System of Particles and Rotational Motion

For a body of uniform mass density, the centre of mass (CM) coincides with the geometric centre of the body.

(i) Sphere: The CM is at the geometric centre of the sphere, i.e., at the centre point equidistant from all points on the surface.

(ii) Cylinder: The CM is at the midpoint of the axis of the cylinder, i.e., at the geometric centre on the axis of symmetry.

(iii) Ring: The CM is at the geometric centre of the ring, which lies at the centre of the circle (not on the material of the ring itself).

(iv) Cube: The CM is at the geometric centre of the cube, i.e., at the point of intersection of the body diagonals.

Does the CM necessarily lie inside the body?
No, the centre of mass does not necessarily lie inside the body. For example:
- For a ring, the CM lies at the centre of the ring, which is not part of the material of the ring.
- For a hollow sphere or a horseshoe-shaped object, the CM lies outside the material of the body.

Thus, the CM can lie outside the body.

Given:
- Separation between H and Cl nuclei: $d = 1.27\,\text{Å} = 1.27 \times 10^{-10}\,\text{m}$
- Mass of Cl atom: $m_{\text{Cl}} = 35.5\,m$ where $m$ = mass of H atom
- Mass of H atom: $m_{\text{H}} = m$

Setting up coordinates:
Let the H nucleus be at the origin ($x = 0$) and the Cl nucleus be at $x = 1.27\,\text{Å}$.

Formula for CM:
$$x_{\text{CM}} = \frac{m_{\text{H}} \cdot x_{\text{H}} + m_{\text{Cl}} \cdot x_{\text{Cl}}}{m_{\text{H}} + m_{\text{Cl}}}$$

Substituting values:
$$x_{\text{CM}} = \frac{m \times 0 + 35.5m \times 1.27}{m + 35.5m}$$

$$x_{\text{CM}} = \frac{35.5 \times 1.27}{36.5}\,\text{Å}$$

$$x_{\text{CM}} = \frac{45.085}{36.5}\,\text{Å}$$

$$\boxed{x_{\text{CM}} \approx 1.235\,\text{Å} \approx 1.24 \times 10^{-10}\,\text{m}}$$

Conclusion: The CM of the HCl molecule is located approximately $1.24\,\text{Å}$ from the hydrogen atom (or $0.03\,\text{Å}$ from the Cl nucleus), i.e., very close to the heavier chlorine atom.

Given:
- Initial speed of the system (trolley + child) = $V$
- The floor is smooth (frictionless), so no external horizontal force acts on the system.
- The child runs about on the trolley — this is an internal action.

Concept: When no external force acts on a system, the velocity of the centre of mass remains constant (Newton's first law applied to the system).

$$\mathbf{F}_{\text{ext}} = 0 \implies \frac{d\mathbf{P}}{dt} = 0 \implies \mathbf{P} = \text{constant}$$

Since $\mathbf{P} = M\mathbf{V}_{\text{CM}}$ and total mass $M$ is constant:
$$\mathbf{V}_{\text{CM}} = \text{constant} = V$$

The running of the child on the trolley is an internal action and cannot change the momentum of the system.

Answer: The speed of the CM of the (trolley + child) system remains $V$, unchanged.

To Show: Area of triangle formed by vectors $\mathbf{a}$ and $\mathbf{b}$ $= \dfrac{1}{2}|\mathbf{a} \times \mathbf{b}|$

Proof:

Let vectors $\mathbf{a}$ and $\mathbf{b}$ be represented by two sides of a triangle (with the same tail/origin), and let $\theta$ be the angle between them.

The two vectors $\mathbf{a}$ and $\mathbf{b}$ form a parallelogram. The triangle formed between $\mathbf{a}$ and $\mathbf{b}$ is exactly half of this parallelogram.

Area of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$:
$$A_{\text{parallelogram}} = |\mathbf{a}||\mathbf{b}|\sin\theta = |\mathbf{a} \times \mathbf{b}|$$

This is because the cross product $\mathbf{a} \times \mathbf{b}$ has magnitude $ab\sin\theta$, which equals the area of the parallelogram with sides $a$ and $b$ and included angle $\theta$.

Area of the triangle = $\dfrac{1}{2} \times$ base $\times$ height

The base of the triangle is $|\mathbf{a}|$ and the height is $|\mathbf{b}|\sin\theta$.

$$A_{\text{triangle}} = \frac{1}{2}|\mathbf{a}||\mathbf{b}|\sin\theta = \frac{1}{2}|\mathbf{a} \times \mathbf{b}|$$

Since the triangle occupies exactly half the area of the parallelogram:
$$\boxed{A_{\text{triangle}} = \frac{1}{2}|\mathbf{a} \times \mathbf{b}|}$$

Hence proved.

To Show: $|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$ = Volume of the parallelepiped formed by $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$.

Proof:

Let $\mathbf{b}$ and $\mathbf{c}$ be two sides of the base of the parallelepiped, and $\mathbf{a}$ be the third edge.

Step 1: Area of the base

The cross product $\mathbf{b} \times \mathbf{c}$ gives a vector whose:
- Magnitude $= |\mathbf{b}||\mathbf{c}|\sin\phi$ = Area of the parallelogram base (where $\phi$ is the angle between $\mathbf{b}$ and $\mathbf{c}$)
- Direction is perpendicular to the base (along the normal to the base)

So, $|\mathbf{b} \times \mathbf{c}|$ = Area of base $= A$

Step 2: Height of the parallelepiped

Let $\theta$ be the angle between $\mathbf{a}$ and the vector $\mathbf{b} \times \mathbf{c}$ (i.e., the normal to the base).

The height of the parallelepiped:
$$h = |\mathbf{a}|\cos\theta$$

Step 3: Volume
$$V = A \times h = |\mathbf{b} \times \mathbf{c}| \times |\mathbf{a}|\cos\theta$$

Step 4: Scalar triple product
$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = |\mathbf{a}||\mathbf{b} \times \mathbf{c}|\cos\theta$$

Comparing:
$$|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = |\mathbf{a}||\mathbf{b} \times \mathbf{c}|\cos\theta = V$$

$$\boxed{|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = \text{Volume of the parallelepiped}}$$

Hence proved.

Given:
- Position vector: $\mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}$
- Momentum: $\mathbf{p} = p_x\hat{i} + p_y\hat{j} + p_z\hat{k}$

Angular momentum:
$$\mathbf{l} = \mathbf{r} \times \mathbf{p}$$

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

$$\mathbf{l} = \hat{i}(yp_z - zp_y) - \hat{j}(xp_z - zp_x) + \hat{k}(xp_y - yp_x)$$

Components of angular momentum:
$$\boxed{l_x = yp_z - zp_y}$$
$$\boxed{l_y = zp_x - xp_z}$$
$$\boxed{l_z = xp_y - yp_x}$$

Special case: Motion in the $x$-$y$ plane

If the particle moves only in the $x$-$y$ plane, then:
- $z = 0$ (particle stays in the plane)
- $p_z = 0$ (no momentum component perpendicular to the plane)

Substituting $z = 0$ and $p_z = 0$:
$$l_x = y(0) - (0)p_y = 0$$
$$l_y = (0)p_x - x(0) = 0$$
$$l_z = xp_y - yp_x \neq 0 \text{ (in general)}$$

Therefore, when the particle moves in the $x$-$y$ plane, the angular momentum has only a $z$-component:
$$\mathbf{l} = (xp_y - yp_x)\hat{k}$$

Hence proved.

Given:
- Two particles, each of mass $m$ and speed $v$
- They travel in opposite directions along parallel lines
- The parallel lines are separated by a distance $d$

Setup:
Let particle 1 move along the line $y = 0$ in the $+x$ direction, and particle 2 move along the line $y = d$ in the $-x$ direction.

Let the reference point $P$ be at coordinates $(a, b)$.

Angular momentum of particle 1 about P:

Position of particle 1 relative to $P$: $\mathbf{r}_1 = (x_1 - a)\hat{i} + (0 - b)\hat{j}$

Momentum of particle 1: $\mathbf{p}_1 = mv\hat{i}$

$$\mathbf{l}_1 = \mathbf{r}_1 \times \mathbf{p}_1 = [(x_1-a)\hat{i} - b\hat{j}] \times mv\hat{i}$$
$$= mv(x_1-a)(\hat{i}\times\hat{i}) - mvb(\hat{j}\times\hat{i})$$
$$= 0 - mvb(-\hat{k}) = mvb\hat{k}$$

Angular momentum of particle 2 about P:

Position of particle 2 relative to $P$: $\mathbf{r}_2 = (x_2 - a)\hat{i} + (d - b)\hat{j}$

Momentum of particle 2: $\mathbf{p}_2 = -mv\hat{i}$

$$\mathbf{l}_2 = \mathbf{r}_2 \times \mathbf{p}_2 = [(x_2-a)\hat{i} + (d-b)\hat{j}] \times (-mv\hat{i})$$
$$= -mv(x_2-a)(\hat{i}\times\hat{i}) - mv(d-b)(\hat{j}\times\hat{i})$$
$$= 0 - mv(d-b)(-\hat{k}) = mv(d-b)\hat{k}$$

Total angular momentum:
$$\mathbf{L} = \mathbf{l}_1 + \mathbf{l}_2 = mvb\hat{k} + mv(d-b)\hat{k}$$
$$= mv[b + d - b]\hat{k}$$
$$\boxed{\mathbf{L} = mvd\hat{k}}$$

The total angular momentum $\mathbf{L} = mvd\hat{k}$ is independent of $a$ and $b$ (the coordinates of the reference point $P$).

Hence, the angular momentum of the two-particle system is the same whatever be the point about which it is calculated. $\blacksquare$

Given:
- Weight of bar = $W$
- Angle of left string with vertical: $\theta_1 = 36.9°$
- Angle of right string with vertical: $\theta_2 = 53.1°$
- Length of bar = $2\,\text{m}$
- Bar is in static equilibrium

Let $T_1$ = tension in left string, $T_2$ = tension in right string.

Condition 1: Translational equilibrium (vertical)

The strings are attached to the ceiling and make angles with the vertical. The horizontal components of tension must balance each other, and vertical components must support $W$.

For the bar in equilibrium:
$$T_1\cos 36.9° + T_2\cos 53.1° = W \quad \text{...(1)}$$

Condition 2: Translational equilibrium (horizontal)
$$T_1\sin 36.9° = T_2\sin 53.1° \quad \text{...(2)}$$

Using $\sin 36.9° = 0.6$, $\cos 36.9° = 0.8$, $\sin 53.1° = 0.8$, $\cos 53.1° = 0.6$:

From equation (2):
$$T_1 \times 0.6 = T_2 \times 0.8$$
$$T_1 = \frac{4}{3}T_2 \quad \text{...(3)}$$

Substituting in equation (1):
$$\frac{4}{3}T_2 \times 0.8 + T_2 \times 0.6 = W$$
$$\frac{3.2}{3}T_2 + 0.6T_2 = W$$
$$T_2\left(\frac{3.2 + 1.8}{3}\right) = W$$
$$T_2 \times \frac{5}{3} = W$$
$$T_2 = \frac{3W}{5} = 0.6W$$

From (3): $T_1 = \frac{4}{3} \times 0.6W = 0.8W$

Condition 3: Rotational equilibrium (torques about left end)

Taking torques about the left end of the bar:
- Torque due to $T_1$ (at left end) = 0
- Torque due to $W$ (at distance $d$, clockwise) = $W \cdot d$
- Torque due to $T_2$ (at right end, i.e., at $2\,\text{m}$, anticlockwise) = $T_2 \times 2$

For rotational equilibrium:
$$T_2 \times 2 = W \times d$$
$$0.6W \times 2 = W \times d$$
$$d = 1.2\,\text{m}$$

The centre of gravity of the bar is at $d = 1.2\,\text{m}$ from the left end.

Given:
- Mass of car: $m = 1800\,\text{kg}$
- Distance between axles: $L = 1.8\,\text{m}$
- Distance of CG from front axle: $l_f = 1.05\,\text{m}$
- Distance of CG from back axle: $l_b = 1.8 - 1.05 = 0.75\,\text{m}$
- $g = 10\,\text{m/s}^2$

Weight of car:
$$W = mg = 1800 \times 10 = 18000\,\text{N}$$

Let $F_f$ = total reaction force on both front wheels, $F_b$ = total reaction force on both back wheels.

Condition 1: Translational equilibrium (vertical)
$$F_f + F_b = W = 18000\,\text{N} \quad \text{...(1)}$$

Condition 2: Rotational equilibrium

Taking torques about the front axle:
$$F_b \times 1.8 = W \times 1.05$$
$$F_b = \frac{18000 \times 1.05}{1.8} = \frac{18900}{1.8} = 10500\,\text{N}$$

From equation (1):
$$F_f = 18000 - 10500 = 7500\,\text{N}$$

Force on each wheel:

Since there are 2 front wheels and 2 back wheels:
$$\text{Force on each front wheel} = \frac{F_f}{2} = \frac{7500}{2} = \boxed{3750\,\text{N}}$$

$$\text{Force on each back wheel} = \frac{F_b}{2} = \frac{10500}{2} = \boxed{5250\,\text{N}}$$

Given:
- Equal torque $\tau$ applied to both
- Same mass $m$ and radius $R$
- Same time $t$

Moment of inertia:

For a hollow cylinder about its axis of symmetry:
$$I_{\text{cylinder}} = mR^2$$

For a solid sphere about an axis through its centre:
$$I_{\text{sphere}} = \frac{2}{5}mR^2$$

Clearly: I_{\text{sphere}} < I_{\text{cylinder}}

Angular acceleration:

Using $\tau = I\alpha$:
$$\alpha = \frac{\tau}{I}$$

$$\alpha_{\text{cylinder}} = \frac{\tau}{mR^2}$$

$$\alpha_{\text{sphere}} = \frac{\tau}{\frac{2}{5}mR^2} = \frac{5\tau}{2mR^2}$$

Since \frac{5}{2} > 1:
\alpha_{\text{sphere}} > \alpha_{\text{cylinder}}

Angular speed after time $t$ (starting from rest):
$$\omega = \alpha t$$

\omega_{\text{sphere}} = \frac{5\tau t}{2mR^2} > \omega_{\text{cylinder}} = \frac{\tau t}{mR^2}

Conclusion: The solid sphere will acquire a greater angular speed after a given time, because its moment of inertia is smaller ($\frac{2}{5}mR^2$) compared to the hollow cylinder ($mR^2$), and hence it has a larger angular acceleration for the same applied torque.

Given:
- Mass: $m = 20\,\text{kg}$
- Angular speed: $\omega = 100\,\text{rad s}^{-1}$
- Radius: $R = 0.25\,\text{m}$

Moment of inertia of solid cylinder about its axis:
$$I = \frac{1}{2}mR^2 = \frac{1}{2} \times 20 \times (0.25)^2$$
$$I = \frac{1}{2} \times 20 \times 0.0625 = 0.625\,\text{kg m}^2$$

Kinetic energy of rotation:
$$K = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.625 \times (100)^2$$
$$K = \frac{1}{2} \times 0.625 \times 10000$$
$$\boxed{K = 3125\,\text{J}}$$

Angular momentum:
$$L = I\omega = 0.625 \times 100$$
$$\boxed{L = 62.5\,\text{J s} = 62.5\,\text{kg m}^2\text{s}^{-1}}$$

(a) Finding the new angular speed:

Given:
- Initial angular speed: $\omega_1 = 40\,\text{rev/min}$
- New moment of inertia: $I_2 = \dfrac{2}{5}I_1$
- No external torque (frictionless turntable)

Applying conservation of angular momentum:
$$I_1\omega_1 = I_2\omega_2$$

$$\omega_2 = \frac{I_1}{I_2}\omega_1 = \frac{I_1}{\frac{2}{5}I_1} \times 40$$

$$\omega_2 = \frac{5}{2} \times 40$$

$$\boxed{\omega_2 = 100\,\text{rev/min}}$$

(b) Comparing kinetic energies:

Initial kinetic energy:
$$K_1 = \frac{1}{2}I_1\omega_1^2$$

Final kinetic energy:
$$K_2 = \frac{1}{2}I_2\omega_2^2 = \frac{1}{2} \times \frac{2}{5}I_1 \times (100)^2$$

Ratio:
$$\frac{K_2}{K_1} = \frac{\frac{1}{2} \times \frac{2}{5}I_1 \times (100)^2}{\frac{1}{2} \times I_1 \times (40)^2}$$

$$\frac{K_2}{K_1} = \frac{2}{5} \times \frac{10000}{1600} = \frac{2}{5} \times \frac{25}{4} = \frac{50}{20} = 2.5$$

$$K_2 = 2.5\,K_1$$

Thus K_2 > K_1, i.e., the new kinetic energy is 2.5 times the initial kinetic energy.

Account for the increase: The increase in kinetic energy comes from the work done by the child in folding his arms against the centrifugal force (in the rotating frame) or equivalently, the child uses his internal (muscular) energy to pull his arms inward. This internal energy is converted into the additional kinetic energy of rotation. Angular momentum is conserved, but kinetic energy is not, since an internal force does work on the system.

Given:
- Mass of hollow cylinder: $m = 3\,\text{kg}$
- Radius: $R = 40\,\text{cm} = 0.4\,\text{m}$
- Force applied on rope: $F = 30\,\text{N}$

Moment of inertia of hollow cylinder about its axis:
$$I = mR^2 = 3 \times (0.4)^2 = 3 \times 0.16 = 0.48\,\text{kg m}^2$$

Torque applied:
$$\tau = F \times R = 30 \times 0.4 = 12\,\text{N m}$$

Angular acceleration:
Using $\tau = I\alpha$:
$$\alpha = \frac{\tau}{I} = \frac{12}{0.48}$$
$$\boxed{\alpha = 25\,\text{rad s}^{-2}}$$

Linear acceleration of the rope:

Since there is no slipping, the linear acceleration of the rope equals the tangential acceleration at the rim:
$$a = R\alpha = 0.4 \times 25$$
$$\boxed{a = 10\,\text{m s}^{-2}}$$

Given:
- Angular speed: $\omega = 200\,\text{rad s}^{-1}$
- Torque: $\tau = 180\,\text{N m}$
- Efficiency = 100%

Formula for power in rotational motion:
$$P = \tau \cdot \omega$$

Substituting values:
$$P = 180 \times 200$$

$$\boxed{P = 36000\,\text{W} = 36\,\text{kW}}$$

Note: The torque is needed to overcome friction. Since the engine is 100% efficient, all the power input goes into maintaining the angular speed against frictional torque.

Given:
- Original disk: radius $R$, uniform surface mass density $\sigma$
- Hole: radius $R/2$, centre at distance $R/2$ from centre of original disk

Method: Treat the removed portion as a negative mass.

Mass of original disk:
$$M = \sigma \pi R^2$$

Mass of removed circular portion:
$$m = \sigma \pi \left(\frac{R}{2}\right)^2 = \frac{\sigma \pi R^2}{4} = \frac{M}{4}$$

Mass of remaining body:
$$M' = M - m = M - \frac{M}{4} = \frac{3M}{4}$$

Setting up coordinates:
Let the centre of the original disk be at the origin. Let the centre of the hole be at $x = +R/2$ (on the positive $x$-axis).

Centre of mass of original disk: $x_1 = 0$

Centre of mass of removed portion: $x_2 = +R/2$

Using the formula for CM of composite body:

The original disk = remaining body + removed portion

$$M \cdot x_{\text{original}} = M' \cdot x_{\text{CM}} + m \cdot x_2$$

$$M \times 0 = \frac{3M}{4} \times x_{\text{CM}} + \frac{M}{4} \times \frac{R}{2}$$

$$0 = \frac{3M}{4} x_{\text{CM}} + \frac{MR}{8}$$

$$\frac{3M}{4} x_{\text{CM}} = -\frac{MR}{8}$$

$$x_{\text{CM}} = -\frac{R}{8} \times \frac{4}{3} = -\frac{R}{6}$$

$$\boxed{x_{\text{CM}} = -\frac{R}{6}}$$

The centre of gravity of the resulting flat body is at a distance $R/6$ from the centre of the original disk, on the side opposite to the hole (i.e., at $x = -R/6$).

Given:
- Metre stick initially balanced at 50 cm (centre)
- Two coins, each of mass 5 g, placed at 12.0 cm mark
- New balance point: 45.0 cm
- Total mass of coins: $m = 2 \times 5 = 10\,\text{g}$

Concept: When balanced at 45.0 cm, the net torque about the new pivot (45.0 cm) is zero.

Let $M$ = mass of the metre stick. The CM of the metre stick is at 50.0 cm.

Taking torques about the new pivot at 45.0 cm:

- Distance of coins (at 12.0 cm) from pivot: $45.0 - 12.0 = 33.0\,\text{cm}$ (on the left side)
- Distance of CM of stick (at 50.0 cm) from pivot: $50.0 - 45.0 = 5.0\,\text{cm}$ (on the right side)

Rotational equilibrium:

Anticlockwise torque = Clockwise torque

$$m \times 33.0 = M \times 5.0$$

$$10 \times 33.0 = M \times 5.0$$

$$M = \frac{10 \times 33.0}{5.0} = \frac{330}{5}$$

$$\boxed{M = 66\,\text{g}}$$

The mass of the metre stick is 66 g.

Given:
- Mass of O₂ molecule: $m = 5.30 \times 10^{-26}\,\text{kg}$
- Moment of inertia: $I = 1.94 \times 10^{-46}\,\text{kg m}^2$
- Mean speed: $v = 500\,\text{m/s}$
- $K_{\text{rot}} = \dfrac{2}{3}K_{\text{trans}}$

Kinetic energy of translation:
$$K_{\text{trans}} = \frac{1}{2}mv^2 = \frac{1}{2} \times 5.30 \times 10^{-26} \times (500)^2$$
$$= \frac{1}{2} \times 5.30 \times 10^{-26} \times 2.5 \times 10^5$$
$$= \frac{1}{2} \times 1.325 \times 10^{-20}$$
$$= 6.625 \times 10^{-21}\,\text{J}$$

Kinetic energy of rotation:
$$K_{\text{rot}} = \frac{2}{3}K_{\text{trans}} = \frac{2}{3} \times 6.625 \times 10^{-21}$$
$$= 4.417 \times 10^{-21}\,\text{J}$$

Finding angular velocity:
$$K_{\text{rot}} = \frac{1}{2}I\omega^2$$

$$\omega^2 = \frac{2K_{\text{rot}}}{I} = \frac{2 \times 4.417 \times 10^{-21}}{1.94 \times 10^{-46}}$$

$$\omega^2 = \frac{8.834 \times 10^{-21}}{1.94 \times 10^{-46}} = 4.55 \times 10^{25}\,\text{rad}^2\text{s}^{-2}$$

$$\omega = \sqrt{4.55 \times 10^{25}}$$

$$\omega = \sqrt{4.55} \times 10^{12.5} = 2.133 \times 10^{12.5}$$

$$\omega = 2.133 \times 3.162 \times 10^{12}$$

$$\boxed{\omega \approx 6.7 \times 10^{12}\,\text{rad s}^{-1}}$$

The average angular velocity of the oxygen molecule is approximately $6.7 \times 10^{12}\,\text{rad s}^{-1}$.