Systems Of Particles And Rotational MotionClass 11 Physics NCERT Solutions

Given: Let the mass of the hydrogen atom be $m_H = m$. Mass of the chlorine atom, $m_{Cl} = 35.5m$. Separation between the atoms, $d = 1.27 \AA$.

To Find: The location of the Centre of Mass (CM) of the HCl molecule.

Calculation: Let us place the hydrogen atom at the origin of a coordinate system. So, its position is $x_H = 0$. The position of the chlorine atom will be $x_{Cl} = 1.27 \AA$.

The position of the centre of mass, $X_{CM}$, is given by the formula: $X_{CM} = \frac{m_H x_H + m_{Cl} x_{Cl}}{m_H + m_{Cl}}$

Substituting the values: $X_{CM} = \frac{m(0) + 35.5m(1.27 \AA)}{m + 35.5m}$ $X_{CM} = \frac{35.5m \times 1.27 \AA}{36.5m}$ $X_{CM} = \frac{35.5 \times 1.27}{36.5} \AA$ $X_{CM} \approx 1.235 \AA$

Final Answer: The centre of mass of the HCl molecule is located at approximately $1.235 \AA$ from the hydrogen atom, along the line joining the two atoms.

The system consists of the trolley and the child. The forces involved when the child runs on the trolley (like the force of friction between the child's feet and the trolley) are internal forces to this system.

According to Newton's first law for a system of particles, if the net external force on the system is zero, the velocity of its centre of mass remains constant.

In this case, the trolley is moving on a smooth horizontal floor. This means there is no external horizontal force (like friction) acting on the (trolley + child) system. The forces of gravity and the normal reaction from the floor are vertical forces, which cancel each other out. Thus, the net external force on the system is zero.

Since there is no net external force, the velocity of the centre of mass of the (trolley + child) system will not change. Initially, the child is stationary relative to the trolley, and the entire system moves with a uniform speed $V$. Therefore, the speed of the centre of mass of the system is $V$.

Even when the child runs about on the trolley, the velocity of the centre of mass of the system will remain unchanged.

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

Let two vectors $\mathbf{a}$ and $\mathbf{b}$ represent the adjacent sides of a parallelogram, say OP and OQ.

The magnitude of the vector product (cross product) of these two vectors is given by: $|\mathbf{a} \times \mathbf{b}| = ab \sin\theta$ where $a$ and $b$ are the magnitudes of vectors $\mathbf{a}$ and $\mathbf{b}$ respectively, and $\theta$ is the angle between them.

Geometrically, the magnitude $|\mathbf{a} \times \mathbf{b}|$ represents the area of the parallelogram with adjacent sides given by vectors $\mathbf{a}$ and $\mathbf{b}$. Area of parallelogram OPRQ = $ab \sin\theta = |\mathbf{a} \times \mathbf{b}|$.

The triangle contained between the vectors $\mathbf{a}$ and $\mathbf{b}$ can be considered as half of this parallelogram. The diagonal OR divides the parallelogram into two triangles of equal area, one of which is triangle OPQ.

Therefore, the area of the triangle with sides $\mathbf{a}$ and $\mathbf{b}$ is: Area of $\triangle OPQ = \frac{1}{2} \times (\text{Area of parallelogram OPRQ})$ $\text{Area of } \triangle OPQ = \frac{1}{2} |\mathbf{a} \times \mathbf{b}|$

Thus, the area of the triangle contained between the vectors $\mathbf{a}$ and $\mathbf{b}$ is one half of the magnitude of $\mathbf{a} \times \mathbf{b}$. Hence proved.

Let the three vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ represent the adjacent edges of a parallelepiped.

The scalar triple product of these three vectors is given by $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$.

First, consider the vector product $\mathbf{b} \times \mathbf{c}$. The magnitude of this vector is: $|\mathbf{b} \times \mathbf{c}| = bc \sin\phi$ where $\phi$ is the angle between vectors $\mathbf{b}$ and $\mathbf{c}$. This magnitude represents the area of the parallelogram forming the base of the parallelepiped. Area of base = $|\mathbf{b} \times \mathbf{c}|$. The direction of the vector $\mathbf{d} = \mathbf{b} \times \mathbf{c}$ is perpendicular to the plane containing $\mathbf{b}$ and $\mathbf{c}$.

Now, consider the dot product of vector $\mathbf{a}$ with the vector $\mathbf{d} = (\mathbf{b} \times \mathbf{c})$: $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{a} \cdot \mathbf{d} = |\mathbf{a}| |\mathbf{d}| \cos\theta$ where $\theta$ is the angle between vector $\mathbf{a}$ and vector $\mathbf{d}$.

Geometrically, $|\mathbf{a}| \cos\theta$ is the component of vector $\mathbf{a}$ along the direction of $\mathbf{d}$. Since $\mathbf{d}$ is normal to the base of the parallelepiped, $|\mathbf{a}| \cos\theta$ represents the height ($h$) of the parallelepiped. So, $h = a \cos\theta$.

Substituting this into the expression: $|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = (\text{Area of base}) \times (\text{Height})$ $|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = |\mathbf{b} \times \mathbf{c}| \times (a \cos\theta)$ This is the formula for the volume of the parallelepiped.

Final Answer: The magnitude of the scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ is equal to the volume of the parallelepiped whose adjacent edges are represented by the vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$. Hence proved.

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

To Find: Components of angular momentum $\mathbf{l}$ along the x, y, and z axes.

Formula: The angular momentum $\mathbf{l}$ is defined as: $\mathbf{l} = \mathbf{r} \times \mathbf{p}$

Calculation: We can calculate the cross product using the determinant form: $\mathbf{l} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ x & y & z \ p_x & p_y & p_z \end{vmatrix}$ Expanding the determinant: $\mathbf{l} = \hat{\mathbf{i}}(yp_z - zp_y) - \hat{\mathbf{j}}(xp_z - zp_x) + \hat{\mathbf{k}}(xp_y - yp_x)$ $\mathbf{l} = (yp_z - zp_y)\hat{\mathbf{i}} + (zp_x - xp_z)\hat{\mathbf{j}} + (xp_y - yp_x)\hat{\mathbf{k}}$

The components of the angular momentum $\mathbf{l}$ are: $l_x = yp_z - zp_y$ $l_y = zp_x - xp_z$ $l_z = xp_y - yp_x$

Condition for motion in the x-y plane: If the particle moves only in the x-y plane, its z-coordinate is always zero, so $z=0$. Also, its velocity has no z-component, which means its momentum has no z-component, so $p_z = 0$.

Substituting $z=0$ and $p_z=0$ into the component equations: $l_x = y(0) - (0)p_y = 0$ $l_y = (0)p_x - x(0) = 0$ $l_z = xp_y - yp_x$

So, the angular momentum vector becomes: $\mathbf{l} = 0\hat{\mathbf{i}} + 0\hat{\mathbf{j}} + (xp_y - yp_x)\hat{\mathbf{k}} = (xp_y - yp_x)\hat{\mathbf{k}}$

Conclusion: When the particle moves only in the x-y plane, the x and y components of its angular momentum are zero. The angular momentum has only a z-component. Hence proved.

Let the two particles be $P_1$ and $P_2$. Let their masses be $m_1=m$ and $m_2=m$. Let their velocities be $\mathbf{v}_1 = v\hat{\mathbf{j}}$ and $\mathbf{v}_2 = -v\hat{\mathbf{j}}$. Let the parallel lines be along the y-axis. Let the line for $P_1$ be at $x=0$ and the line for $P_2$ be at $x=d$. So, their position vectors are $\mathbf{r}_1 = y_1\hat{\mathbf{j}}$ and $\mathbf{r}_2 = d\hat{\mathbf{i}} + y_2\hat{\mathbf{j}}$.

Let's choose an arbitrary point O with position vector $\mathbf{r}_O = x_O\hat{\mathbf{i}} + y_O\hat{\mathbf{j}} + z_O\hat{\mathbf{k}}$ as the origin for calculating angular momentum.

The position vector of particle $P_1$ with respect to O is $\mathbf{R}_1 = \mathbf{r}_1 - \mathbf{r}_O = -x_O\hat{\mathbf{i}} + (y_1-y_O)\hat{\mathbf{j}} - z_O\hat{\mathbf{k}}$. The position vector of particle $P_2$ with respect to O is $\mathbf{R}_2 = \mathbf{r}_2 - \mathbf{r}_O = (d-x_O)\hat{\mathbf{i}} + (y_2-y_O)\hat{\mathbf{j}} - z_O\hat{\mathbf{k}}$.

The angular momentum of the system about O is $\mathbf{L} = \mathbf{L}_1 + \mathbf{L}_2 = \mathbf{R}_1 \times (m\mathbf{v}_1) + \mathbf{R}_2 \times (m\mathbf{v}_2)$.

$\mathbf{L}_1 = m(\mathbf{R}_1 \times \mathbf{v}_1) = m [(-x_O\hat{\mathbf{i}} + (y_1-y_O)\hat{\mathbf{j}} - z_O\hat{\mathbf{k}}) \times (v\hat{\mathbf{j}})]$ $\mathbf{L}_1 = m [-x_O v (\hat{\mathbf{i}} \times \hat{\mathbf{j}}) - z_O v (\hat{\mathbf{k}} \times \hat{\mathbf{j}})] = m[-x_O v \hat{\mathbf{k}} + z_O v \hat{\mathbf{i}}]$

$\mathbf{L}_2 = m(\mathbf{R}_2 \times \mathbf{v}_2) = m [((d-x_O)\hat{\mathbf{i}} + (y_2-y_O)\hat{\mathbf{j}} - z_O\hat{\mathbf{k}}) \times (-v\hat{\mathbf{j}})]$ $\mathbf{L}_2 = m [-(d-x_O)v (\hat{\mathbf{i}} \times \hat{\mathbf{j}}) + z_O v (\hat{\mathbf{k}} \times \hat{\mathbf{j}})] = m[-(d-x_O)v \hat{\mathbf{k}} - z_O v \hat{\mathbf{i}}]$

Total angular momentum $\mathbf{L} = \mathbf{L}_1 + \mathbf{L}_2$ $\mathbf{L} = m[-x_O v \hat{\mathbf{k}} + z_O v \hat{\mathbf{i}}] + m[-(d-x_O)v \hat{\mathbf{k}} - z_O v \hat{\mathbf{i}}]$ $\mathbf{L} = m[(-x_O v - dv + x_O v)\hat{\mathbf{k}} + (z_O v - z_O v)\hat{\mathbf{i}}]$ $\mathbf{L} = m(-dv)\hat{\mathbf{k}} = -mvd\hat{\mathbf{k}}$

The magnitude of the angular momentum is $mvd$ and its direction is along the negative z-axis.

This final expression for $\mathbf{L}$ is independent of the coordinates $(x_O, y_O, z_O)$ of the arbitrary point O. Therefore, the angular momentum vector of the two-particle system is the same whatever be the point about which the angular momentum is taken. Hence proved.

Given: Length of the bar, $L = 2 \text{ m}$. Angle with vertical at left end, $\theta_1 = 36.9^\circ$. Angle with vertical at right end, $\theta_2 = 53.1^\circ$. Weight of the bar = $W$. Let $T_1$ and $T_2$ be the tensions in the left and right strings respectively. Let $d$ be the distance of the centre of gravity (CG) from the left end.

To Find: The distance $d$ of the CG from the left end.

Calculation: The bar is in equilibrium. This means the net force and net torque on the bar are zero.

Condition for Translational Equilibrium: Resolving forces horizontally: $T_1 \sin\theta_1 - T_2 \sin\theta_2 = 0$ $T_1 \sin(36.9^\circ) = T_2 \sin(53.1^\circ)$ $T_1 (0.6) = T_2 (0.8)$ (Since $\sin(36.9^\circ) \approx 0.6$ and $\sin(53.1^\circ) \approx 0.8$) $T_1 = \frac{0.8}{0.6} T_2 = \frac{4}{3} T_2 \quad ... (i)$

Resolving forces vertically: $T_1 \cos\theta_1 + T_2 \cos\theta_2 - W = 0$ $T_1 \cos(36.9^\circ) + T_2 \cos(53.1^\circ) = W$ $T_1 (0.8) + T_2 (0.6) = W \quad ... (ii)$

Substitute (i) into (ii): $(\frac{4}{3} T_2)(0.8) + T_2(0.6) = W$ $\frac{3.2}{3} T_2 + 0.6 T_2 = W$ $(\frac{3.2 + 1.8}{3}) T_2 = W$ $\frac{5}{3} T_2 = W \implies T_2 = \frac{3}{5} W$ From (i), $T_1 = \frac{4}{3} (\frac{3}{5} W) = \frac{4}{5} W$.

Condition for Rotational Equilibrium: Let's take torques about the left end of the bar. The sum of torques must be zero. Torque due to $T_1$ is zero as it passes through the pivot. Anticlockwise torque due to $T_2$ = Clockwise torque due to weight $W$. $(T_2 \cos\theta_2) \times L = W \times d$ We must consider the perpendicular components of forces. Let's take torques about the CG. Torque about CG = 0 $(T_1 \cos\theta_1) \times d - (T_2 \cos\theta_2) \times (L-d) = 0$ $T_1 \cos\theta_1 \times d = T_2 \cos\theta_2 \times (L-d)$ $(\frac{4}{5} W) (0.8) d = (\frac{3}{5} W) (0.6) (2-d)$ $0.64 d = 0.36 (2-d)$ $0.64 d = 0.72 - 0.36 d$ $0.64 d + 0.36 d = 0.72$ $1.00 d = 0.72$ $d = 0.72 \text{ m}$

Final Answer: The distance of the centre of gravity of the bar from its left end is $0.72 \text{ m}$.

Given: Mass of the car, $M = 1800 \text{ kg}$. Weight of the car, $W = Mg = 1800 \times 9.8 = 17640 \text{ N}$. Distance between axles, $L = 1.8 \text{ m}$. Distance of centre of gravity (CG) from front axle, $d_f = 1.05 \text{ m}$. Distance of CG from back axle, $d_b = 1.8 - 1.05 = 0.75 \text{ m}$. Let $R_f$ be the total reaction force on the two front wheels and $R_b$ be the total reaction force on the two back wheels.

To Find: Force on each front wheel and each back wheel.

Calculation: The car is in equilibrium.

Condition for Translational Equilibrium: The net upward force equals the net downward force. $R_f + R_b = W = 17640 \text{ N} \quad ... (i)$

Condition for Rotational Equilibrium: The net torque about any point is zero. Let's take torques about the front axle. Anticlockwise torque = Clockwise torque $R_b \times L = W \times d_f$ $R_b \times 1.8 = 17640 \times 1.05$ $R_b = \frac{17640 \times 1.05}{1.8} = 10290 \text{ N}$

Now, substitute the value of $R_b$ in equation (i): $R_f + 10290 = 17640$ $R_f = 17640 - 10290 = 7350 \text{ N}$

This is the total force on the two front wheels and two back wheels. Force on each front wheel = $\frac{R_f}{2} = \frac{7350}{2} = 3675 \text{ N}$. Force on each back wheel = $\frac{R_b}{2} = \frac{10290}{2} = 5145 \text{ N}$.

Final Answer: The force exerted by the level ground on each front wheel is $3675 \text{ N}$ and on each back wheel is $5145 \text{ N}$.

Given: Same torque $\tau$ is applied to both. Same mass $M$ and same radius $R$ for both.

To Find: Which body will acquire a greater angular speed $\omega$ after a time $t$.

Formula: The relation between torque, moment of inertia, and angular acceleration is: $\tau = I\alpha$ where $I$ is the moment of inertia and $\alpha$ is the angular acceleration. From this, $\alpha = \frac{\tau}{I}$.

The kinematic equation for angular motion is: $\omega = \omega_0 + \alpha t$ Assuming both start from rest, $\omega_0 = 0$. So, $\omega = \alpha t = (\frac{\tau}{I})t$.

Since $\tau$ and $t$ are the same for both bodies, the angular speed $\omega$ is inversely proportional to the moment of inertia $I$. $\omega \propto \frac{1}{I}$ The body with the smaller moment of inertia will acquire a greater angular speed.

Comparison: Comparing the two moments of inertia: $I_{sphere} = 0.4 MR^2 < MR^2 = I_{cylinder}$ So, the moment of inertia of the solid sphere is less than that of the hollow cylinder.

Since $\omega \propto \frac{1}{I}$, the solid sphere, having a smaller moment of inertia, will have a greater angular acceleration and thus will acquire a greater angular speed after a given time.

Final Answer: The solid sphere will acquire a greater angular speed.

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

To Find: (a) Rotational kinetic energy, $K_R$. (b) Magnitude of angular momentum, $L$.

Calculation: First, calculate the moment of inertia $I$: $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 20 \text{ kg} \times (0.25 \text{ m})^2$ $I = 10 \times 0.0625 = 0.625 \text{ kg m}^2$

(a) Kinetic Energy: $K_R = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.625 \text{ kg m}^2 \times (100 \text{ rad s}^{-1})^2$ $K_R = \frac{1}{2} \times 0.625 \times 10000 = 3125 \text{ J}$

(b) Angular Momentum: $L = I\omega = 0.625 \text{ kg m}^2 \times 100 \text{ rad s}^{-1}$ $L = 62.5 \text{ J s}$

Final Answer: The kinetic energy associated with the rotation of the cylinder is $3125 \text{ J}$. The magnitude of the angular momentum of the cylinder about its axis is $62.5 \text{ J s}$.

Given: Initial angular speed, $\omega_1 = 40 \text{ rev/min}$. Let the initial moment of inertia be $I_1$. Final moment of inertia, $I_2 = \frac{2}{5}I_1$.

Principle: Since the turntable rotates without friction, there is no external torque acting on the system (child + turntable). Therefore, the angular momentum of the system is conserved. $L_1 = L_2$ $I_1\omega_1 = I_2\omega_2$

Calculation: $I_1 \times (40 \text{ rev/min}) = (\frac{2}{5}I_1) \times \omega_2$ $\omega_2 = \frac{5}{2} \times 40 \text{ rev/min}$ $\omega_2 = 100 \text{ rev/min}$

Final Answer for (a): The new angular speed of the child is $100 \text{ rev/min}$.

Formula: Rotational kinetic energy, $K = \frac{1}{2}I\omega^2$. Initial kinetic energy, $K_1 = \frac{1}{2}I_1\omega_1^2$. Final kinetic energy, $K_2 = \frac{1}{2}I_2\omega_2^2$.

Calculation: Substitute $I_2 = \frac{2}{5}I_1$ and $\omega_2 = \frac{5}{2}\omega_1$ into the expression for $K_2$. $K_2 = \frac{1}{2} (\frac{2}{5}I_1) (\frac{5}{2}\omega_1)^2$ $K_2 = \frac{1}{2} (\frac{2}{5}I_1) (\frac{25}{4}\omega_1^2)$ $K_2 = (\frac{2 \times 25}{5 \times 4}) \times (\frac{1}{2}I_1\omega_1^2)$ $K_2 = (\frac{50}{20}) K_1 = \frac{5}{2} K_1 = 2.5 K_1$

Since $K_2 = 2.5 K_1$, the final kinetic energy is greater than the initial kinetic energy.

Accounting for the Increase in Kinetic Energy: The increase in kinetic energy comes from the internal work done by the child. To pull his arms in, the child has to do work against the centrifugal force acting on his arms. This muscular work done by the child is converted into rotational kinetic energy, causing the system to spin faster and have more kinetic energy. The energy is supplied internally by the child's body.

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

To Find: (a) Angular acceleration, $\alpha$. (b) Linear acceleration of the rope, $a$.

Calculation: First, calculate the moment of inertia $I$: $I = MR^2 = 3 \text{ kg} \times (0.4 \text{ m})^2 = 3 \times 0.16 = 0.48 \text{ kg m}^2$

Next, calculate the torque $\tau$ produced by the force: $\tau = F \times R = 30 \text{ N} \times 0.4 \text{ m} = 12 \text{ Nm}$

(a) Angular Acceleration ($\alpha$): Using $\tau = I\alpha$: $\alpha = \frac{\tau}{I} = \frac{12 \text{ Nm}}{0.48 \text{ kg m}^2} = 25 \text{ rad/s}^2$

(b) Linear Acceleration ($a$): Since there is no slipping, the linear acceleration of the rope is equal to the tangential acceleration at the rim of the cylinder. $a = R\alpha = 0.4 \text{ m} \times 25 \text{ rad/s}^2 = 10 \text{ m/s}^2$

Final Answer: The angular acceleration of the cylinder is $25 \text{ rad/s}^2$. The linear acceleration of the rope is $10 \text{ m/s}^2$.

Given: Uniform angular speed, $\omega = 200 \text{ rad s}^{-1}$. Torque transmitted by the engine, $\tau = 180 \text{ N m}$. Engine efficiency is 100%.

To Find: The power required by the engine, $P$.

Formula: The power ($P$) in rotational motion is given by the product of torque ($\tau$) and angular velocity ($\omega$): $P = \tau \omega$

Calculation: Substituting the given values into the formula: $P = 180 \text{ N m} \times 200 \text{ rad s}^{-1}$ $P = 36000 \text{ W}$ $P = 36 \text{ kW}$

Since the engine is 100% efficient, the power required by the engine is equal to the power delivered.

Final Answer: The power required by the engine is $36000 \text{ W}$ or $36 \text{ kW}$.

Let the original uniform disk have its centre at the origin (0, 0). Let $\sigma$ be the mass per unit area of the disk.

Original Disk: Radius = $R$. Mass, $M_1 = \sigma \times (\pi R^2)$. Centre of gravity, $C_1 = (0, 0)$.

Removed Portion (Hole): This can be treated as a body with negative mass. Radius = $R/2$. Mass, $M_2 = -\sigma \times \pi (R/2)^2 = -\frac{\sigma \pi R^2}{4} = -\frac{M_1}{4}$. Centre of the hole is at $R/2$ from the origin. Let's assume it is along the x-axis. So, its centre of gravity is $C_2 = (R/2, 0)$.

Resulting Body: The resulting body is the combination of the original disk and the removed portion. Let the centre of gravity of the resulting body be at $(X_{CG}, Y_{CG})$.

Formula for Centre of Gravity: $X_{CG} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2}$ $Y_{CG} = \frac{M_1 y_1 + M_2 y_2}{M_1 + M_2}$

Calculation: By symmetry, the centre of gravity of the resulting body will lie on the x-axis, so $Y_{CG} = 0$.

Let's calculate $X_{CG}$: $X_{CG} = \frac{(M_1)(0) + (-\frac{M_1}{4})(\frac{R}{2})}{M_1 + (-\frac{M_1}{4})}$ $X_{CG} = \frac{-\frac{M_1 R}{8}}{\frac{3M_1}{4}}$ $X_{CG} = -\frac{M_1 R}{8} \times \frac{4}{3M_1}$ $X_{CG} = -\frac{4R}{24} = -\frac{R}{6}$

The negative sign indicates that the centre of gravity shifts to the side opposite to where the hole was cut.

Final Answer: The centre of gravity of the resulting flat body is located at a distance of $R/6$ from the centre of the original disc, on the side opposite to the centre of the hole.

Given: Length of the metre stick = $100 \text{ cm}$. Initial balance point (fulcrum) = $50 \text{ cm}$ mark (centre of the stick). Mass of two coins, $m_c = 2 \times 5 \text{ g} = 10 \text{ g}$. Position of the coins = $12.0 \text{ cm}$ mark. New balance point (fulcrum) = $45.0 \text{ cm}$ mark. Let $M$ be the mass of the metre stick.

To Find: The mass of the metre stick, $M$.

Principle: For the stick to be in rotational equilibrium, the sum of clockwise moments about the new fulcrum must equal the sum of anticlockwise moments about the new fulcrum (Principle of Moments).

The weight of the metre stick acts at its centre of gravity, which is at the $50 \text{ cm}$ mark (since it was initially balanced there). The weight of the coins acts at the $12.0 \text{ cm}$ mark. The new fulcrum is at the $45.0 \text{ cm}$ mark.

Calculation: Anticlockwise Moment: The coins are at the $12.0 \text{ cm}$ mark, which is to the left of the new fulcrum at $45.0 \text{ cm}$. This creates an anticlockwise torque. Distance of coins from fulcrum = $45.0 \text{ cm} - 12.0 \text{ cm} = 33.0 \text{ cm}$. Anticlockwise moment = $m_c \times g \times (33.0)$.

Clockwise Moment: The centre of gravity of the stick is at the $50.0 \text{ cm}$ mark, which is to the right of the new fulcrum. This creates a clockwise torque. Distance of CG from fulcrum = $50.0 \text{ cm} - 45.0 \text{ cm} = 5.0 \text{ cm}$. Clockwise moment = $M \times g \times (5.0)$.

Equating moments: Anticlockwise moment = Clockwise moment $m_c \times g \times 33.0 = M \times g \times 5.0$ (The acceleration due to gravity, $g$, cancels out from both sides) $10 \text{ g} \times 33.0 = M \times 5.0$ $330 = 5M$ $M = \frac{330}{5} = 66 \text{ g}$

Final Answer: The mass of the metre stick is $66 \text{ g}$.

Given: Mass of the oxygen molecule, $M = 5.30 \times 10^{-26} \text{ kg}$. Moment of inertia, $I = 1.94 \times 10^{-46} \text{ kg m}^2$. Mean speed (translational), $v = 500 \text{ m/s}$. Relation between kinetic energies: $K_{rotation} = \frac{2}{3} K_{translation}$.

To Find: The average angular velocity of the molecule, $\omega$.

Calculation: First, calculate the translational kinetic energy: $K_{translation} = \frac{1}{2} \times (5.30 \times 10^{-26} \text{ kg}) \times (500 \text{ m/s})^2$ $K_{translation} = \frac{1}{2} \times 5.30 \times 10^{-26} \times 250000$ $K_{translation} = 6.625 \times 10^{-21} \text{ J}$

Next, calculate the rotational kinetic energy using the given relation: $K_{rotation} = \frac{2}{3} K_{translation} = \frac{2}{3} \times (6.625 \times 10^{-21} \text{ J})$ $K_{rotation} \approx 4.417 \times 10^{-21} \text{ J}$

Now, use the formula for rotational kinetic energy to find $\omega$: $K_{rotation} = \frac{1}{2}I\omega^2$ $\omega^2 = \frac{2 \times K_{rotation}}{I}$ $\omega^2 = \frac{2 \times (4.417 \times 10^{-21} \text{ J})}{1.94 \times 10^{-46} \text{ kg m}^2}$ $\omega^2 \approx 4.55 \times 10^{25} \text{ (rad/s)}^2$ $\omega = \sqrt{4.55 \times 10^{25}} \text{ rad/s}$ $\omega \approx 6.75 \times 10^{12} \text{ rad/s}$

Final Answer: The average angular velocity of the molecule is approximately $6.75 \times 10^{12} \text{ rad/s}$.