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Hello, Class 11 Physics enthusiasts! We're excited to provide you with comprehensive System of Particles and Rotational Motion class 11 notes. These notes, created by our experts, are intended to provide a thorough understanding of each concept, taking into account both theoretical knowledge and practical applications.
Physics is more than just formulas; it is about comprehending how these concepts apply in the real world. Our class 11 physics chapter 7 notes not only explain the theories, but also explore the practical applications of each concept. In today's exam environment, where application-based questions are common, these CBSE class 11 physics ch 7 notes will not only help you prepare for exams, but will also foster a genuine interest in the subject.
Don't miss out on this opportunity to improve your understanding and excel in your exams. Download the physics class 11 chapter 7 notes pdf now and take a journey of discovery through Chapter 7.
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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.
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 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;
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.
Differentiating w.r.t. time, we get,
Again, differentiating w.r.t. time, we get,
Ma = Fext
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.
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
P = p1 + p2 + …. + pn
= m1v1 + m2v2 + …. + mnvn
P = MV;
Where P = Momentum particles and V = Velocity of COM
Newton’s Second Law in reference to the system of particles is represented by the following expression:
dP/dt = Fext
In case when the total external force acting on a system of particles is zero (Fext = 0), then, in that case, the total linear momentum of the system is constant and can be represented as:
(dP/dt = 0; P = constant).
The Centre of mass having velocity also remains constant (Since P = mv = 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.
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.
The rotational analogue of force is a moment of force which is also known as Torque).
Torque is a vector quantity.
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.
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 (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 = M k2, where k is the radius of gyration.
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.
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.
We can derive equations of motion similar to translational motion.
Translational motion
v=v0+at
x=x0+v0t+0.5at2
v2=v02+2as
Rotational Motion
ω=ω0+αt
θ=θ0+ω0t+0.5αt2
ω2=ω02+2αθ
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.
Rolling motion may be defined as a combination of rotation and translation motion.
Particles of a rolling body have two velocities.
Transnational, which is the velocity of COM.
The linear velocity of rotational motion.
In the figure, we observe that every point has two velocities.
One Velocity is found in the direction COM and the other velocity is found to be perpendicular to the line joining centre and the point.
At Point Po have velocities and both are in opposite direction to each other.
For no-slipping condition: Velocity is found has zero value, so
Vcom = ω R
There is an instantaneous axis of rotation that can be seen through PO and parallel to ‘w’ .
At Point, P1 Velocities are added up.
K.E.rolling = K.E.translation + K.E.rotation
Kinetic energy of a rolling body
where k is radius of gyration of body
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Ans- The centre of gravity is the geometric centre whereas the centre of mass is the mass of the point where the entire mass of the body is considered. The object is said to be small when the vertical height of the object is close to the earth’s surface. If the distance between the object and the surface of the earth is large then the object is an extended object.
Buildings and ponds are considered to be small objects while a mountain and a lake are considered to be extended objects.
Every topic is important in the chapter no particular topic is, therefore, marked to be an important one hence proper revision of all the topics through NCERT notes for Class 11 Physics chapter 7 needs to be done to secure good marks in CBSE Board examinations.
a) a pencil
b) a shotput
c) a dice
d) a bangle
Ans-
The correct answer is d) a bangle
NCERT Class 11 Physics chapter 7 notes the centre of gravity of a body is the point on the body where the total gravitational torque on the body is zero.
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