NCERT Class 11 Physics Chapter 7 notes System of Particles and Rotational Motion - Download PDF

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NCERT Notes for Class 11 Physics Chapter 6

The chapter 6 of Physics in NCERT Class 11 on System of Particles and rotational motion provides a compact and organized overview of important concepts such as the torque, angular momentum, and moment of inertia. These notes can be used to revise within very short time and prepare exams in CBSE, JEE, and NEET.

Rotational Motion

• Rotational motion describes the movement of an object as it spins or revolves around an axis. Consider a spinning top, a rotating wheel, or the Earth turning on its axis; these are all examples of rotational motion.
• In this type of motion, each point in the object follows a circular path around a fixed line known as the axis. The motion is a rotation rather than a linear translation.

Centre of Mass

• The centre of mass is an imaginary point in a system where all of the system's mass is concentrated. When two bodies are joined, their centre of mass is along the line that connects them.

(Here m₁ & m₂ are two bodies in which m₁ is at a distance x₁, & m₂ at a distance x₂ from the origin O).

• The coordinates of a system's centre of mass (X, Y, Z) are generally given by $\sum \frac{m_i \vec{r_i}}{m}$
• Considering a thin rod of length $l$, the origin is at the geometric centre of the rod. and the X-axis is along the length of the rod, and for every element dm of the rod at x, there is an element of the same mass dm located at x.

• The total contribution of each such integral and the integral x dm is zero. Therefore, the COM coincides with its geometric centre.

• The centre of mass coincides with the geometric centre and with the same geometry is applicable to homogenous rings, discs, spheres, or even thick rods of circular or rectangular cross-sections;

Motion of COM

• The centre of mass of a system of particles moves in a manner such that the entire mass of the system is considered to be concentrated at the centre, and all external forces are treated as if they act solely on that point.

$M \vec{R}=\sum m_i \vec{r_i}=m_1 \vec{r_1}+m_2 \vec{r_2}+\ldots \ldots m_n \vec{r_n}$

Differentiating w.r.t. time, we get,

$\begin{aligned} & M \frac{\vec{d R}}{d t}=m_1 \frac{\vec{d r_1}}{d t}+m_2 \frac{\vec{d r_2}}{d t}+\ldots \ldots m_n \frac{\vec{d r_n}}{d t} \\ & M \vec{v}=m_1 \vec{v_1}+m_2 \vec{v_2} \ldots \ldots .+m_n \vec{v_n}\end{aligned}$

Again, differentiating w.r.t. time, we get,

$M \vec{a}=m_1 \vec{a_1}+m_2 \vec{a_2} \ldots \ldots . .+m_n \vec{a_n}$

M$\vec{a}$ = $ \vec{F_{ext}}$

• Instead of considering extended bodies as single particles, we can consider them as systems of particles.

• The translational component of their motion could be attained by taking the mass of the whole system and all the external forces on the system to be acting at the centre of mass.

Linear Momentum of System of Particles

• The linear momentum of a system of particles is the product of the system's total mass and the velocity of its centre of mass. This relationship is expressed as follows:
• Linear Momentum=Total Mass×Velocity of Center of Mass

$\vec{p}$ = $\vec{p}_1$+ $\vec{p}_2$+ …. + $\vec{p}_n$

$\vec{p}$= m₁$\vec{v}_1$+ m₂ $\vec{v}_2$+ …. + m_n $\vec{v}_n$

$\vec{p} = M\vec{v}$;

Where $\vec{p}$ = Momentum particles and $\vec{v}$ = Velocity of COM

• Newton's Second Law of Motion in reference to the system of particles is represented by the following expression:

$\frac{\vec{dp}}{dt} $= $\vec{F}_{ext}$

• In case when the total external force acting on a system of particles is zero ($\vec{F}_{ext}$ = 0), then, in that case, the total linear momentum of the system is constant and can be represented as:

($\frac{\vec{dp}}{dt}$ = 0; $\vec{p}$ = constant).

• The Centre of mass having velocity also remains constant (Since $\vec{p} = m\vec{v}$ = Constant).

• If the total external force on a body is zero, then the internal forces cause complex trajectories of individual particles in spite of the COM moving with a constant velocity.

Angular Velocity & Linear Velocity

• Each particle of a rotating body is considered to move in a circle. We define the angular displacement of a given particle about its centre in unit time as angular velocity.

• Average angular velocity =∆θ/∆t
• Instantaneous angular velocity, ω=dθ/dt
• In a pure rotation motion, all the parts of the moving body have the same angular velocity.
• It is a vector quantity
• $\vec{v} = \vec{ω} \times \vec{r}$; where $\vec{v}$ is the linear velocity of a particle and r is the radius of the circle
• Angular acceleration is outlined as the rate of amendment of angular speed with relevance time. $\vec{α} = \frac{\vec{dω}} {dt}$

Torque & Angular Momentum

• The rotational analogue of force is a moment of force which is also known as Torque.

• Torque is a vector quantity.

Equilibrium of Rigid Body

• The translational state of the motion of the rigid body is due to the changes in force, i.e. its total linear momentum changes.

• The rotational state of motion of the rigid body is due to the changes in torque i.e. the total angular momentum of the body changes.

• Unless stated otherwise, only external forces and torques should be dealt.

• A rigid body is said to be in mechanical equilibrium when both its linear momentum and angular momentum are constant with time. This means

Total force zero = Translational Equilibrium

Total torque zero = Rotational Equilibrium

• A couple or torque can be defined as a pair of equal and opposite forces with different lines of action is termed as couple or torque.
• Without translation, the couple can produce the rotation. For example, when you open the lid of a jar, you put a couple on it.

Centre of Gravity

• The centre of gravity of a body is the point on the body where the total gravitational torque on the body is zero.
• In uniform gravity or gravity-free space, the centre of gravity of the body coincides with the centre of mass.
• The condition for the centre of gravity and centre of mass will not coincide is that if the value of g will vary from part to part of the body.

Moment of Inertia

• Moment of inertia (I) is the analogue of the mass in rotational motion.
• A change in rotational motion is resisted by the moment of inertia about a given axis of rotation. It can be regarded as a measure of rotational inertia of the body.
• It is a measure of how the different parts of the body are distributed at different distances from the axis.
• The moment of inertia of a rigid body depends on the mass of the body, its form and size, and therefore the distribution of mass concerning the axis of rotation.

• The radius of gyration of a body concerning association in axis is outlined as the distance from the axis of a mass whose mass is up to the mass of the total body and whose moment of inertia is up to the instant of inertia of the body concerning the axis.

I = Mk², where k is the radius of gyration.

The Theorem of Perpendicular Axis

• Perpendicular Axis Theorem: The moment of inertia of a two-dimensional body (lamina) regarding an axis perpendicular to its plane is the sum of its moments of inertia regarding two perpendicular axes synchronous with perpendicular axis and lying within the plane of the body.

• It is applicable only to planar bodies.

The Theorem of Parallel Axis

• Parallel Axis Theorem: The moment of inertia of a body about an axis parallel to an axis passing through the center of mass is equal to the sum of the moment of inertia of body about an axis passing through center of mass and product of mass & square of the distance between two axes. This theorem is applicable to a body of any form.

Kinematics of Rotational Motion for a Fixed Axis

We can derive equations of motion similar to translational motion.

Translational motion

v=v_o+at

x=x_o+v_ot+0.5at²

v²=v_o²+2as

Rotational Motion

ω=ω_o+αt

θ=θ_o+ω_ot+0.5αt²

ω²=ω_o²+2αθ

Dynamics of Rotational Motion for a Fixed Axis

• The components of torques are considered along the direction of the fixed axis and this is so because the component of the torque is perpendicular to the axis of rotation and this will tend to turn the axis from its position.
• The forces that lie in planes perpendicular to the axis are only considered. Forces parallel to the axis give torques perpendicular to the axis.
• Also, the components of the position vectors which are perpendicular to the axis are only considered. Components of position vectors on the axis can lead to torques perpendicular to the axis.

System of Particles and Rotational Motion Previous year Question and Answer

Q1: The net external torque on a system of particle about an axis is zero. Which of the following are compatible with it?

a) the forces may be acting radially from a point on the axis

b) the forces may be acting on the axis of rotation

c) the forces may be acting parallel to the axis of rotation

d) the torque caused by some forces may be equal and opposite to that caused by other forces

Answer:

Torque is given by $\vec{\tau}=\vec{r} \times \vec{F}$
Or
$\tau=r F \sin \theta \widehat{n}$ where $\theta$ is the angle between both the vectors and $\widehat{n}$ is the unit vector perpendicular to the plane of $\vec{r}$ and $\vec{F}$
1. When the force is acting radially in the direction of $\vec{r}, \theta=0$

$$
\tau=r F \sin 0^{\circ}=0
$$

2. When the forces are acting on the axis of rotation then $\theta=0$ and thus $\tau=0$
3. The component of forces in the plane of $\vec{r}$ and $\vec{F}$ when they are parallel to the axis of rotation is $F \cos 90^{\circ}$ which is equal to 0 . Hence $\tau=0$
4. When torques are equal in magnitude but opposite in direction, the net resultant is 0 .

Hence, the answers are option (a) and (b).

Q2: A Merry-go-round, made of a ring-like platform of radius $R$ and mass $M$, is revolving with angular speed $\omega$. A person of mass $M$ is standing on it. At one instant, the person jumps off the round, radially away from the center of the round (as seen from the round). The speed of the round afterwards is

Answer:

As no external torque acts on the system, angular momentum should be conserved.
Hence. $I \omega=$ constant
where, $I$ is moment of inertia of the system and $\omega$ is angular
velocity of the system.
From Eq. (i), $I_1 \omega_1=I_2 \omega_2$
(where, $\omega_1$ and $\omega_2$ are angular velocities before and after jumping)

$
\Rightarrow \quad I \omega=\frac{I}{2} \times \omega_2
$

(As mass reduced to half, hence moment of inertia also reduced to half)

$
\Rightarrow \quad \omega_2=2 \omega
$

Q3: Which of the following points A,B and C is the likely position of the center of mass of the system shown in Fig. 7.1?

Answer:

The volume of the hollow sphere is occupied equally by the air and sand. Since the pressure exerted by the air region is less than that of sand, the mass of sand is greater than the air. The centre of mass shifts towards a heavier portion which in this case is sand. B is exactly at the line dividing air and sand, so it is incorrect. D is at the farthest end and divides the sand area very unequally, so it also cancels out. C is almost at the center with regard to the mass.

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