Chapter 6: System of Particles and Rotational Motion

Set up a coordinate system in the plane of the triangle. Place the vertex with mass $$m_1 = 100\,\mathrm{g}$$ at the origin $$O$$, the vertex with mass $$m_2 = 150\,\mathrm{g}$$ at point $$B$$ on the positive $$x$$-axis at $$(0.5,\,0)\,\mathrm{m}$$, and the vertex with mass $$m_3 = 200\,\mathrm{g}$$ at point $$C$$ above the $$x$$-axis. Since the triangle is equilateral with side $$0.5\,\mathrm{m}$$, the coordinates of $$C$$ are $$\left(0.25,\,0.25\sqrt{3}\right)\,\mathrm{m}$$.

The $$x$$-coordinate of the centre of mass is

$$x_{CM} = \dfrac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \dfrac{100\cdot 0 + 150\cdot 0.5 + 200\cdot 0.25}{100 + 150 + 200}\,\mathrm{m}.$$

$$x_{CM} = \dfrac{0 + 75 + 50}{450}\,\mathrm{m} = \dfrac{125}{450}\,\mathrm{m} = \dfrac{5}{18}\,\mathrm{m}.$$

The $$y$$-coordinate is

$$y_{CM} = \dfrac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \dfrac{100\cdot 0 + 150\cdot 0 + 200\cdot 0.25\sqrt{3}}{450}\,\mathrm{m}.$$

$$y_{CM} = \dfrac{50\sqrt{3}}{450}\,\mathrm{m} = \dfrac{\sqrt{3}}{9}\,\mathrm{m}.$$

Hence the centre of mass lies at $$\left(\dfrac{5}{18},\,\dfrac{\sqrt{3}}{9}\right)\,\mathrm{m}$$ from the chosen origin.

Consider a uniform triangular lamina $$ABC$$. To find its centre of mass, we exploit the symmetry of the lamina by splitting it into a large number of thin strips parallel to the base.

Each such strip is a slim rectangle (essentially a one-dimensional rod), so its centre of mass lies at its midpoint. The locus of all these midpoints is the line joining vertex $$A$$ to the midpoint $$D$$ of the opposite side $$BC$$ — that is, the median $$AD$$.

Therefore the centre of mass of the entire lamina must lie on the median $$AD$$. By repeating the argument with strips parallel to side $$AC$$ (or $$AB$$), the centre of mass must also lie on the median from $$B$$ (and from $$C$$). The three medians intersect at a single point, the centroid $$G$$ of the triangle.

The centroid divides every median in the ratio $$2:1$$ from the vertex. Hence

$$AG : GD = 2 : 1.$$

So the centre of mass of a uniform triangular lamina lies at the centroid — the intersection of its medians.

Choose the corner of the L-shape as the origin, with the $$x$$- and $$y$$-axes along the two outer edges (as in Fig. 6.11). The L-shape can be divided into three identical squares, each of side $$1\,\mathrm{m}$$. Since the lamina is uniform, each square has the same mass

$$m = \dfrac{3\,\mathrm{kg}}{3} = 1\,\mathrm{kg}.$$

The centre of each square (where its mass may be considered to be concentrated) lies at its geometric centre:

• Square 1 (bottom-left): centre at $$C_1 = (0.5,\,0.5)\,\mathrm{m}$$.
• Square 2 (bottom-right): centre at $$C_2 = (1.5,\,0.5)\,\mathrm{m}$$.
• Square 3 (top-left): centre at $$C_3 = (0.5,\,1.5)\,\mathrm{m}$$.

Treating the lamina as three point masses at $$C_1,\,C_2,\,C_3$$:

$$x_{CM} = \dfrac{m\cdot 0.5 + m\cdot 1.5 + m\cdot 0.5}{3m} = \dfrac{2.5}{3}\,\mathrm{m} = \dfrac{5}{6}\,\mathrm{m}.$$

$$y_{CM} = \dfrac{m\cdot 0.5 + m\cdot 0.5 + m\cdot 1.5}{3m} = \dfrac{2.5}{3}\,\mathrm{m} = \dfrac{5}{6}\,\mathrm{m}.$$

So the centre of mass of the L-shaped lamina lies at $$\left(\dfrac{5}{6},\,\dfrac{5}{6}\right)\,\mathrm{m}$$ from the corner. Note that this point lies inside the lamina (within square 1), as expected.

Scalar product. Using the component formula $$\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$$,

$$\mathbf{a}\cdot\mathbf{b} = (3)(-2) + (-4)(1) + (5)(3) = -6 - 4 + 15 = 5.$$

Vector product. Using the determinant form

$$\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -4 & 5 \\ -2 & 1 & 3 \end{vmatrix}.$$

Expanding along the first row,

$$\mathbf{a}\times\mathbf{b} = \hat{\mathbf{i}}\bigl[(-4)(3) - (5)(1)\bigr] - \hat{\mathbf{j}}\bigl[(3)(3) - (5)(-2)\bigr] + \hat{\mathbf{k}}\bigl[(3)(1) - (-4)(-2)\bigr].$$

$$\mathbf{a}\times\mathbf{b} = \hat{\mathbf{i}}(-12 - 5) - \hat{\mathbf{j}}(9 + 10) + \hat{\mathbf{k}}(3 - 8).$$

$$\mathbf{a}\times\mathbf{b} = -17\hat{\mathbf{i}} - 19\hat{\mathbf{j}} - 5\hat{\mathbf{k}}.$$

Check: $$\mathbf{a}\cdot(\mathbf{a}\times\mathbf{b}) = 3(-17) + (-4)(-19) + 5(-5) = -51 + 76 - 25 = 0$$, confirming that $$\mathbf{a}\times\mathbf{b}$$ is perpendicular to $$\mathbf{a}$$.

The torque about the origin is $$\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}$$, where

$$\mathbf{r} = \hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}},\qquad \mathbf{F} = 7\hat{\mathbf{i}} + 3\hat{\mathbf{j}} - 5\hat{\mathbf{k}}.$$

Compute the cross product using the determinant form:

$$\boldsymbol{\tau} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 7 & 3 & -5 \end{vmatrix}.$$

$$\boldsymbol{\tau} = \hat{\mathbf{i}}\bigl[(-1)(-5) - (1)(3)\bigr] - \hat{\mathbf{j}}\bigl[(1)(-5) - (1)(7)\bigr] + \hat{\mathbf{k}}\bigl[(1)(3) - (-1)(7)\bigr].$$

$$\boldsymbol{\tau} = \hat{\mathbf{i}}(5 - 3) - \hat{\mathbf{j}}(-5 - 7) + \hat{\mathbf{k}}(3 + 7).$$

$$\boldsymbol{\tau} = 2\hat{\mathbf{i}} + 12\hat{\mathbf{j}} + 10\hat{\mathbf{k}}.$$

The magnitude is

$$|\boldsymbol{\tau}| = \sqrt{2^2 + 12^2 + 10^2} = \sqrt{4 + 144 + 100} = \sqrt{248} \approx 15.75\,\mathrm{N\,m}.$$

Let a particle of mass $$m$$ move with constant velocity $$\mathbf{v}$$. Pick any fixed reference point $$O$$ and let $$\mathbf{r}(t)$$ be the position vector of the particle relative to $$O$$ at time $$t$$.

The angular momentum about $$O$$ is

$$\mathbf{l} = \mathbf{r}\times \mathbf{p} = \mathbf{r}\times (m\mathbf{v}).$$

Method 1 (geometric). The magnitude of $$\mathbf{l}$$ is $$|\mathbf{l}| = m v\, d$$, where $$d$$ is the perpendicular distance from $$O$$ to the straight line along which the particle moves. Since the particle moves with constant velocity, this line is fixed in space, so $$d$$ is constant. The direction of $$\mathbf{l}$$ is also fixed (perpendicular to the plane containing $$O$$ and the line of motion). Hence $$\mathbf{l}$$ is constant.

Method 2 (by differentiation). Differentiate $$\mathbf{l}$$ with respect to time:

$$\dfrac{d\mathbf{l}}{dt} = \dfrac{d\mathbf{r}}{dt}\times m\mathbf{v} + \mathbf{r}\times m\dfrac{d\mathbf{v}}{dt}.$$

Since $$\dfrac{d\mathbf{r}}{dt} = \mathbf{v}$$, the first term is $$\mathbf{v}\times m\mathbf{v} = m(\mathbf{v}\times\mathbf{v}) = \mathbf{0}$$. The second term is also zero because $$\mathbf{v}$$ is constant, so $$\dfrac{d\mathbf{v}}{dt} = \mathbf{0}$$. Therefore

$$\dfrac{d\mathbf{l}}{dt} = \mathbf{0},$$

so $$\mathbf{l}$$ is constant in time. Hence proved.

A couple consists of two equal and opposite forces $$\mathbf{F}$$ and $$-\mathbf{F}$$ acting at two distinct points $$A$$ and $$B$$ of a body. Let $$O$$ be an arbitrary reference point, and let $$\mathbf{r}_1$$ and $$\mathbf{r}_2$$ be the position vectors of $$A$$ and $$B$$ relative to $$O$$.

The total moment of the couple about $$O$$ is

$$\boldsymbol{\tau} = \mathbf{r}_1\times \mathbf{F} + \mathbf{r}_2\times (-\mathbf{F}).$$

$$\boldsymbol{\tau} = (\mathbf{r}_1 - \mathbf{r}_2)\times \mathbf{F}.$$

The vector $$\mathbf{r}_1 - \mathbf{r}_2$$ is the position vector of $$A$$ relative to $$B$$, which depends only on the relative geometry of the two points where the couple acts. It does not depend on the choice of $$O$$.

Therefore the moment of a couple is independent of the reference point $$O$$. Its magnitude equals $$F\cdot d$$ where $$d$$ is the perpendicular distance between the two parallel lines of action.

Set the origin at the left end of the bar. Then:

• The two knife-edges (supports) are at $$x_1 = 10\,\mathrm{cm}$$ (left support, reaction $$R_1$$) and $$x_2 = 70 - 10 = 60\,\mathrm{cm}$$ (right support, reaction $$R_2$$).
• The weight of the bar, $$W_{\text{bar}} = 4.00 \times 9.8 = 39.2\,\mathrm{N}$$, acts at the centre of the uniform bar, $$x = 35\,\mathrm{cm}$$.
• The load, $$W_{\text{load}} = 6.00 \times 9.8 = 58.8\,\mathrm{N}$$, hangs at $$x = 30\,\mathrm{cm}$$.

The bar is in equilibrium, so the net upward force equals the total downward force:

$$R_1 + R_2 = W_{\text{bar}} + W_{\text{load}} = 39.2 + 58.8 = 98.0\,\mathrm{N}.$$

Now take moments about the left knife-edge (at $$x_1 = 10\,\mathrm{cm}$$), so $$R_1$$ contributes no moment. Taking clockwise moments due to the weights and balancing them with the counter-clockwise moment of $$R_2$$:

$$R_2\,(60 - 10) = W_{\text{load}}\,(30 - 10) + W_{\text{bar}}\,(35 - 10).$$

$$R_2 \times 50 = 58.8 \times 20 + 39.2 \times 25.$$

$$50\,R_2 = 1176 + 980 = 2156.$$

$$R_2 = \dfrac{2156}{50} = 43.12\,\mathrm{N}.$$

Then

$$R_1 = 98.0 - 43.12 = 54.88\,\mathrm{N}.$$

Thus the left knife-edge supports about $$54.9\,\mathrm{N}$$, while the right knife-edge supports about $$43.1\,\mathrm{N}$$. The left support carries more weight because the load is closer to it.

Let $$A$$ be the foot of the ladder on the floor and $$B$$ its top end against the wall. The length is $$AB = 3\,\mathrm{m}$$ and the horizontal distance of $$A$$ from the wall is $$1\,\mathrm{m}$$. The vertical height of $$B$$ above the floor is

$$BC = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}\,\mathrm{m}.$$

The weight of the ladder is $$W = 20\times 9.8 = 196\,\mathrm{N}$$, acting downward at the midpoint $$D$$ of $$AB$$.

Forces on the ladder:

• $$\mathbf{W} = 196\,\mathrm{N}$$ downward at $$D$$.
• $$\mathbf{F}_w$$: reaction of the wall, perpendicular to the wall (so horizontal), at $$B$$. Wall is frictionless, hence no vertical component.
• $$\mathbf{N}$$: normal reaction of floor on $$A$$, vertical (upward).
• $$\mathbf{f}$$: frictional force of floor on $$A$$, horizontal, directed toward the wall (preventing the foot from slipping outward).

Translational equilibrium.

Vertical: $$N = W = 196\,\mathrm{N}.$$

Horizontal: $$f = F_w.$$

Rotational equilibrium (take moments about $$A$$, so $$N$$ and $$f$$ contribute zero):

$$F_w\cdot BC - W\cdot\dfrac{1}{2}\cdot 1 = 0.$$

(The wall reaction $$F_w$$ acts horizontally at height $$BC = 2\sqrt{2}\,\mathrm{m}$$; the weight acts vertically with horizontal lever arm equal to half the horizontal distance, i.e. $$0.5\,\mathrm{m}$$.)

$$F_w \times 2\sqrt{2} = 196 \times 0.5.$$

$$F_w = \dfrac{98}{2\sqrt{2}} = \dfrac{49}{\sqrt{2}} = \dfrac{49\sqrt{2}}{2} \approx 34.6\,\mathrm{N}.$$

Therefore $$f = F_w \approx 34.6\,\mathrm{N}$$.

The reaction of the floor has two components: $$N = 196\,\mathrm{N}$$ (vertical) and $$f \approx 34.6\,\mathrm{N}$$ (horizontal). Its magnitude is

$$F_{\text{floor}} = \sqrt{N^2 + f^2} = \sqrt{196^2 + 34.6^2} \approx 199.0\,\mathrm{N}.$$

It makes an angle $$\alpha$$ with the horizontal given by

$$\tan\alpha = \dfrac{N}{f} = \dfrac{196}{34.6} \approx 5.66,\quad \alpha \approx 80^\circ.$$

Equation (6.36) gives the kinetic energy of rotation of a rigid body about a fixed axis,

$$K = \dfrac{1}{2}\,I\,\omega^2,$$

where $$I$$ is the moment of inertia about the axis and $$\omega$$ is the angular speed. We derive it from first principles as follows.

Model the rigid body as a collection of $$N$$ point particles. Let the $$i$$-th particle have mass $$m_i$$ and be located at a perpendicular distance $$r_i$$ from the axis of rotation. Since the body is rigid and rotates about the axis with angular speed $$\omega$$, every particle moves in a circle of radius $$r_i$$ centred on the axis, with linear speed

$$v_i = \omega\,r_i.$$

The kinetic energy of the $$i$$-th particle is

$$K_i = \dfrac{1}{2}\,m_i v_i^2 = \dfrac{1}{2}\,m_i\,\omega^2\,r_i^2.$$

Summing over all particles of the body,

$$K = \sum_{i=1}^{N} K_i = \sum_{i=1}^{N} \dfrac{1}{2}\,m_i\,\omega^2\,r_i^2 = \dfrac{1}{2}\,\omega^2\sum_{i=1}^{N} m_i r_i^2.$$

The quantity $$\displaystyle\sum_{i=1}^{N} m_i r_i^2$$ is by definition the moment of inertia $$I$$ of the body about the axis. Hence

$$K = \dfrac{1}{2}\,I\,\omega^2,$$

which is Eq. (6.36). Notice the close analogy with translational kinetic energy $$\dfrac{1}{2}mv^2$$: mass $$m$$ is replaced by moment of inertia $$I$$, and linear speed $$v$$ is replaced by angular speed $$\omega$$.

For a body of uniform mass density, the centre of mass coincides with the body's geometric centre. The geometric centre is located using the body's symmetries:

• (i) Sphere — the geometric centre is the centre of the sphere (the point equidistant from every point on its surface).
• (ii) Cylinder — the centre of the cylinder, i.e. the midpoint of its axis.
• (iii) Ring — the geometric centre of the ring, which is the centre of the circular loop.
• (iv) Cube — the centre of the cube, i.e. the point at which the four body diagonals meet.

The centre of mass does not always lie inside the body. The ring is an explicit example: its centre of mass is at the geometric centre of the loop, a point that lies in empty space, not in any material of the ring. Other examples include a hollow sphere or a horseshoe-shaped body.

Take a coordinate axis along the line joining the two nuclei. Place the hydrogen nucleus at the origin and the chlorine nucleus at $$x = 1.27\,\mathrm{\AA}$$. Let the mass of a hydrogen atom be $$m$$. Then the mass of a chlorine atom is $$35.5\,m$$, and the nuclear masses can be taken as point masses since nearly all the atomic mass is concentrated in the nucleus.

The $$x$$-coordinate of the centre of mass is

$$x_{CM} = \dfrac{m_H\,x_H + m_{Cl}\,x_{Cl}}{m_H + m_{Cl}} = \dfrac{m\cdot 0 + 35.5\,m\cdot 1.27}{m + 35.5\,m}.$$

$$x_{CM} = \dfrac{35.5 \times 1.27}{36.5}\,\mathrm{\AA} = \dfrac{45.085}{36.5}\,\mathrm{\AA} \approx 1.235\,\mathrm{\AA}.$$

The centre of mass therefore lies on the line joining the two nuclei, at a distance of about $$1.235\,\mathrm{\AA}$$ from the hydrogen nucleus (equivalently, about $$0.035\,\mathrm{\AA}$$ from the chlorine nucleus). This is consistent with our intuition: because chlorine is so much heavier than hydrogen, the CM lies very close to the chlorine nucleus.

The centre of mass of the (trolley + child) system obeys

$$M\,\mathbf{a}_{CM} = \mathbf{F}_{\text{ext}},$$

where $$M$$ is the total mass and $$\mathbf{F}_{\text{ext}}$$ is the resultant of all external forces on the system.

The forces between the child and the trolley (the child's push on the floor of the trolley, friction between his shoes and the trolley, etc.) are internal to the chosen system and do not change the motion of the CM.

The external forces are gravity (downward) and the normal reaction from the smooth floor (upward); these cancel. Because the floor is smooth, no horizontal external force acts on the system. Hence the horizontal acceleration of the CM is zero, and its horizontal velocity remains whatever it was before the child started running.

Initially the trolley and the (stationary) child both moved horizontally at speed $$V$$, so the CM moved at $$V$$. It continues to move at this speed regardless of how the child runs about on the trolley.

$$\boxed{v_{CM} = V.}$$

Place the two vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ with a common starting point $$O$$. They define a triangle with vertices $$O$$, the tip of $$\mathbf{a}$$, and the tip of $$\mathbf{b}$$. Let $$\theta$$ be the angle between $$\mathbf{a}$$ and $$\mathbf{b}$$, with $$0\le\theta\le\pi$$.

Take $$\mathbf{a}$$ as the base of the triangle. The base length is $$|\mathbf{a}|$$. The height of the triangle (the perpendicular distance from the tip of $$\mathbf{b}$$ to the line containing $$\mathbf{a}$$) is

$$h = |\mathbf{b}|\sin\theta.$$

Therefore the area of the triangle is

$$A_{\triangle} = \dfrac{1}{2}\cdot \text{base}\cdot \text{height} = \dfrac{1}{2}|\mathbf{a}||\mathbf{b}|\sin\theta.$$

By the definition of the vector (cross) product,

$$|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta.$$

Comparing the two expressions,

$$A_{\triangle} = \dfrac{1}{2}|\mathbf{a}\times\mathbf{b}|.$$

This is the required result. (Geometrically, $$|\mathbf{a}\times\mathbf{b}|$$ is the area of the parallelogram with $$\mathbf{a}$$ and $$\mathbf{b}$$ as adjacent sides, and the triangle is half of this parallelogram.)

Place $$\mathbf{a}$$, $$\mathbf{b}$$ and $$\mathbf{c}$$ with a common origin. They form a parallelepiped whose three edges meeting at the origin are these vectors.

Take the face containing $$\mathbf{b}$$ and $$\mathbf{c}$$ as the base of the parallelepiped. The base is a parallelogram with area

$$A = |\mathbf{b}\times\mathbf{c}|.$$

The direction of $$\mathbf{b}\times\mathbf{c}$$ is normal to this base. Let $$\phi$$ be the angle between $$\mathbf{a}$$ and the normal direction $$\mathbf{b}\times\mathbf{c}$$. The height of the parallelepiped — the perpendicular distance of the tip of $$\mathbf{a}$$ from the base — is

$$h = |\mathbf{a}|\,|\cos\phi|.$$

Volume of the parallelepiped:

$$V = A\,h = |\mathbf{b}\times\mathbf{c}|\cdot |\mathbf{a}|\,|\cos\phi|.$$

On the other hand, by the definition of the scalar product,

$$\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = |\mathbf{a}|\,|\mathbf{b}\times\mathbf{c}|\cos\phi.$$

Taking the magnitude,

$$|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})| = |\mathbf{a}|\,|\mathbf{b}\times\mathbf{c}|\,|\cos\phi| = V.$$

Thus the scalar triple product $$\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$$ has magnitude equal to the volume of the parallelepiped, as required. (The sign of the triple product encodes the handedness of the three vectors.)

The angular momentum of a particle about the origin is

$$\mathbf{l} = \mathbf{r}\times\mathbf{p}.$$

Writing $$\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}$$ and $$\mathbf{p} = p_x\hat{\mathbf{i}} + p_y\hat{\mathbf{j}} + p_z\hat{\mathbf{k}}$$, evaluate the cross product as a determinant:

$$\mathbf{l} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix}.$$

Expanding,

$$\mathbf{l} = \hat{\mathbf{i}}\,(y p_z - z p_y) + \hat{\mathbf{j}}\,(z p_x - x p_z) + \hat{\mathbf{k}}\,(x p_y - y p_x).$$

Hence the components of $$\mathbf{l}$$ along the three axes are

$$l_x = y p_z - z p_y,\qquad l_y = z p_x - x p_z,\qquad l_z = x p_y - y p_x.$$

Planar motion. If the particle is confined to the $$x$$–$$y$$ plane, then its $$z$$-coordinate is zero throughout the motion, $$z = 0$$, and its velocity (and hence momentum) has no $$z$$-component, $$p_z = 0$$. Substituting $$z = 0$$ and $$p_z = 0$$ in the expressions above,

$$l_x = y\cdot 0 - 0\cdot p_y = 0,$$

$$l_y = 0\cdot p_x - x\cdot 0 = 0,$$

$$l_z = x p_y - y p_x,$$

which is in general non-zero. Therefore for motion confined to the $$x$$–$$y$$ plane, the angular momentum has only a $$z$$-component (perpendicular to the plane of motion), as required.

Choose coordinates so that the two parallel lines lie in the $$x$$–$$y$$ plane and are parallel to the $$x$$-axis. Take line $$L_1$$ as $$y = +d/2$$ (the particle here moves in the $$+x$$ direction with momentum $$\mathbf{p}_1 = mv\,\hat{\mathbf{i}}$$), and line $$L_2$$ as $$y = -d/2$$ (the particle here moves in the $$-x$$ direction with momentum $$\mathbf{p}_2 = -mv\,\hat{\mathbf{i}}$$).

Let $$P$$ be an arbitrary reference point with position vector $$\mathbf{r}_P = (x_0, y_0, 0)$$. Let the instantaneous positions of the two particles be $$(x_1, d/2, 0)$$ and $$(x_2, -d/2, 0)$$.

The position vectors of the two particles relative to $$P$$ are

$$\mathbf{r}'_1 = (x_1 - x_0,\; d/2 - y_0,\; 0),\qquad \mathbf{r}'_2 = (x_2 - x_0,\; -d/2 - y_0,\; 0).$$

The total angular momentum about $$P$$ is

$$\mathbf{L} = \mathbf{r}'_1\times\mathbf{p}_1 + \mathbf{r}'_2\times\mathbf{p}_2.$$

Since both momenta lie along the $$x$$-axis and both position vectors lie in the $$x$$–$$y$$ plane, each cross product is along the $$z$$-axis. Using $$(\mathbf{r}\times\mathbf{p})_z = x\,p_y - y\,p_x$$:

$$L_z = -(d/2 - y_0)(mv) - (-d/2 - y_0)(-mv).$$

$$L_z = -mv\,(d/2 - y_0) - mv\,(d/2 + y_0).$$

$$L_z = -mv\bigl[(d/2 - y_0) + (d/2 + y_0)\bigr] = -mv\,d.$$

The $$x$$- and $$y$$- components of $$\mathbf{L}$$ are clearly zero. Hence

$$\mathbf{L} = -m v d\,\hat{\mathbf{k}}.$$

The coordinates $$x_0,\,y_0$$ of the reference point $$P$$ have dropped out completely. Therefore the angular momentum vector of the two-particle system is independent of the point about which it is computed.

(This is consistent with the result for a couple in Example 6.7: when the total momentum of a system is zero — as it is here, since $$\mathbf{p}_1 + \mathbf{p}_2 = 0$$ — its angular momentum is the same about every reference point.)

Let the bar $$AB$$ hang horizontally in equilibrium, with the left end $$A$$ at the origin and the right end $$B$$ at distance $$L = 2\,\mathrm{m}$$. Let $$T_1$$ be the tension in the string attached to $$A$$ (which makes angle $$\theta_1 = 36.9^\circ$$ with the vertical) and $$T_2$$ the tension in the string attached to $$B$$ (which makes angle $$\theta_2 = 53.1^\circ$$ with the vertical). The centre of gravity $$G$$ of the bar lies on $$AB$$ at distance $$d$$ from $$A$$.

Note $$\theta_1 + \theta_2 = 90^\circ$$ — the two angles are complementary. Using the 3–4–5 right-triangle ratios:

$$\sin 36.9^\circ \approx 0.6,\;\cos 36.9^\circ \approx 0.8,\;\sin 53.1^\circ \approx 0.8,\;\cos 53.1^\circ \approx 0.6.$$

Horizontal force balance. The horizontal components of the two tensions are equal and opposite:

$$T_1\sin\theta_1 = T_2\sin\theta_2.$$

$$0.6\,T_1 = 0.8\,T_2 \quad\Longrightarrow\quad T_1 = \dfrac{4}{3}T_2.$$

Vertical force balance.

$$T_1\cos\theta_1 + T_2\cos\theta_2 = W.$$

$$0.8\,T_1 + 0.6\,T_2 = W.$$

$$0.8\cdot\dfrac{4}{3}T_2 + 0.6\,T_2 = W \quad\Longrightarrow\quad \dfrac{32}{30}T_2 + \dfrac{18}{30}T_2 = W \quad\Longrightarrow\quad \dfrac{5}{3}T_2 = W.$$

$$T_2 = 0.6\,W,\qquad T_1 = \dfrac{4}{3}(0.6\,W) = 0.8\,W.$$

Rotational equilibrium. Take moments about $$A$$ (so $$T_1$$ contributes no torque). Since the bar is horizontal, only the vertical components of the forces produce a torque about $$A$$:

$$T_2\cos\theta_2\cdot L = W\cdot d.$$

$$d = \dfrac{T_2\cos\theta_2\cdot L}{W} = \dfrac{(0.6\,W)(0.6)(2)}{W}\,\mathrm{m} = 0.72\,\mathrm{m}.$$

The centre of gravity of the bar is therefore located at a distance $$d = 0.72\,\mathrm{m}$$ from the left end.

Total weight of the car

$$W = M\,g = 1800 \times 9.8 = 17{,}640\,\mathrm{N}.$$

Let $$N_f$$ be the total normal reaction at the two front wheels and $$N_b$$ at the two back wheels. The wheelbase is $$L = 1.8\,\mathrm{m}$$ and the CG lies $$a = 1.05\,\mathrm{m}$$ behind the front axle, so it is at distance $$L - a = 0.75\,\mathrm{m}$$ ahead of the back axle.

Force balance (vertical).

$$N_f + N_b = W = 17{,}640\,\mathrm{N}.$$

Torque balance about the back axle. Only $$N_f$$ and $$W$$ produce moments about this axis (the moment arm of $$N_b$$ is zero):

$$N_f\cdot L = W\cdot(L - a).$$

$$N_f = \dfrac{W(L - a)}{L} = \dfrac{17{,}640 \times 0.75}{1.8} = 7350\,\mathrm{N}.$$

Then

$$N_b = W - N_f = 17{,}640 - 7350 = 10{,}290\,\mathrm{N}.$$

By symmetry left/right, the load on each pair of wheels is shared equally:

$$\text{Force on each front wheel} = \dfrac{N_f}{2} = \dfrac{7350}{2} = 3675\,\mathrm{N}.$$

$$\text{Force on each back wheel} = \dfrac{N_b}{2} = \dfrac{10{,}290}{2} = 5145\,\mathrm{N}.$$

The back wheels carry more weight because the CG is closer to the back axle.

Let each body have mass $$M$$ and radius $$R$$, and let the same torque $$\tau$$ be applied to each (starting from rest).

Moments of inertia about the chosen axes:

• Hollow cylinder about its symmetry axis: $$I_{\text{cyl}} = M R^2$$.
• Solid sphere about an axis through its centre: $$I_{\text{sph}} = \dfrac{2}{5}M R^2$$.

From $$\tau = I\alpha$$, the angular acceleration of each is

$$\alpha_{\text{cyl}} = \dfrac{\tau}{M R^2},\qquad \alpha_{\text{sph}} = \dfrac{\tau}{(2/5)\,M R^2} = \dfrac{5\,\tau}{2\,M R^2}.$$

Comparing,

$$\dfrac{\alpha_{\text{sph}}}{\alpha_{\text{cyl}}} = \dfrac{5}{2} = 2.5.$$

Both start from rest, so after a time $$t$$ their angular speeds are $$\omega_{\text{cyl}} = \alpha_{\text{cyl}}\,t$$ and $$\omega_{\text{sph}} = \alpha_{\text{sph}}\,t$$. Hence

$$\dfrac{\omega_{\text{sph}}}{\omega_{\text{cyl}}} = \dfrac{5}{2} = 2.5.$$

The solid sphere acquires a greater angular speed (in fact $$2.5$$ times that of the hollow cylinder) because it has a smaller moment of inertia.

For a solid cylinder rotating about its symmetry axis, the moment of inertia is

$$I = \dfrac{1}{2}M R^2 = \dfrac{1}{2}(20)(0.25)^2 = \dfrac{1}{2}(20)(0.0625) = 0.625\,\mathrm{kg\,m^2}.$$

Kinetic energy of rotation.

$$K = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}(0.625)(100)^2 = \dfrac{1}{2}(0.625)(10{,}000) = 3125\,\mathrm{J}.$$

Angular momentum about the axis.

$$L = I\omega = 0.625 \times 100 = 62.5\,\mathrm{kg\,m^2\,s^{-1}}.$$

Take $$M = 3\,\mathrm{kg}$$, $$R = 0.40\,\mathrm{m}$$, $$F = 30\,\mathrm{N}$$.

Moment of inertia of a hollow cylinder (thin shell) about its symmetry axis:

$$I = M R^2 = 3 \times (0.40)^2 = 3 \times 0.16 = 0.48\,\mathrm{kg\,m^2}.$$

The rope is wound on the rim, so the pull on it provides a tangential force at distance $$R$$ from the axis. Torque about the axis:

$$\tau = F\,R = 30 \times 0.40 = 12\,\mathrm{N\,m}.$$

From $$\tau = I\alpha$$,

$$\alpha = \dfrac{\tau}{I} = \dfrac{12}{0.48} = 25\,\mathrm{rad/s^2}.$$

Because the rope does not slip, the linear acceleration of the rope equals the tangential acceleration of the rim:

$$a = R\,\alpha = 0.40 \times 25 = 10\,\mathrm{m/s^2}.$$

The instantaneous power delivered by a torque to a rotating body is the rotational analogue of $$P = Fv$$:

$$P = \tau\,\omega.$$

Substituting $$\tau = 180\,\mathrm{N\,m}$$ and $$\omega = 200\,\mathrm{rad/s}$$,

$$P = 180 \times 200 = 36{,}000\,\mathrm{W} = 36\,\mathrm{kW}.$$

This power is exactly what the engine must supply to overcome frictional torque and keep the angular speed constant (since the engine is taken to be 100% efficient).

Set the origin $$O$$ at the centre of the original disc, with the $$x$$-axis along the line joining $$O$$ to the centre $$O'$$ of the hole. Then $$O' = (R/2,\,0)$$.

Let $$\sigma$$ be the (uniform) mass per unit area. Let $$M$$ be the mass of the original (full) disc, so

$$M = \sigma\,\pi R^2.$$

The mass of the removed disc (the hole) is

$$m = \sigma\,\pi(R/2)^2 = \sigma\,\pi R^2/4 = M/4.$$

Use the standard trick of treating the holed disc as (full disc) $$-$$ (small disc). By symmetry, the full disc has its CG at $$O = (0,0)$$ and the small (removed) disc has its CG at $$O' = (R/2,\,0)$$.

If $$x_{CM}$$ is the $$x$$-coordinate of the CG of the remaining (holed) body, the requirement that the CG of full disc equals the weighted average of (remaining body CM) and (hole CM) gives

$$M\cdot 0 = (M - m)\,x_{CM} + m\cdot\dfrac{R}{2}.$$

$$x_{CM} = -\dfrac{m\cdot R/2}{M - m} = -\dfrac{(M/4)(R/2)}{M - M/4} = -\dfrac{M R/8}{3M/4} = -\dfrac{R}{6}.$$

The negative sign indicates that the centre of gravity of the holed disc lies on the side of $$O$$ opposite to the hole. So the new CG is at distance $$R/6$$ from the original centre $$O$$, along the line joining $$O$$ to $$O'$$ but on the opposite side from the hole.

If the metre stick balances on a knife edge at its centre when no coins are placed, the centre of gravity of the (uniform) stick must lie at the $$50.0\,\mathrm{cm}$$ mark, so $$M$$ (the mass of the stick) can be treated as concentrated at that point.

After two coins of total mass $$m = 5 + 5 = 10\,\mathrm{g}$$ are stacked at the $$12.0\,\mathrm{cm}$$ mark, the new balance point is at the $$45.0\,\mathrm{cm}$$ mark. For equilibrium about this new pivot, the moments of the two weights about the pivot must cancel:

$$m\,g\,(45.0 - 12.0) = M\,g\,(50.0 - 45.0).$$

The factor $$g$$ cancels. Both moment arms are in centimetres; we can keep them so since they will appear in a ratio:

$$10\,\mathrm{g}\times 33.0\,\mathrm{cm} = M \times 5.0\,\mathrm{cm}.$$

$$M = \dfrac{10 \times 33.0}{5.0}\,\mathrm{g} = \dfrac{330}{5}\,\mathrm{g} = 66\,\mathrm{g}.$$

So the metre stick weighs $$66\,\mathrm{g}$$.

Given:

• Mass of molecule $$m = 5.30\times 10^{-26}\,\mathrm{kg}$$.
• Moment of inertia $$I = 1.94\times 10^{-46}\,\mathrm{kg\,m^2}$$.
• Mean translational speed $$v = 500\,\mathrm{m/s}$$.
• $$K_{\text{rot}} = \dfrac{2}{3}K_{\text{trans}}$$.

The translational kinetic energy is

$$K_{\text{trans}} = \dfrac{1}{2}m v^2,$$

and the rotational kinetic energy is

$$K_{\text{rot}} = \dfrac{1}{2}I\omega^2.$$

Using the given ratio,

$$\dfrac{1}{2}I\omega^2 = \dfrac{2}{3}\left(\dfrac{1}{2}m v^2\right).$$

$$I\omega^2 = \dfrac{2}{3}m v^2 \quad\Longrightarrow\quad \omega^2 = \dfrac{2 m v^2}{3 I}.$$

Substituting numerical values,

$$\omega^2 = \dfrac{2 \times 5.30\times 10^{-26} \times (500)^2}{3 \times 1.94\times 10^{-46}}.$$

$$\omega^2 = \dfrac{2 \times 5.30\times 10^{-26}\times 2.5\times 10^{5}}{5.82\times 10^{-46}}.$$

$$\omega^2 = \dfrac{2.65\times 10^{-20}}{5.82\times 10^{-46}} \approx 4.553\times 10^{25}\,\mathrm{rad^2/s^2}.$$

$$\omega \approx \sqrt{4.553\times 10^{25}} \approx 6.75\times 10^{12}\,\mathrm{rad/s}.$$

So the average angular speed of an oxygen molecule under these conditions is about $$6.75\times 10^{12}\,\mathrm{rad/s}$$.