System of Particles and Rotational Motion Class 11 Notes: CBSE Physics Chapter 6

Summary - Class 11 Physics System of Particles and Rotational Motion Notes (Chapter 6) 

System of particles and rotational motion comes under the fifth unit, Motion of system and particles. This along with Unit IV and Unit VI have a total weightage of 17 marks, which implies students will definitely find questions from this chapter. This is purely an understanding-based chapter which talks about the system of particles and rotational motion. The topics extensively covered in the notes of Physics Class 11 Chapter 6 are:

• Centre of mass and its motion.

• Centre of mass of a 2 particle system, rigid body and a uniform rod.

• The momentum of force and momentum conservation.

• Torque.

• Moment of Inertia.

• Angular momentum and laws of conservation of momentum.

• Radius of gyration.

• Parallel and perpendicular axis theorem.

Notes of System of Particles and Rotational Motion

Given below are brief explanations of some important concepts and topics covered in the chapter. For an in-depth understanding of the same, refer to notes on the system of particles and rotational motion in class 11.

Centre of Mass

The centre of mass of a body is the point at which the entire mass of the body is said to be concentrated. It is also defined as the balancing point of the system. If any external force is to be applied at the centre of mass, the body is said to remain unaffected. It means that the body will stay at rest if at rest and there will be no change in the velocity of the body in motion.

For the centre of mass of a two-particle system at motion,

\[\overline{v} = \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}}\]

\[\overline{a} = \frac{m_{1}a_{1} + m_{2}a_{2}}{m_{1} + m_{2}}\]

Where,

\[\overline{v}\] = Velocity of the centre of mass.

\[\overline{a}\] = Acceleration of the centre of mass.

Motion of Centre of Mass

The motion of the centre of mass is governed by Newton’s laws of motion. If external forces act on a system, the centre of mass will accelerate according to these forces. The key takeaway here is that the motion of the centre of mass simplifies the analysis of the entire system, allowing us to treat it as if all the mass is concentrated at this point. In the absence of external forces, the centre of mass moves with a constant velocity, demonstrating the conservation of momentum.

• Linear Momentum of a System of Particles

For a system of particles, the total linear momentum is the sum of the linear momenta of the individual particles. It is given by:

P=$ \sum m_i v_i$​

Where $m_i$​ is the mass and $v_i$​ is the velocity of the i-th particle. The motion of the centre of mass helps simplify this concept, as the total momentum of the system is the product of the total mass and the velocity of the centre of mass. This section helps explain how forces affect the motion of systems of particles and how external forces change the total momentum.

• Vector Product of Two Vectors

The vector product (or cross product) of two vectors is essential in understanding rotational dynamics, especially for quantities like torque and angular momentum. If two vectors A and B are multiplied using the cross product, the resulting vector is perpendicular to both A and B and is given by:

$A \times B = |A||B|\sin\theta \hat{n}$

Where $\theta$ is the angle between the two vectors and $\hat{n}$is a unit vector perpendicular to the plane formed by A and B. The magnitude of the vector product depends on the sine of the angle between the two vectors. The cross-product plays a crucial role in calculating rotational quantities like torque and angular momentum.

• Angular Velocity and its Relation with Linear Velocity

Angular velocity ($\omega$) tells us how fast an object is rotating. It is related to linear velocity through the equation:

$v = r\omega$

Where v is the linear velocity of a point on a rotating object, r is the distance of the point from the axis of rotation, and $\omega$ is the angular velocity. This relationship is essential when dealing with objects in circular motion, as it links the rotational motion to the linear speed of points on the object.

Torque

The turning effect of force about a fixed axis is defined as Torque. It can also be defined as the Moment of force. The SI unit of Torque is Nm.

\[\tau = \overline{r} \times \overline{F} = rF Sin \theta\]

Where,

τ = Moment of force or torque

r̅ = Perpendicular distance

F̅ = Force

θ = Angle between the two vectors r and F

Torque can also be calculated in terms of angular moment. The relationship between torque and angular momentum is defined by:

τ = dL/dt

Equilibrium of a Rigid Body

A rigid body is in equilibrium when both the net force and the net torque acting on it are zero. This condition ensures that the body is neither accelerating translationally nor rotating. For translational equilibrium:

$\sum F = 0$

And for rotational equilibrium:

$\sum \tau = 0$

Moment of Inertia

Moment of Inertia is the phenomenon by which the body in motion opposes the change in its rotational motion. Mathematically, it is defined as the product of the mass of particles and their distance from the axis of rotation. It is also called the rotational inertia of the body.

Kinematics of Rotational Motion About a Fixed Axis

Just like in linear motion, rotational motion can be described using kinematic equations. For a rotating object, the angular displacement ($\theta$), angular velocity ($\omega$), and angular acceleration ($\alpha$) are related through kinematic equations similar to those for linear motion:

$\omega = \omega_0 + \alpha t$

These equations help in solving problems involving objects rotating about a fixed axis.

Dynamics of Rotational Motion About a Fixed Axis

Rotational dynamics involves the forces that cause objects to rotate. Newton’s second law for rotation states that the torque acting on a body is equal to the product of its moment of inertia and angular acceleration:

$\tau = I\alpha$

This equation is the rotational analogue of F = ma and helps describe how forces cause changes in rotational motion.

Angular Momentum in Case of Rotation About a Fixed Axis

Angular momentum is a measure of the rotational motion of an object. For a body rotating about a fixed axis, the angular momentum is related to the moment of inertia and angular velocity:

$L = I\omega$

Angular momentum is conserved in the absence of external torques, just like linear momentum is conserved in the absence of external forces.

System of Particles and Rotational Motion Class 11 Notes  Physics - Basic Subjective Questions

Section-A (1 Mark Questions)

1. A wheel 0.5m in radius is moving with a speed of 12m/s. find its angular speed?

Ans. $v=r\omega$ 

$\omega =\dfrac{v}{r}=\dfrac{12}{0\cdot 5}$

$\omega =24\;rad/s$ .

2. State the condition for mechanical equilibrium of a body?

Ans. For mechanical equilibrium of a body the vector sum of all the forces and moments (torques) acting on the body must be zero.

3. How is angular momentum related to linear momentum? 

Ans. $\vec{\vec{L}}=\vec{r}\times \vec{p}$

Or L = rp sin θ

Where θ is the angle between $\vec{r}$ and $\vec{p}$ .

4. What is the position of the centre of mass of a uniform triangular lamina?

Ans. Position of the centre of mass of a uniform triangular lamina at the centroid of the triangular lamina.

5. What is the moment of inertia of a sphere of mass 20 kg and radius $\dfrac{1}{4}m$ about its diameter?

Ans. $I=\dfrac{2}{5}MR^{2}$

$I=\dfrac{2}{5}\times 20\times \left ( \dfrac{1}{4} \right )^{2}$

$I=0\cdot 5\;kgm^{2}$

6. What are the factors on which moment of inertia of a body depends?

Ans.

(i) Mass of the body

(ii) Shape and size of the body

(iii) Position of the axis of rotation

7. Two particles in an iAnsated system undergo head on collision. What is the acceleration of the centre of mass of the system?

Ans. Acceleration of center of mass is zero as all forces are internal forces.

8. Which component of a force does not contribute towards torque? 

Ans. The radial component of a force does not contribute towards torque.

9. What is the position of centre of mass of a rectangular lamina? 

Ans. The centre of mass of a rectangular lamina is the point of intersection of diagonals.

10. Does the centre of mass of a body necessarily lie inside the body?

Ans. The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc., lies outside the body.

Section-B (2 Marks Questions)

11. A planet revolves around on massive star in a highly elliptical orbit is its angular momentum constant over the entire orbit. Give reason?

Ans. A planet revolves around the star under the effect of gravitational force since the force is radial and does not contribute towards torque. Thus in the absence of an external torque angular momentum of the planet remains constant.

12. Prove the relation $vec{\tau }=\dfrac{d\vec{L}}{dt}$

Ans. We know $\vec{L}=I\vec{\omega }$

Differentiating wrt. Time

$\dfrac{d\vec{L}}{dt}=\dfrac{d}{dt}\left ( I\vec{\omega } \right )=\dfrac{Id\vec{\omega }}{dt}=1\vec{\alpha }$ …(1)

We know that

$\vec{\tau }=I\vec{\alpha }$ …(2)

From (1) and (2) $\vec{\tau }=\dfrac{d\vec{L}}{dt}$

13. What is the torque of the force $\vec{F}=\hat{2}i-\hat{3}j+\hat{4}k$ acting at the point about the $\vec{r}=\left (\hat{3}i-\hat{2}j+\hat{3}k  \right )m$ origin?

Ans. $\vec{\tau }=\vec{r}\times \vec{F}$

$\vec{\tau }=\begin{vmatrix} \hat{}i & \hat{}j & \hat{}k\\ 2 & -3 & 4\\ 3& 2 & 3 \end{vmatrix}$

$\vec{\tau }=\left ( -\hat{17}i+\hat{6}j+\hat{13}k \right )NM$

14. What is the value of linear velocity if $\vec{\omega }=\hat{3}i-\hat{4}j+\hat{}k$ and $\vec{r}=\hat{}5i-\hat{6}j+\hat{6}k$ ?

Ans. $\vec{v}=\vec{\omega }\times \vec{r}$

$\vec{v}=\left ( \hat{3}i-\hat{4}j+\hat{}k \right )\times \left ( \hat{}5i-\hat{6}j+\hat{6}k \right )$

$\vec{t}=\begin{vmatrix} \hat{}i & \hat{}j & \hat{}k\\ 3& -4 & 1\\ 5& -6 & 6 \end{vmatrix}$

$\vec{v}=\left ( -\hat{18}i-\hat{13}j+\hat{2}k \right )m/s$

15. Find the expression for radius of gyration of a Ansid sphere about one of its diameters?

Ans. M.I. of a Ansid sphere about its diameter $=\dfrac{2}{5}MR^{2}$

K = Radius of Gyration

$I=MK^{2}=\dfrac{2}{5}MR^{2}$

$K^{2}=\dfrac{2}{5}R^{2}$

$K=\sqrt{\dfrac{2}{5}}R$

16. Prove that the centre of mass of two particles divides the line joining the particles in the inverse ratio of their masses? 

Ans. $\vec{r}_{cm}=\dfrac{m_{1}+\vec{r}_{1}+m_{2}\vec{r}_{2}}{m_{1}+m_{2}}$

If centre of mass is at the origin

$\vec{r}_{cm}=0$

$\Rightarrow m_{1}\vec{r}_{1}+m_{2}\vec{r}_{2}=0$

$m_{1}\vec{r}_{1}=-m_{2}\vec{r}_{2}$

In terms of magnitude $m_{1}\left | \vec{r}_{1} \right |=m_{2}\left | \vec{r}_{2} \right |$

$\Rightarrow \dfrac{m_{1}}{m_{2}}=\dfrac{r_{2}}{r_{1}}$

Important formula in Class 9 Science Chapter 6 System Of Particles And Rotational Motion

1. Centre of Mass:

The position of the centre of mass for a system of particles is given by:
$R_{\text{cm}} = \frac{m_1r_1 + m_2r_2 + \dots + m_nr_n}{m_1 + m_2 + \dots + m_n}$

Where:

• $R_{\text{cm}}$​ is the position of the centre of mass

• $m_1, m_2, \dots, m_n$​ are the masses of the particles

• $r_1, r_2, \dots, r_n$​ are the positions of the particles

2. Torque (τ\tauτ):

Torque is the rotational equivalent of force and is given by:
$\tau = r \times F$
Where:

• $\tau$ is the torque

• r is the distance from the axis of rotation

• F is the applied force

3. Moment of Inertia (III):

The moment of inertia for a point mass is given by:
$I = mr^2$
Where:

• I is the moment of inertia

• m is the mass of the object

• r is the distance from the axis of rotation

4 . Angular Momentum (L):

Angular momentum is the rotational equivalent of linear momentum:
$L = I \omega$
Where:

• L is the angular momentum

• I is the moment of inertia

• $\omega$ is the angular velocity

5. Kinetic Energy of Rotation:

The kinetic energy for an object in rotational motion is:
$KE = \frac{1}{2} I \omega^2$
Where:

• KE is the kinetic energy

• I is the moment of inertia

• $\omega$ is the angular velocity

6. Relation Between Linear and Angular Quantities:

$v = r\omega$
Where:

• v is the linear velocity

• r is the radius

• $\omega$ is the angular velocity.

Important Topics of Class 11 Physics Chapter 6 System of Particles and Rotational Motion

Importance of Physics Class 11 Physics Chapter 6 System Of Particles And Rotational Motion Revision Notes 

• Class 11 Physics Chapter 6 Notes simplify complex topics, helping students break down the concepts of rotation, center of mass, and system of particles easily.

• They provide clear explanations of key formulas and laws, such as torque, angular momentum, and the moment of inertia, which are crucial for solving problems related to rotational motion.

• The notes highlight practical applications of rotational dynamics in real-world systems like machinery, planets, and rotating objects, making the concepts more relatable.

• Well-organised notes help students revise quickly before exams, summarizing all key points to ensure nothing is missed.

• Solved examples in the notes guide students through typical problems, showing them how to apply concepts to various scenarios.

• System Of Particles And Rotational Motion Notes PDF support learning by simplifying complex derivations and helping students focus on the core principles needed for competitive exams and CBSE board exams.

Tips for Learning the Class 11 Chapter 6 Physics System Of Particles And Rotational Motion

• Start by understanding the basic concepts of rotational motion, such as angular velocity, torque, and moment of inertia, and how they relate to linear motion.

• Draw free-body diagrams to visually represent forces and torques acting on objects, which will simplify the process of solving problems.

• Break down complex problems into smaller, manageable parts and solve them step by step.

• Practice vector cross products to better understand how torque and angular momentum are calculated.

• Relate the concepts of rotational motion to real-world examples such as spinning wheels, rotating fans, or machinery to make the concepts more tangible.

• Regularly practice a variety of problems, including those with multiple steps, to gain confidence in applying the concepts.

• Review key concepts and formulas frequently to reinforce your understanding of the subject.

Conclusion 

Vedatu’s FREE Revision notes PDF download for CBSE Class 11 Physics Chapter 6 - "Systems of Particles and Rotational Motion" is very useful for students. These notes provide a comprehensive and concise resource that simplifies complex concepts in rotational dynamics. Understanding the principles discussed in this chapter is crucial not only for academic success but also for real-world applications in engineering and physics. These notes offer students a structured pathway to grasp topics like angular momentum, moment of inertia, and rotational equilibrium. They not only aid in exam preparation but also empower students to apply these fundamental principles in various scientific and practical scenarios. Overall, these downloadable notes are an invaluable tool for enhancing students' understanding and competence in the realm of rotational motion.

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