System of Particles and Rotational Motion | Class 11 Physics Chapter 6 One Shot | New NCERT CBSE

LearnoHub - Class 11, 12・2 minutes read

The text discusses rotational motion, center of mass, torque, angular momentum, and mechanical equilibrium, emphasizing key concepts in physics. It explains the application of forces, principles of moments, and calculations of torque and mechanical advantage, concluding with discussions on rolling motion and kinetic energy.

Insights

Extended bodies are systems of particles with considerable size, contrasting with rigid bodies that maintain a fixed shape and do not deform under force.
Rotational motion involves different velocities in different particles in a rigid body, contrasting with translational motion where all particles move with the same velocity.
Center of mass coordinates are crucial in determining the mass distribution within a system of particles or bodies, calculated based on the total mass and individual masses.
Torque, angular momentum, and mechanical equilibrium play significant roles in understanding rotational motion, with torque depending on the position vector and force, and angular momentum being the cross product of radius and linear momentum.

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Recent questions

What is rotational motion in physics?
Rotational motion involves objects moving around a fixed axis.

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Summary

00:00

"Understanding Rotational Motion in Physics"

The topic discussed is rotational motion in physics, specifically focusing on the system of particles and rotational motion in class 11 physics.
The text is presented by Roshni from Learn Hub, a free learning platform offering education in Physics, Chemistry, Maths, and Biology.
It distinguishes between particles and extended bodies, explaining that extended bodies are systems of particles with considerable size.
Rigid bodies are introduced as bodies that maintain a fixed shape and do not deform under force, contrasting with idealized extended bodies.
The concept of rigid bodies is defined by the constant distance between particles, emphasizing the unchanging shape.
Translational motion is described as the movement of all particles in a rigid body with the same velocity, while rotational motion involves different velocities in different particles.
The text demonstrates rotational motion using a block and a ball, showcasing how rotation about a fixed axis leads to translational motion.
The difference between circular motion and rotational motion is explained, with rotational motion involving all particles moving about a fixed axis.
Examples like a spinning top and a table fan illustrate rotational motion around a fixed point or axis.
Concepts like center of mass, angular velocity, angular momentum, torque, and moment of inertia are introduced as essential for understanding particle motion.

15:33

Center of Mass Calculation for Various Shapes

The center of mass position is given by m1 x1 + m2 x2 / m1 + m2, known as the mass weighted mean of m1 and m2.
In a system of two particles, knowing their masses and coordinates allows for easy determination of the center of mass.
If the masses of two particles are equal (m1 = m2), the center of mass will be at x1 + x2 / 2, exactly midway between the particles.
In three dimensions, the center of mass coordinates are x, y, and z, calculated as the sum of m x, m y, and m z divided by the total mass.
For a rigid body with continuous mass distribution, the center of mass position vector can be written as r = 1 / m ∫ r dm.
Homogeneous bodies have uniform mass distribution, with their center of mass coinciding with their geometric center.
The center of mass of a sphere, cylinder, ring, and cube with uniform mass density is at their geometric centers.
To find the center of mass of an L-shaped lamina with three squares, the coordinates of the squares' centers (c1, c2, c3) are determined.
The coordinates of the centers of the squares are found to be c1 (1/2, 1/2), c2 (3/2, 1/2), and c3 (1/2, 3/2).
To find the center of mass of the entire lamina, assume the center of mass coordinates as x, y.

32:41

Calculate Masses and Coordinates for Center of Mass

Find the value of p and y, then calculate x.
The summation of mii3 for laminas with i as 1, 2, and 3.
Given m1, m2, m3, total mass of lamba is 3 kg.
Determine the mass of the square if the total mass is 3 kg.
Calculate the coordinates x1, x2, x3 as 1/2, 3/2, and 1/2 respectively.
Calculate the center of mass coordinates as 5/6.
Calculate the y coordinates y1, y2, y3 as 1/2, 1/2, and 3/2 respectively.
The center of mass coordinates are 5/6.
Understand the motion of particles and the center of mass.
Newton's second law relates to the total external force acting on the system.

50:40

"Vector Products and Angular Motion Simplified"

When moving from point a to point b, ensure that the direction of rotation of your fingers remains consistent.
Rotate your fingers in the direction from a to b, focusing on the smaller angle between the vectors a and b.
The orientation of the vectors a and b determines the direction of vector c.
Properties of vector products include non-commutativity, reflection, and distribution with respect to vector addition.
Unit vectors' cross product results can be positive or negative, with a memory tip provided for clarity.
The mathematical form of vector products involves multiplying corresponding components of the vectors and taking the difference.
A shortcut method for finding the cross product of vectors involves multiplying corresponding components and taking the difference.
Angular displacement is measured in terms of angles, with angular velocity denoted by omega and determined by the right-hand screw rule.
The relationship between linear and angular velocity is expressed as v = omega cross r, with v = r omega applicable for tangential velocity.
Angular acceleration, denoted by alpha, is the rate of change of angular velocity, analogous to linear acceleration in translational motion.

01:08:49

Understanding Torque and Angular Momentum

Torque is equal to the cross product of radius and force.
The angle between the forces is crucial in determining torque.
Torque can be calculated as the perpendicular component of the radius multiplied by force.
Torque depends on the position vector, force, and the angle between them.
If the angle between force and position vector is 0°, torque becomes zero.
If the angle is 90°, torque is at its maximum value.
Torque is a vector quantity with SI unit Newton meter.
Angular momentum is denoted by 'l' and is the cross product of radius and linear momentum.
The time rate of change of angular momentum of a particle equals the torque acting on it.
The total angular momentum of a system of particles is the sum of individual angular momenta.

01:27:29

Forces, Moments, and Mechanical Advantage Explained

The text discusses the concept of forces and gives an example of opening a bottle lid by rotating it.
It explains the application of forces using fingers and how they result in opposite forces causing rotation.
The text delves into the principle of moments, defining a lever as a rod pivoting on a point.
It details the conditions for mechanical equilibrium, emphasizing translational and rotational equilibrium.
The text introduces terms like load and effort, associating them with specific distances from the pivot point.
It highlights the relationship between load arm, effort arm, and the principle of moments for mechanical equilibrium.
The concept of mechanical advantage is explained as the ratio of load to effort or distance from the pivot point.
The text elaborates on how mechanical advantage influences the ease of lifting loads based on the lengths of effort and load arms.
It transitions to discussing the center of gravity, explaining how it relates to balancing objects and the concept of gravitational torque.
The text concludes by connecting the center of gravity to the distribution of mass and the uniformity of acceleration due to gravity in a body.

01:44:10

Equilibrium and Torque in Car Dynamics

At the center of gravity, the body is in equilibrium when the net gravitational torque is zero.
Equilibrium involves translational equilibrium, where the net upward force equals the net downward force.
The mass of the entire car is represented as 'm', with the net downward force acting on it.
Rotational equilibrium requires that the net torque on the body is zero.
Torque due to the back wheel is calculated by multiplying the force by the perpendicular distance.
The torque due to the back wheel is found to be 1.8 - 1.05, resulting in 1.8 - 1.05 = 0.75.
The torque due to the front wheel is calculated similarly, resulting in 1.05.
The value of RF is determined to be 7350 Newtons, leading to the value of RB as 10290 Newtons.
The force exerted on each front wheel is 3675 Newtons, while on each back wheel it is 5140 Newtons.
Moment of inertia is the rotational analog of mass, calculated by summing the mass elements' distances squared.

02:01:25

Rotational motion dynamics and conservation principles explained.

Perpendicular axis theorem has been removed from the syllabus.
Discussed calculating moment of inertia for a disc with different axes of rotation.
Encouraged self-exercise to understand and calculate moments of inertia for various shapes like cylinders and spheres.
Differentiated between quantities in motion like angular velocity, displacement, and acceleration.
Introduced equations of rotational motion analogous to translational motion equations.
Explained dynamics of rotational motion, focusing on torque as the cause of rotation.
Compared quantities of motion in rotational and translational motion, emphasizing differences in displacement, velocity, acceleration, mass, and moment of inertia.
Explored conservation of angular momentum, highlighting the constant nature of angular momentum in rotating systems.
Demonstrated practical application through solving kinetic energy and angular momentum problems for rotating objects.
Introduced the concept of rolling motion, emphasizing the distinction between rolling and slipping without motion.

02:18:31

"Rolling Motion: Velocity, Energy, and Work"

Velocity at the center of mass moves parallel to the surface, indicating the direction of motion.
In rolling without slipping, the velocity of the center of mass should be equal to the product of the radius and angular velocity.
Rolling without slipping occurs when the point of contact with the surface is at rest, ensuring motion only through rolling.
Kinetic energy in rolling motion is the sum of translational and rotational kinetic energies, calculated using specific formulas.
Work required to stop a rolling object is equal to its total energy, which is the sum of translational and rotational kinetic energies.

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