at the rim.

1. \(\dfrac{3R}{8} = 3\) cm.

2. \(\dfrac{R}{2} = 4\) cm.

3. \(\dfrac{h}{4} = 3\) cm.

\(y_{\text{cm}} = \dfrac{1}{M}\int y\,dm = \dfrac{1}{M}\int_0^\pi R\sin\theta \cdot \dfrac{M}{\pi}d\theta = \dfrac{R}{\pi}[-\cos\theta]_0^\pi = \dfrac{R}{\pi}(1 + 1) = \dfrac{2R}{\pi}\).

1. \(A\): \(2 \times 6 = 12\) cm\(^2\), centre \((1, 3)\). \(B\): \(4 \times 2 = 8\) cm\(^2\), centre \((4, 1)\).

2. \(x = \dfrac{12 \times 1 + 8 \times 4}{20} = 2.2\) cm; \(y = \dfrac{12 \times 3 + 8 \times 1}{20} = 2.2\) cm.

3. The lamina is uniform and of the same thickness everywhere, so mass is proportional to area and the ratio is what matters.

The rod's own centre is \(1\) m from the ball. \(x_{\text{cm}} = \dfrac{1 \times 0 + 2 \times 1}{3} = \dfrac{2}{3}\) m from the ball.

Mass \(\propto\) area \(\propto\) radius\(^2\): the masses are in the ratio \(1 : 4\). The centres are \(3R\) apart. From the larger disc's centre: \(x = \dfrac{1 \times 3R + 4 \times 0}{5} = 0.6R\), towards the small disc.

Masses \(\propto\) volume: \(1 : 8\). Centres at heights \(\dfrac{a}{2}\) and \(a\). \(y = \dfrac{1 \times a/2 + 8 \times a}{9} = \dfrac{17a}{18} \approx 0.94a\).

With the origin shifted by \(d\) every position becomes \(x_i' = x_i - d\). Then \(x_{\text{cm}}' = \dfrac{m_1(x_1 - d) + m_2(x_2 - d)}{m_1 + m_2} = x_{\text{cm}} - d\). The number changes by exactly the shift, so it describes the same physical point.

1. Area \(\propto\) radius\(^2\): the hole is \(\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}\) of the disc's mass \(M\).

2. Remaining mass \(\dfrac{3M}{4}\). Taking \(O\) as origin and \(C\) at \(+\dfrac{R}{2}\): \(x_{\text{cm}} = \dfrac{M \times 0 - \frac{M}{4} \times \frac{R}{2}}{3M/4} = -\dfrac{R}{6} = -2\) cm.

3. The minus sign puts it \(2\) cm from \(O\) on the side away from the hole, where more material remains.

Volume \(\propto\) radius\(^3\): the cavity is \(\dfrac{1}{8}\) of the mass. \(x_{\text{cm}} = \dfrac{-\frac{M}{8} \times \frac{R}{2}}{7M/8} = -\dfrac{R}{14} = -1\) cm: \(1\) cm from the centre, away from the cavity.

1. Plate \(36\) cm\(^2\), hole \(4\) cm\(^2\), remainder \(32\) cm\(^2\).

2. \(x_{\text{cm}} = \dfrac{36 \times 0 - 4 \times 2}{32} = -0.25\) cm: a quarter centimetre to the left of \(O\), away from the hole.

The full plate is the remainder plus the piece cut out. So \(M\vec r_{\text{plate}} = m_{\text{rem}}\vec r_{\text{rem}} + m_{\text{hole}}\vec r_{\text{hole}}\). Solving for the remainder moves the hole's term to the other side with a minus sign. The hole enters exactly like a mass of \(-m_{\text{hole}}\) at its own centre.

Directly: what remains is a uniform rod of length \(\dfrac{3L}{4}\) that starts at \(\dfrac{L}{4}\); its centre is at \(\dfrac{L}{4} + \dfrac{3L}{8} = \dfrac{5L}{8}\).

Negative mass: full rod \(M\) at \(\dfrac{L}{2}\), removed piece \(\dfrac{M}{4}\) at \(\dfrac{L}{8}\). \(x = \dfrac{M \cdot L/2 - (M/4)(L/8)}{3M/4} = \dfrac{15L/32}{3/4} = \dfrac{5L}{8}\).

No external horizontal force acts, so the centre of mass stays put: \(60\,\Delta x_{\text{man}} = 90\,\Delta x_{\text{boat}}\) in magnitude, opposite in direction. With \(\Delta x_{\text{man}} + \Delta x_{\text{boat}} = 3\) m relative to each other: the boat moves \(\dfrac{60}{150} \times 3 = 1.2\) m backward and the man \(1.8\) m forward over the water.

From the Earth's centre: \(x = \dfrac{1 \times 3.84 \times 10^{5}}{81 + 1} \approx 4.7 \times 10^{3}\) km. This is less than \(6400\) km, so the centre of mass of the pair lies inside the Earth.

No. The centre of mass of a uniform ring is at its centre, where there is no material. That of an L-shaped plate can fall in the empty corner. It is a weighted average position, not a point of the body.