NCERT Solutions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion - Param Himalaya
6.1 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 :
(i) The centre of mass of sphere is located at its centre.
(ii) The centre of mass of cylinder is at the centre of its axis of symmetry.
(iii) The centre of mass of a ring is at centre of the ring.
(iv) The centre of mass of a cube is at its geometrical centre.
The centre of mass may lie outside the body also, e.g., in case a ring, a hollow sphere, a hollow cylinder, a hollow cube etc.
6.2 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 :
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* We'll assume the hydrogen atom is located at the origin (0 m).
* The chlorine atom is located at a distance of 1.27 Å (1.27 * 10⁻¹⁰ m) from the hydrogen atom.
* The mass of the hydrogen atom is 1 unit.
Calculations:
* Mass of chlorine atom: 35.5 units (given)
* Position of CM:
The CM of a two-particle system is given by:
$CM_{position} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$
where:
* m1 and m2 are the masses of the hydrogen and chlorine atoms, respectively.
* x1 and x2 are their respective positions.
In this case:
* m1 = 1 unit
* x1 = 0 m
* m2 = 35.5 units
* x2 = 1.27 * 10⁻¹⁰ m
Therefore,
$CM_{position} = \frac{1 \times 0 + 35.5 \times 1.27 \times 10^{-10}}{1 + 35.5}$
$CM_{position} = \frac{35.5 \times 1.27 \times 10^{-10}}{36.5}$
$CM_{position} \approx 1.24 \times 10^{-10} \, \text{m}$
Therefore, the approximate location of the CM of the HCl molecule is 1.24 × 10⁻¹⁰ meters from the hydrogen atom.
This result is expected as the chlorine atom is much heavier than the hydrogen atom, so the CM will be much closer to the chlorine atom.
6.3 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 :
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It is necessary that an external force is applied on a system to change its motion. In this case, the boy is running arbitrarily on trolley with velocity v, which produces only internal forces in the system. The internal force cannot affect the motion of the system. As there is no external force on the system, so the speed of the centre of mass of the trolley and child system will remain same.
Certainly, let's derive the proof that the speed of the center of mass of the (trolley + child) system remains constant.
Assumptions:
* Let the mass of the trolley be M.
* Let the mass of the child be m.
* Let the initial velocity of the trolley and the child (which is the same) be V.
Initial Momentum of the System:
The initial momentum of the system (trolley + child) is given by:
$P = m_{1} .V_{1} + m_{2} . V_{2}$
$P_{initial} = M.V + m.V$
$P_{initial} = (M + m) \times V$
Internal Forces and Momentum Conservation:
* When the child starts running on the trolley, internal forces (like friction between the child's feet and the trolley) come into play.
* However, according to Newton's Third Law, these internal forces always occur in equal and opposite pairs.
* This means that the net external force acting on the system remains zero.
* Since the net external force is zero, the total momentum of the system must be conserved.
Final Momentum of the System:
* Let the velocity of the trolley after the child starts running be $V_{trolley}$
* Let the velocity of the child relative to the trolley be $v_{child/trolley}$.
* Therefore, the velocity of the child with respect to the ground is $V_{trolley} + v_{child/trolley}$.
* The final momentum of the system is:
$P = m_{1} .V_{1} + m_{2} . V_{2}$
$P_{final}=M.V_{tr} + m.(V_{tr} + v_{ch/tr})$
Conservation of Momentum:
Since the net external force is zero, we have:
$P_{initial} = P_{final}$
$(M + m).V=M.V_{tr} + m.(V_{tr} + v_{ch/tr})$
$(M + m).V=M.V_{tr} + m.V_{tr} +m.v_{ch/tr})$
$(M + m).V=(M + m) . V_{tr} + m.v_{ch/tr}$
$V=\frac{(M + m) .V_{tr}+ m. v_{ch/tr}}{M + m}$
$V=V_{tr} +\frac{m}{(M + m)}. v_{ch/tr}$
Velocity of the Center of Mass:
The velocity of the center of mass $(V_{CM})$ is given by:
$V_{CM}=\frac{m_{1} . V_{1} + m_{2}.V_{2}}{m_{1}+ m_{2}}$
$V_{CM}=\frac{M . V_{tr} + m ( V_{tr}+ v_{ch/tr})}{M + m}$
$V_{CM}=\frac{(M + m) .V_{tr}+ m. v_{ch/tr}}{M + m}$
$V_{CM}=V_{tr} +\frac{m}{(M + m)}. v_{ch/tr}$
Comparing this with the equation derived from momentum conservation, we see that:
$V_{CM} = V$
Therefore, we have mathematically proven that the speed of the center of mass of the (trolley + child) system remains constant at V, regardless of the child's motion on the trolley.
This demonstrates the fundamental principle that internal forces do not affect the motion of the center of mass of a system.
6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
Solution :
The following figure shows two vectors a and b at an angle θ extended to form a parallelogram.
From the above figure in ΔOMN,
sinθ=$\frac{MN}{OM} = \frac{MN}{| b |}$
MN=| b |sinθ ---- (1)
The magnitude for cross product of a and b is, | a×b |=| a || b |sinθ
From the equation (1),
| a×b |=( OK )( MN )
Multiply and divide by 2.
| a×b |= $2 \times \frac{1}{2} \times ( OK )( MN )$
=2×Area of ΔOMK
Area of ΔOMK= 1/2 | a×b |
Thus, the area of the triangle contained between the vectors a and b is one half the magnitude of the a×b.
6.5 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 vector $\left| \overrightarrow{b} \times \overrightarrow{c} \right|$ is perpendicular to the plane containing vectors $\left| \overrightarrow{b} \right |$ and $\left| \overrightarrow{c} \right|$.
Let θ be the angle between vector $\left| \overrightarrow{a} \right|$ and the vector$\left| \overrightarrow{b} \times \overrightarrow{c} \right|$.
The height of the parallelepiped is given by $\left| \overrightarrow{a} \right| cos \theta$, which is the component of vector a perpendicular to the base formed by $\left| \overrightarrow{b} \right |$ and $\left| \overrightarrow{c} \right |$
The base area is given by the magnitude of the vector product $|\left| \overrightarrow{b} \times \overrightarrow{c} \right|$
$\overrightarrow{a}.(\overrightarrow{b} \times \overrightarrow{c}) =\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \times \overrightarrow{c} \right| cos \theta$
$\overrightarrow{a}.(\overrightarrow{b} \times \overrightarrow{c})= \left| \overrightarrow{a} \right| cos \theta |\left| \overrightarrow{b} \times \overrightarrow{c} \right|$
$\overrightarrow{a}.(\overrightarrow{b} \times \overrightarrow{c}) $= Base area × height of the parallelepiped
$\overrightarrow{a}.(\overrightarrow{b} \times \overrightarrow{c}) $= Volume
Therefore, the magnitude of the scalar triple product $\overrightarrow{a}.(\overrightarrow{b} \times \overrightarrow{c}) $ is indeed equal to the volume of the parallelepiped formed by the vectors $\left| \overrightarrow{a} \right |$ , $\left| \overrightarrow{b} \right |$ and $\left| \overrightarrow{c} \right |$.
6.6 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:
Angular momentum :
$\vec{L} = L_{x}\hat{i}+L_{y}\hat{j}+L_{z}\hat{k}$
Linear momentum
$\vec{p} = p_{x}\hat{i}+p_{y}\hat{j}+p_{z}\hat{k}$
Position vector of the body
$\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}$
Angular momentum,
$\vec{L} = \vec{r} \times \vec{p}$
=$(x\hat{i}+y\hat{j}+z\hat{k}) \times (p_{x}\hat{i}+p_{y}\hat{j}+p_{z}\hat{k})$
=$\begin{array}{l}\begin{bmatrix} \hat{i} & \hat{j} & \hat{k}\\ x & y & z\\ p_{x}& p_{y} & p_{z} \end{bmatrix}\end{array}$
=$\hat{i}(yp_{z}-zp_{y})+\hat{j}(zp_{x}-xp_{z})+\hat{k}(xp_{y}-yp_{x})$
So $\vec{L}=L_{x}\hat{i}+L_{y}\hat{j}+L_{z}\hat{k} =\hat{i}(yp_{z}-zp_{y})+\hat{j}(zp_{x}-xp_{z})+\hat{k}(xp_{y}-yp_{x})$
On comparing the coefficient $\hat{i} , \hat{j} ,\hat{k}$
$L_{x} = (yp_{z}-zp_{y})$
$L_{y} = (zp_{x}-xp_{z})$
$L_{z} = (xp_{y}-yp_{x})$
If the particle is moving in x-y plane
Then z = 0 , $p_{z}$ = 0
$L_{x} = (y \times 0 - 0 \times hp_{y})=0$
$L_{y} = ( 0 \times p_{x}-x \times 0)=0$
$L_{z} = (xp_{y}-yp_{x})$
This is only z component of $\vec{L}$ exist if particle is moving in x-y plane
6.7 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:
Total angular momentum of the two particles system about O is
$\vec{L} = \vec{L}_{1} + \vec{L}_{2}$
$\vec{L} = \vec{r}_{1} \times \vec{p}_{1} +\vec{r}_{2} \times \vec{p}_{2}$
$= (x \hat{i}) \times mv \hat{-j} + (x+d)\hat{i} \times mv(\hat{j)}$
$= -(mvx) \hat{i} \times \hat{j} +(mvx+mvd) \hat{i}\times \hat{j}$
$= -(mvx) \hat{k} +(mvx+mvd) \hat{k}$
$= -mvx \hat{k} + mvx \hat{k}+mvd \hat{k}$
$\vec{L}= mvd \hat{k}$
Here $\vec{L}$ does not depend on x &Hence on the origin . This the angular momentum of two particles system in same whatever be the point about which the angular momentum is taken.
6.8 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 free body diagram of the bar is shown in the following figure. Length of the bar, l=2m
$T_{1}$ and $T_{2}$ are the tensions produced in the left and right strings respectively.
At translational equilibrium, we have: $T_{1}\sin~36.9^{\circ}=T_{2}\sin~53.1^{\circ}$
$\frac{T_{1}}{T_{2}}=\frac{\sin~53.1^{\circ}}{\sin~36.9^{\circ}}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\Rightarrow T_{1}=\frac{4}{3}T_{2}$
For rotational equilibrium, on taking the torque about the centre of gravity, we have:
$T_{1}\cos~36.9\times d=T_{2}\cos~53.1(2-d)$
$T_{1}\times \frac{4}{5}d=T_{2} \frac{3}{5}(2-d)$
$\frac{4}{3}\times T_{2}\times \frac{4}{5}d=T_{2} \frac{3}{5}[2-d]$
$\frac{16}{15} \times d = \frac{3}{5} \times [2-d]$
$16d = 9[2-d]$
$16d = 18-9d$
$16d+9d= 18$
$25d = 18$
$d=\frac{18}{25}=0.72m$
$d= 72cm$
Hence, the C.G. (centre of gravity) of the given bar lies 0.72 m from its left end.
6.9 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:
various forces acting on the car are shown in this figure.
$R_{f}$ and $R_{b}$ are the forces exerted by the level ground on the front and back wheels respectively.
At translational equilibrium: $R_{f}+R_{b}=W$
$R_{f}+R_{b}=mg$
$R_{f}+R_{b}=1800\times9.8$
$R_{f}+R_{b}=17640 N$ --(i)
For rotational equilibrium, on taking the torque about the C.G., we have:
$R_{f} \times (1.05)=R_{b} \times (1.8-1.05)$
$R_{f} \times 1.05=R_{b} \times 0.75$
$\frac{R_{f}}{R_{b}}=\frac{0.75}{1.05}$
$\frac{R_{f}}{R_{b}}=\frac{5}{7}$
$\frac{R_{b}}{R_{f}}=\frac{7}{5}$
$R_{b}=1.4R_{f}$ --- (ii)
Solving equations (i) and (ii), we get:
$1.4R_{f} + R_{f}= 17640$
$2.4 R_{f} = 17640$
$R_{f}= \frac{17640}{2.4}$
$R_{f}= 7350 N$
$R_{b} = 17640 - 7350$
$R_{b}= 10290N$
Therefore, the force exerted on each front wheel =$\frac{7350}{2} = 3675 N$
The force exerted on each back wheel = $\frac{10290}{2} = 5145 N$
6.10 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 m and r be the respective masses of the hollow cylinder and the solid sphere.
$ \tau=I \alpha$
$\alpha = \frac{\tau}{I}$
$\alpha_{1} = \frac{\tau}{I_{1}} = \frac{\tau}{mr^{2}}$
$\alpha_{2} = \frac{\tau}{I_{2}} = \frac{\tau}{ \frac{2}{5}mr^{2}}$
$\alpha_{2} = \frac{5}{2} \alpha_{1} = 2.5. \alpha_{1}$
So
$\frac{\alpha_{2}}{\alpha_{1}} = 2.5$
$\alpha_{2} > \alpha_{1}$. ____ (i)
Now , using the relation :
$\omega = \omega_{o} + \alpha.t$
Where , $\omega_{o}$ = initial
angular velocity
t = Time of rotation
$\omega$ = Final velocity
For equal $ \omega_{o}$ and t , we have
$\omega \propto \alpha$ ____(ii)
From equation (i) and (ii)
$\omega_{2} > \omega_{1}$
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.
6.11 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:
Mass of the cylinder, m = 20 kg
Angular speed $\omega = 100 rad s^{-1}$
Radius of the cylinder, r = 0.25 m
The moment of inertia of the solid cylinder:
$I= \frac{1}{2}mr^{2}$
$I= \frac{1}{2} \times 20 \times (0.25)^{2}$
$I = 0.625 kg m^{2}$
Kinetic energy, $K = \frac{1}{2}I\omega^{2}$
$K = \frac{1}{2} \times 6.25 \times (100)^{2}$
$K =3125J$
Angular momentum, $L = I.\omega$
$L= 0.625 \times 100 = 62.5 J~S$
6.12 (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 :
(a) Conservation of angular momentum
$I_{1} . \omega_{1} = I_{2}. \omega_{2}$
$I_{1} \times 40 = \frac{2}{5}I_{1} \times \omega_{2}$
$100 = \omega_{2}$
$\omega_{2} = 100rpm$
(b) $Initial K.E = \frac{1}{2}I_{1}.\omega_{1}^{2}$
$Initial K.E = \frac{1}{2}I_{1}.40^{2}$
$Initial K.E = \frac{1}{2}I_{1}. 1600$
$Initial K.E = 800 I_{1}$
$final K.E = \frac{1}{2}I_{2}.\omega_{2}^{2}$
$final K.E= \frac{1}{2} \times \frac{2}{5} I_{1} \times (100)^{2}$
$final K.E= \frac{10000}{5}. I_{1}$
$final K.E= 2000 I_{1}$
So
$\frac{Final~K.E}{Initial~K.E} = \frac{2000I_{1}}{800I_{1}}$
$\frac{Final~K.E}{Initial~K.E} = 2.5$
This shows that the final kinetic energy of rotation is greater than the initial kinetic energy of rotation.
Explanation for the increase in kinetic energy:
The increase in kinetic energy comes from the work done by the child in pulling his arms closer to his body. This work is done against the centripetal force acting on the child's hands. As the child pulls his arms in, his moment of inertia decreases, and his angular speed increases to conserve angular momentum. This increase in angular speed leads to an increase in kinetic energy.
6.13 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:
1. Calculate the Torque:
* Torque (τ) is the force applied multiplied by the radius of the cylinder:
τ = F * r
τ = 30 N * 0.4 m = 12 N·m
2. Calculate the Moment of Inertia:
* The moment of inertia (I) of a hollow cylinder is given by:
$ I = m.r^{2}$
$ I = 3 kg . (0.4 m)^{2} = 0.48 kg·m^{2}$
3. Calculate the Angular Acceleration:
* The relationship between torque, moment of inertia, and angular acceleration (α) is:
τ = I * α
* Rearranging to solve for α:
$\alpha = \frac{I}{tau}$
$\alpha = \frac{12 N·m} { 0.48 kg·m^{2}}= 25 rad/s^{2}$
4. Calculate the Linear Acceleration:
* Linear acceleration (a) is related to angular acceleration by:
a = α * r
$a = 25 rad/s^{2} * 0.4 m = 10 m/s^{2}$
Therefore:
* The angular acceleration of the cylinder is $25 rad/s^{2}$.
* The linear acceleration of the rope is $10 m/s^{2}$.
6.14 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 :
1. Understand the Concept
* Power: Power is the rate at which work is done. In rotational motion, it's the product of torque and angular velocity.
* Torque: Torque is a twisting force that causes rotation. In this case, the engine needs to provide torque to overcome friction and maintain a constant angular speed.
* Angular Velocity: Angular velocity is the rate of change of angular displacement. It's measured in radians per second (rad/s).
2. Formula
* Power (P) = Torque (τ) × Angular Velocity (ω)
3. Given Values
* Angular Velocity (ω) = 200 rad/s
* Torque (τ) = 180 N·m
4. Calculation
* Power (P) = 180 N·m × 200 rad/s
* Power (P) = 36000 W = 36KW
5. Answer
The power required by the engine is 36,000 Watts or 36 kilowatts (kW).
6.15 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:
Centre of mass of original disk = O
Centre of mass of circular hole = P
Centre of mass of Remaining pattern = Q
Let m be the mass per unit area.
Mass of original disk , $M = \pi R^{2}m$
Mass of circular hole cut , $m_{1} = \pi (\frac{R}{2})^{2}m$
Mass of Remaining portion ,$m_{2}= M-m_{1}$
= $\pi R^{2}m - \frac{\pi}{4} R^{2} m$
$m_{2}= \frac{3}{4} \pi R^{2} m$
After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:
$x = \frac{m_{1}x_{1}+ m_{2}x_{2}}{m_{1}+m_{2}}$
$0 = \frac{ \pi (\frac{R}{2})^{2}m . \frac{R}{2}+ \frac{3}{4} \pi R^{2} m . (-OQ}{m_{1}+m_{2}}$
$\pi (\frac{R}{2})^{2}m . \frac{R}{2}+ \frac{3}{4} \pi R^{2} m .(- OQ) =0$
$ \pi (\frac{R}{2})^{2}m . \frac{R}{2}=\frac{3}{4} \pi R^{2} m . OQ$
$\pi (\frac{R^{2}}{4})m . \frac{R}{2}=\frac{3}{4} \pi R^{2} m . OQ$
$OQ = \frac{R}{6}$
6.16 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 :
Torque: Torque is the rotational force that tends to cause rotation. It's calculated as the force multiplied by the perpendicular distance from the pivot point.
Equilibrium: For the meter stick to be balanced, the net torque acting on it must be zero.
Torque due to the coins:
Distance from the pivot: 45 cm - 12 cm = 33 cm
Torque due to the meter stick:
Distance from the pivot: 50 cm - 45 cm = 5 cm
Since the meter stick is balanced:
Torque due to the coins = Torque due to the meter stick
(Weight of coins) × (Distance from pivot of coins) = (Weight of meter stick) × (Distance from pivot of meter stick)
10g × 33 cm = Weight of meter stick × 5 cm
Calculate Weight of Meter Stick:
Weight of meter stick = (10g × 33 cm) / 5 cm
Weight of meter stick = 66g
Mass × g = 66g
The mass of the meter stick is 66 grams
6.17 The oxygen molecule has a mass of $5.30 \times 10^{-26} kg$and a moment of inertia of $1.94 \times 10^{-46} kg m^{2}$ 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:
Calculate Translational Kinetic Energy:
$KE_{trans} = \frac{1}{2}mv^2$
$KE_{trans} = \frac{1}{2} \times (5.30 \times 10^{-26} \, \text{kg}) \times (500 \, \text{m/s})^2$
$KE_{trans} = 6.625 \times 10^{-21} \, \text{J}$
Calculate Rotational Kinetic Energy:
$KE_{rot} = \frac{2}{3} KE_{trans}$
$KE_{rot} = \frac{2}{3} \times 6.625 \times 10^{-21} \, \text{J}$
$KE_{rot} = 4.417 \times 10^{-21} \, \text{J}$
Calculate Angular Velocity:
$KE_{rot} = \frac{1}{2} I \omega^2$
$\omega^2 = \frac{2 \times KE_{rot}}{I}$
$\omega = \sqrt{\frac{2 \times KE_{rot}}{I}}$
$\omega = \sqrt{\frac{2 \times 4.417 \times 10^{-21} \, \text{J}}{1.94 \times 10^{-46} \, \text{kg m}^2}}$
$\omega = 6.75 \times 10^{12} \, \text{rad/s}$
Therefore, the average angular velocity of the oxygen molecule is approximately 6.75 × 10¹² rad/s.