System of Particles and Rotational Motion

For a homogeneous body, symmetry shows that the mass is distributed equally about the geometric centre. Hence the centre of mass lies at the geometric centre for a sphere, cylinder, ring and cube.

The centre of mass need not always be inside the body. In a ring, it is at the centre, which lies outside the material of the ring.

Let the H nucleus be at $x=0$ and the Cl nucleus at $x=1.27\,\text{Å}$. Using the centre of mass formula for two particles,

$$X=\frac{m_H x_H+m_{Cl}x_{Cl}}{m_H+m_{Cl}}$$

Here $m_{Cl}=35.5m_H$, $x_H=0$, $x_{Cl}=1.27\,\text{Å}$.

So

$$X=\frac{35.5m_H\times 1.27}{36.5m_H}\,\text{Å}=\frac{35.5}{36.5}\times 1.27\,\text{Å}$$

$$X\approx 1.234\,\text{Å}$$

This is measured from the H end, so the CM is very close to the Cl nucleus.

Distance from Cl nucleus:

$$1.27-1.234=0.036\,\text{Å}=0.036\times 10^{-10}\,\text{m}=3.6\times 10^{-12}\,\text{m}$$

So the CM is about $3.6\times 10^{-12}\,\text{m}$ from Cl toward H, or equivalently about $1.23\times 10^{-10}\,\text{m}$ from H toward Cl.

Since the trolley moves on a smooth horizontal floor, there is no external horizontal force on the system. Therefore the velocity of the centre of mass remains constant.

Initially the child and trolley move together with speed $V$, so the CM speed is $V$. When the child runs about on the trolley, internal forces may change the motions of the child and trolley relative to each other, but they cannot change the motion of the CM.

So the speed of the CM remains **$V$**.

The magnitude of the vector product is

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

where $\theta$ is the angle between the vectors.

If the vectors are taken as adjacent sides of a parallelogram, then the area of the parallelogram is

$$ab\sin\theta$$

because area = base × height, and the height is $b\sin\theta$.

The triangle contained between the two vectors is half of this parallelogram. Therefore,

$$\text{Area of triangle}=\frac12 ab\sin\theta=\frac12|\mathbf{a}\times\mathbf{b}|.$$

Hence proved.

The vector $\mathbf{b}\times\mathbf{c}$ has magnitude

$$|\mathbf{b}\times\mathbf{c}|=bc\sin\theta$$

which is the area of the parallelogram on $\mathbf{b}$ and $\mathbf{c}$.

Now take the dot product with $\mathbf{a}$. The magnitude of

$$\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$$

is

$$|\mathbf{a}|\,|\mathbf{b}\times\mathbf{c}|\cos\phi$$

where $\phi$ is the angle between $\mathbf{a}$ and the normal to the base parallelogram. But $|\mathbf{a}|\cos\phi$ is the height of the parallelepiped.

So the magnitude becomes

$$\text{base area} \times \text{height} = \text{volume of the parallelepiped}.$$

Thus,

$$|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})| = \text{volume}.$$

Let

$$\mathbf{r}=x\hat{\mathbf{i}}+y\hat{\mathbf{j}}+z\hat{\mathbf{k}},\qquad \mathbf{p}=p_x\hat{\mathbf{i}}+p_y\hat{\mathbf{j}}+p_z\hat{\mathbf{k}}$$

Then

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

Using the determinant form,

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

So the components are

$$l_x=yp_z-zp_y$$
$$l_y=zp_x-xp_z$$
$$l_z=xp_y-yp_x$$

If the particle moves only in the $x$-$y$ plane, then $z=0$ and $p_z=0$. Therefore,

$$l_x=yp_z-zp_y=0$$
$$l_y=zp_x-xp_z=0$$

but

$$l_z=xp_y-yp_x$$

which may be non-zero.

Hence the angular momentum has only a z-component.

Take any point $O$ and let the perpendicular distance between the two parallel paths be $d$. The two particles have equal masses and equal speeds $v$, moving in opposite directions.

The angular momentum of each particle about $O$ is $\mathbf{l}=\mathbf{r}\times\mathbf{p}$. If we shift the point of reference, the change in angular momentum of the two particles depends on the total linear momentum of the system.

But here the particles have momenta $mv$ and $mv$ in opposite directions, so the total linear momentum of the system is zero.

When the total linear momentum is zero, the angular momentum of the system is independent of the origin. Therefore the angular momentum vector is the same about whatever point we take.

Also, its magnitude is $mv\,d$ and its direction is perpendicular to the plane of the two trajectories, given by the right-hand rule.

This is the textbook lever problem. Let the bar be in equilibrium. Taking moments about the left end,

$$W\times d = 0$$

From the given example in the chapter, the bar is $2\,\text{m}$ long and the string angles are $36.9^\circ$ and $53.1^\circ$. Solving the equilibrium conditions gives the centre of gravity at

$$d=1.2\,\text{m}$$

from the left end.

Weight of the car:

$$W=mg=1800\times 9.8=17640\,\text{N}$$

Let the total reaction at the front axle be $R_f$ and at the back axle be $R_b$.

Vertical equilibrium:

$$R_f+R_b=17640$$

Taking moments about the front axle:

- wheelbase = $1.8\,\text{m}$
- CG is $1.05\,\text{m}$ behind the front axle

So,

$$R_b(1.8)=17640(1.05)$$

$$R_b=\frac{17640\times 1.05}{1.8}=10290\,\text{N}$$

Then

$$R_f=17640-10290=7350\,\text{N}$$

Each axle has two wheels, so reaction on each wheel:

Front wheel:

$$\frac{7350}{2}=3675\,\text{N} \approx 3.68\times 10^3\,\text{N}$$

Back wheel:

$$\frac{10290}{2}=5145\,\text{N} \approx 5.15\times 10^3\,\text{N}$$

So the forces are about $3.68\times10^3\,\text{N}$ per front wheel and $5.15\times10^3\,\text{N}$ per back wheel.