SYSTEMS OF PARTICLES AND ROTATIONAL MOTION-Exercise
Physics - Exercise
Q1.Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ?
Solution
For objects of uniform mass density, the centre of mass coincides with their geometric centre due to symmetry. Consider each shape individually to locate this point precisely.
(i) Solid Sphere: A uniform sphere exhibits complete spherical symmetry. The centre of mass lies at its geometric centre, which serves as the origin for any chosen coordinate system.
(ii) Cylinder: For a uniform cylinder (solid or hollow), rotational symmetry exists about its central axis. Thus, the centre of mass is located at the midpoint of the axis of symmetry, equidistant from both ends.
(iii) Ring: A thin ring possesses circular symmetry. The centre of mass is at the geometric centre of the ring, precisely at the intersection point of its diameters, though this point lies outside the material of the ring itself.
(iv) Cube: A uniform cube demonstrates symmetry along all three dimensions. The centre of mass occupies the geometric centre, found at the intersection of the body diagonals or midway along each edge dimension.
The centre of mass of a body does not necessarily lie inside the body. For instance, in a ring or hollow sphere, it resides at the empty geometric centre where no mass exists, demonstrating that external positioning is possible for certain configurations.
Q2. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Solution
In the HCl molecule, the nuclei separation is 1.27 Å, with chlorine atom mass approximately 35.5 times that of hydrogen. Place hydrogen nucleus at origin to locate centre of mass along the internuclear axis.
Solution:
Let mass of hydrogen atom \( m_H = m \) and mass of chlorine atom \( m_{Cl} = 35.5m \).
Separation between nuclei \( d = 1.27 \) Å, with hydrogen at \( x = 0 \) Å and chlorine at \( x = 1.27 \) Å.
$$\begin{aligned} x_{CM} &= \frac{m_H \cdot 0 + m_{Cl} \cdot 1.27}{m_H + m_{Cl}} \\\\ &= \frac{m \cdot 0 + 35.5m \cdot 1.27}{m + 35.5m} \\\\ &= \frac{35.5m \cdot 1.27}{36.5m} \\\\ &= \frac{35.5 \times 1.27}{36.5} \\\\ &= 1.24 \, \text{Å} \end{aligned}$$The centre of mass lies 1.24 Å from hydrogen nucleus towards chlorine nucleus, very close to chlorine due to its much larger mass.
Q3. A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?
Solution
The floor provides no external horizontal force since it remains smooth. The trolley moves uniformly, indicating zero net external force on the system.
Initially, both trolley and child move with velocity V. Total momentum equals (M + m)V where M is trolley mass and m is child mass.
With no external force, total momentum remains conserved. Centre of mass velocity stays constant at initial value.
$$\begin{aligned} \vec{v}_{CM} &= \frac{(M + m)V}{M + m} \\ &= V \end{aligned}$$The centre of mass continues moving with uniform speed V, regardless of the child's motion on the trolley. Internal movements cancel out due to action-reaction pairs between child and trolley.
Q4. Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
Solution
Consider the triangle formed by vectors a and b originating from the same point. The area of this triangle relates directly to the cross product magnitude |a × b|.
Take vector a as the base of the triangle with length |a|.
The height equals the perpendicular component of vector b from vector a, given by |b| sin θ where θ is the angle between them.
$$\begin{aligned} \text{Area } \Delta &= \frac{1}{2} \times \text{base} \times \text{height} \\\\ &= \frac{1}{2} \mid a\mid \cdot \mid b\mid \sin \theta \\\\ \text{But }\\ \mid a \times b\mid &= \mid a\mid \mid b\mid \sin \theta \\\\ \therefore \Delta &= \frac{1}{2} |a \times b| \end{aligned}$$Thus the triangle area equals half the magnitude of the cross product vector a × b.
Q5. Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a, b and c.
Solution
The scalar triple product a · (b × c) represents the volume of the parallelepiped formed by vectors a, b, and c.
Consider parallelogram formed by vectors b and c as the base.
Area of base equals magnitude of cross product |b × c|.
Vector a represents height measured along direction perpendicular to base.
$$\begin{aligned} \text{Volume } V &= \text{base area} \times \text{height} \\\\ &= \mid b \times c\mid \cdot h \\\\ \text{But } a \cdot (b \times c) &= \mid a\mid \mid b \times c\mid \cos \theta \\\\ \text{where } \theta &= 0^\circ \text{ (a perpendicular to base)} \\\\ \therefore V &= a \cdot (b \times c) \end{aligned}$$Thus scalar triple product a · (b × c) equals volume of parallelepiped, proving the relation.
Q6. Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components \(p_x,\ p_y\) and \(p_z\). Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Solution
The angular momentum L of a particle equals r × p. Determine its components along x, y, z axes and show that motion confined to x-y plane yields only z-component.
Position vector r = x\hat{i} + y\hat{j} + z\hat{k} and momentum p = p_x\hat{i} + p_y\hat{j} + p_z\hat{k}.
$$\begin{aligned} \vec{L} &= \vec{r} \times \vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix} \\\\ &= (yp_z - zp_y)\hat{i} + (zp_x - xp_z)\hat{j} + (xp_y - yp_x)\hat{k} \end{aligned}$$Components of angular momentum are:
$$\begin{aligned} L_x &= yp_z - zp_y \\ L_y &= zp_x - xp_z \\ L_z &= xp_y - yp_x \end{aligned}$$For motion in x-y plane, z = 0 and p_z = 0.
$$\begin{aligned} L_x &= yp_z - zp_y = y(0) - (0)p_y = 0 \\ L_y &= zp_x - xp_z = (0)p_x - x(0) = 0 \\ L_z &= xp_y - yp_x \quad (\text{unchanged}) \end{aligned}$$Thus only z-component L_z remains non-zero when particle moves solely in x-y plane.
Q7.Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.
Solution
Two particles each of mass m and speed v move in opposite directions along parallel lines separated by distance d. The angular momentum of this system remains constant regardless of reference point choice.
Place particles along parallel lines \(y = +d/2\) and \(y = -d/2\). Velocities: particle 1 has |(\vec{v_1} = v\hat{i}\), particle 2 has \(\vec{v_2} = -v\hat{i}\).
Consider arbitrary point P(x_0, y_0). Position vectors relative to P: \(\vec{r_1} = (x_1 - x_0)\hat{i} + (d/2 - y_0)\hat{j}\), \(\vec{r_2} = (x_2 - x_0)\hat{i} + (-d/2 - y_0)\hat{j}\).
$$\begin{aligned} \vec{L} &= \vec{r_1} \times \vec{p_1} + \vec{r_2} \times \vec{p_2} \\\\ &= m[\vec{r_1} \times \vec{v_1} + \vec{r_2} \times \vec{v_2}] \\\\ L_z &= m[(x_1-x_0)v(d/2-y_0)(0) + (d/2-y_0)(-v)(0)] \\\\ &\quad + m[(x_2-x_0)(-v)(-d/2-y_0)(0) + (-d/2-y_0)(v)(0)] \\\\ &= m v d \end{aligned}$$Only z-components contribute. Notice x-coordinates cancel and \((d/2 - y_0) + (-d/2 - y_0) = -2y_0\) terms vanish, leaving \(L_z = mvd\) independent of \((x_0, y_0)\).
Thus angular momentum vector magnitude and direction remain constant for any reference point, proving the result.
Q8. A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
Solution
The length of the bar is \(2\,\text{m}\) and its weight is \(W\). Let the tensions in the left and right strings be \(T_1\) and \(T_2\) respectively. Since the bar is at rest, it is in both translational and rotational equilibrium.
Resolving the forces horizontally, the horizontal components of the two tensions must be equal.
\[ \begin{aligned} T_1 \sin 36.9^\circ &= T_2 \sin 53.1^\circ \\\\ T_2 &= T_1 \frac{\sin 36.9^\circ}{\sin 53.1^\circ} \\\\ T_2 &= \frac{0.60}{0.80} T_1 \\\\ T_2 &= 0.75\,T_1 \end{aligned} \]
Resolving the forces vertically, the sum of the vertical components of the tensions balances the weight of the bar.
\[ \begin{aligned} T_1 \cos 36.9^\circ + T_2 \cos 53.1^\circ &= W \\\\ T_1(0.80) + (0.75T_1)(0.60) &= W \\\\ T_1(0.80 + 0.45) &= W \\\\ 1.25T_1 &= W \end{aligned} \]
Let the centre of gravity of the bar be at a distance \(d\) from the left end. Taking moments about the left end removes the torque due to \(T_1\). For rotational equilibrium, clockwise and anticlockwise moments must be equal.
\[ \begin{aligned} T_2 \cos 53.1^\circ \times 2 &= W \times d \\\\ (0.75T_1)(0.60) \times 2 &= (1.25T_1)d \\\\ d &= \frac{0.75 \times 0.60 \times 2}{1.25}\\\\ d &= \frac{0.90}{1.25} \\\\ d &= 0.72\,\text{m} \end{aligned} \]
Therefore, the centre of gravity of the bar is located at a distance of \(72\,\text{cm}\) from the left end.
Q9. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Solution
The mass of the car is \(1800\,\text{kg}\). Hence its weight is \(W = 1800 \times 9.8\,\text{N}\). The distance between the front and rear axles is \(1.8\,\text{m}\). The centre of gravity of the car lies \(1.05\,\text{m}\) behind the front axle, so its distance from the rear axle is \(1.8 - 1.05 = 0.75\,\text{m}\).
Let the total normal reaction exerted by the ground on the front axle be \(R_f\) and that on the rear axle be \(R_r\). Since the car is at rest on a level ground, it is in translational and rotational equilibrium.
Taking moments about the centre of gravity ensures rotational equilibrium of the car.
\[ \begin{aligned} R_f \times 1.05 &= R_r \times 0.75 \\ \frac{R_f}{R_r} &= \frac{0.75}{1.05} \\ \frac{R_f}{R_r} &= \frac{5}{7} \\ R_f &= \frac{5}{7} R_r \end{aligned} \]
From vertical force balance, the sum of the reactions must equal the weight of the car.
\[ \begin{aligned} R_f + R_r &= 1800 \times 9.8 \\ \frac{5}{7}R_r + R_r &= 1800 \times 9.8 \\ \frac{12}{7}R_r &= 1800 \times 9.8 \\ R_r &= \frac{1800 \times 9.8 \times 7}{12} \end{aligned} \]
\[ \begin{aligned} R_r &= 10290\,\text{N} \\ R_f &= \frac{5}{7} \times 10290 \\ R_f &= 7350\,\text{N} \end{aligned} \]
Since each axle has two wheels, the reaction force on each wheel is half the total reaction on that axle.
\[ \begin{aligned} \text{Reaction on each rear wheel} &= \frac{10290}{2} = 5145\,\text{N} \\ \text{Reaction on each front wheel} &= \frac{7350}{2} = 3675\,\text{N} \end{aligned} \]
Thus, the level ground exerts a force of \(3675\,\text{N}\) on each front wheel and \(5145\,\text{N}\) on each back wheel.
Q10. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.
Solution
Let a constant torque \(\tau\) be applied to both the hollow cylinder and the solid sphere. Each body has the same mass \(M\) and radius \(R\), and both are free to rotate about an axis passing through their respective centres. Since the torques are equal and act for the same duration, the angular speed acquired by each body depends on its angular acceleration.
According to the rotational form of Newton’s second law, the angular acceleration \(\alpha\) produced by a torque \(\tau\) is given by
\[ \begin{aligned} \tau = I \alpha \\ \alpha = \frac{\tau}{I} \end{aligned} \]
The moment of inertia of a hollow cylinder of mass \(M\) and radius \(R\) about its axis of symmetry is
\[ \begin{aligned} I_{\text{cylinder}} = MR^2 \end{aligned} \]
The moment of inertia of a solid sphere of the same mass and radius about an axis through its centre is
\[ \begin{aligned} I_{\text{sphere}} = \frac{2}{5}MR^2 \end{aligned} \]
Since both bodies experience the same torque, their angular accelerations are inversely proportional to their moments of inertia.
\[ \begin{aligned} \alpha_{\text{cylinder}} &= \frac{\tau}{MR^2} \\ \alpha_{\text{sphere}} &= \frac{\tau}{\tfrac{2}{5}MR^2} = \frac{5\tau}{2MR^2} \end{aligned} \]
Clearly, the angular acceleration of the solid sphere is greater than that of the hollow cylinder. If the torque acts for the same time \(t\), the angular speed acquired by each body is given by \(\omega = \alpha t\).
\[ \begin{aligned} \omega_{\text{cylinder}} &= \alpha_{\text{cylinder}} t \\ \omega_{\text{sphere}} &= \alpha_{\text{sphere}} t \end{aligned} \]
Since \(\alpha_{\text{sphere}} > \alpha_{\text{cylinder}}\), it follows that the solid sphere acquires a greater angular speed than the hollow cylinder after the same time interval.
Q11. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Solution
The mass of the solid cylinder is \(20\,\text{kg}\), its radius is \(0.25\,\text{m}\), and it rotates with an angular speed of \(100\,\text{rad s}^{-1}\). The kinetic energy associated with rotation depends on the moment of inertia of the cylinder about its axis.
For a solid cylinder rotating about its axis of symmetry, the moment of inertia is given by
\[ \begin{aligned} I &= \frac{1}{2}MR^2 \\ &= \frac{1}{2}\times 20 \times (0.25)^2 \\ &= 10 \times 0.0625 \\ &= 0.625\,\text{kg m}^2 \end{aligned} \]
The rotational kinetic energy of the cylinder is
\[ \begin{aligned} KE &= \frac{1}{2}I\omega^2 \\ &= \frac{1}{2}\times 0.625 \times (100)^2 \\ &= 0.3125 \times 10000 \\ &= 3125\,\text{J} \end{aligned} \]
The angular momentum of the cylinder about its axis is given by
\[ \begin{aligned} L &= I\omega \\ &= 0.625 \times 100 \\ &= 62.5\,\text{kg m}^2\text{s}^{-1} \end{aligned} \]
Thus, the kinetic energy associated with the rotation of the cylinder is \(3125\,\text{J}\), and the magnitude of its angular momentum about the axis is \(62.5\,\text{kg m}^2\text{s}^{-1}\).
Q12. (a) A child stands at the centre of a turntable with his two arms outstretched. The
turntable is set rotating with an angular speed of 40 rev/min. How much is the
angular speed of the child if he folds his hands back and thereby reduces his
moment of inertia to 2/5 times the initial value ? Assume that the turntable
rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial
kinetic energy of rotation. How do you account for this increase in kinetic energy?
Solution
The child and the turntable together form an isolated rotating system because the turntable is assumed to be frictionless. Hence, there is no external torque acting on the system and the angular momentum remains conserved throughout the motion.
Let the initial angular speed of the child be \(\omega_1 = 40\,\text{rev min}^{-1}\) and the initial moment of inertia be \(I_1\). When the child folds his hands, his moment of inertia reduces to \(I_2 = \frac{2}{5} I_1\).
Using the law of conservation of angular momentum,
\[ \begin{aligned} I_1 \omega_1 &= I_2 \omega_2 \\ I_1 \omega_1 &= \frac{2}{5} I_1 \omega_2 \\ \omega_2 &= \frac{5}{2} \omega_1 \end{aligned} \]
Substituting the given value of \(\omega_1\),
\[ \begin{aligned} \omega_2 &= \frac{5}{2} \times 40 \\ \omega_2 &= 100\,\text{rev min}^{-1} \end{aligned} \]
Thus, the angular speed of the child increases to \(100\,\text{rev min}^{-1}\) when he folds his hands.
To compare the kinetic energies, the rotational kinetic energy is expressed as \(K = \frac{1}{2} I \omega^2\). The initial kinetic energy of rotation is
\[ \begin{aligned} K_1 &= \frac{1}{2} I_1 \omega_1^2 \end{aligned} \]
The new kinetic energy after folding the hands is
\[ \begin{aligned} K_2 &= \frac{1}{2} I_2 \omega_2^2 \\ &= \frac{1}{2} \left(\frac{2}{5} I_1\right) \left(\frac{5}{2} \omega_1\right)^2 \\ &= \frac{1}{2} I_1 \omega_1^2 \times \frac{5}{2} \end{aligned} \]
This gives
\[ \begin{aligned} K_2 &= \frac{5}{2} K_1 \end{aligned} \]
Hence, the new kinetic energy of rotation is greater than the initial kinetic energy. The increase in kinetic energy arises because the child does work while pulling his arms inward. This work is done by muscular forces against the outward inertial tendency of the arms, and it is converted into additional rotational kinetic energy of the system.
Q13. A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping.
Solution
The rope is of negligible mass and is wound around a hollow cylinder of mass \(3\,\text{kg}\) and radius \(0.40\,\text{m}\). A force of \(30\,\text{N}\) is applied to the rope. Since there is no slipping between the rope and the cylinder, the linear motion of the rope and the rotational motion of the cylinder are directly related.
The applied force produces a torque about the axis of the cylinder. The magnitude of the torque is given by the product of the force and the radius of the cylinder.
\[ \begin{aligned} \tau &= F R \\ &= 30 \times 0.40 \\ &= 12\,\text{N m} \end{aligned} \]
For a hollow cylinder rotating about its axis, the moment of inertia is \(I = MR^2\).
\[ \begin{aligned} I &= 3 \times (0.40)^2 \\ &= 3 \times 0.16 \\ &= 0.48\,\text{kg m}^2 \end{aligned} \]
The angular acceleration of the cylinder is obtained using the rotational equation of motion \(\tau = I\alpha\).
\[ \begin{aligned} \alpha &= \frac{\tau}{I} \\ &= \frac{12}{0.48} \\ &= 25\,\text{rad s}^{-2} \end{aligned} \]
Since there is no slipping, the linear acceleration of the rope is related to the angular acceleration of the cylinder by the relation \(a = \alpha R\).
\[ \begin{aligned} a &= \alpha R \\ &= 25 \times 0.40 \\ &= 10\,\text{m s}^{-2} \end{aligned} \]
Therefore, the angular acceleration of the cylinder is \(25\,\text{rad s}^{-2}\), and the linear acceleration of the rope is \(10\,\text{m s}^{-2}\).
Q14. To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Solution
The rotor is maintained at a constant angular speed of \(200\,\text{rad s}^{-1}\). In ideal conditions with no friction, no torque would be required to keep the rotor rotating uniformly. However, in real situations, frictional forces oppose the motion, and the engine must supply a torque equal to the frictional torque to maintain uniform angular speed.
The torque that must be transmitted by the engine is given as \(180\,\text{N m}\). The power delivered by an engine producing a torque \(\tau\) while rotating with angular speed \(\omega\) is expressed as
\[ \begin{aligned} P &= \tau \omega \end{aligned} \]
Substituting the given values of torque and angular speed,
\[ \begin{aligned} P &= 180 \times 200 \\ &= 36000\,\text{W} \end{aligned} \]
Since the engine is assumed to be \(100\%\) efficient, the mechanical power output is equal to the power required to counter the frictional torque.
Therefore, the power required by the engine to maintain the rotor at a uniform angular speed is \(3.6 \times 10^{4}\,\text{W}\), or \(36\,\text{kW}\).
Q15. From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Solution
Consider the original uniform disc of radius \(R\) and uniform surface mass density. Let the centre of the original disc be taken as the origin \(O\). A circular hole of radius \(R/2\) is cut out, whose centre lies at a distance \(R/2\) from \(O\). Since the body is uniform and flat, the position of the centre of gravity depends only on the areas of the disc and the hole.
Let the mass per unit area be \(\sigma\). The mass of the original disc is proportional to its area, and the same applies to the removed part.
\[ \begin{aligned} \text{Area of original disc} &= \pi R^2 \\\\ \text{Area of hole} &= \pi\left(\frac{R}{2}\right)^2 \\\\ &= \frac{\pi R^2}{4} \end{aligned} \]
The remaining area of the disc is therefore
\[ \begin{aligned} A &= \pi R^2 - \frac{\pi R^2}{4} \\\\ &= \frac{3}{4}\pi R^2 \end{aligned} \]
The centre of mass of the original disc is at the origin, while the centre of mass of the removed hole lies at a distance \(R/2\) from the origin along the chosen axis. Treating the removed part as having negative mass, the position \(x\) of the centre of gravity of the remaining body is given by
\[ \begin{aligned} x &= \frac{(0)\times (\pi R^2) - \left(\frac{\pi R^2}{4}\right)\left(\frac{R}{2}\right)}{\frac{3}{4}\pi R^2} \end{aligned} \]
Simplifying the expression,
\[ \begin{aligned} x &= -\frac{\frac{\pi R^3}{8}}{\frac{3}{4}\pi R^2} \\\\ &= -\frac{R}{6} \end{aligned} \]
The negative sign indicates that the centre of gravity shifts away from the hole. Hence, the centre of gravity of the remaining flat body lies at a distance \(\frac{R}{6}\) from the centre of the original disc, along the line joining the centres, on the side opposite to the hole.
Q16. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Solution
The metre stick is uniform, so its centre of gravity lies at the 50.0 cm mark. Initially, it balances at its centre. When two coins of mass \(5\,\text{g}\) each are placed together at the 12.0 cm mark, the system balances on a knife edge at the 45.0 cm mark. This means that the combined centre of gravity of the stick and the coins lies at 45.0 cm.
Let the mass of the metre stick be \(M\) grams. The total mass of the two coins is \(10\,\text{g}\). For rotational equilibrium about the knife edge at 45.0 cm, the clockwise and anticlockwise moments must be equal.
\[ \begin{aligned} \text{Moment due to coins} &= \text{Moment due to metre stick} \\ 10 \times (45 - 12) &= M \times (50 - 45) \end{aligned} \]
Simplifying the distances,
\[ \begin{aligned} 10 \times 33 &= M \times 5 \end{aligned} \]
Solving for \(M\),
\[ \begin{aligned} M &= \frac{10 \times 33}{5} \\\\ &= 66 \end{aligned} \]
Therefore, the mass of the metre stick is \(66\,\text{g}\).
Q17. The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 ×10-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
Solution
The mass of an oxygen molecule is \(m = 5.30 \times 10^{-26}\,\text{kg}\) and its moment of inertia about the given axis is \(I = 1.94 \times 10^{-46}\,\text{kg m}^2\). The mean speed of the molecule is \(v = 500\,\text{m s}^{-1}\).
The translational kinetic energy of the molecule is
\[ \begin{aligned} K_{\text{trans}} &= \frac{1}{2}mv^2 \end{aligned} \]
It is given that the kinetic energy of rotation is two thirds of the kinetic energy of translation. Hence,
\[ \begin{aligned} K_{\text{rot}} &= \frac{2}{3}K_{\text{trans}} \\\\ &= \frac{2}{3}\left(\frac{1}{2}mv^2\right) \\\\ &= \frac{1}{3}mv^2 \end{aligned} \]
The rotational kinetic energy of the molecule can also be written as
\[ \begin{aligned} K_{\text{rot}} &= \frac{1}{2}I\omega^2 \end{aligned} \]
Equating the two expressions for rotational kinetic energy,
\[ \begin{aligned} \frac{1}{2}I\omega^2 &= \frac{1}{3}mv^2 \\\\ I\omega^2 &= \frac{2}{3}mv^2 \\\\ \omega^2 &= \frac{2mv^2}{3I} \end{aligned} \]
Substituting the given numerical values,
\[ \begin{aligned} \omega^2 &= \frac{2 \times 5.30 \times 10^{-26} \times (500)^2}{3 \times 1.94 \times 10^{-46}} \\\\ &= \frac{2 \times 5.30 \times 10^{-26} \times 2.5 \times 10^{5}}{5.82 \times 10^{-46}} \\\\ &= 4.55 \times 10^{25} \end{aligned} \]
\[ \begin{aligned} \omega &= \sqrt{4.55 \times 10^{25}} \\\\ &\approx 6.7 \times 10^{12}\,\text{rad s}^{-1} \end{aligned} \]
Therefore, the average angular velocity of the oxygen molecule is approximately \(6.7 \times 10^{12}\,\text{rad s}^{-1}\).
