हमारी app डाउनलोड करे और फ्री में पढाई करे Download our app now हमारी app डाउनलोड करे # law of conservation of angular momentum derivation formula state the examples its applications

By   April 13, 2023

get all information law of conservation of angular momentum derivation formula state the examples its applications class 11th physics topic ?

Centre of Mass of Discrete Particles

For a system of particles, the centre of mass is defined as that point where the entire mass of the system is imagined to be concentrated, for considerations of its translational motion. If r1, r2, r3, … rn are the position vectors of masses m1, m2, m3, …mn respectively, the position vector of the centre of mass of the system is where M is the total mass of the system of particles. rCM is the weighted average of all the position vectors of the particles of the system, the contribution of each particle being proportional to its mass. For a system consisting of two particles, the centre of mass is which means that the centre of mass lies exactly in the middle of the line joining the two masses.

Centre of Mass of a Body having Continuous Distribution of Mass

If a body has continuous distribution of mass, the position of its centre of mass is determined by dividing the body into a very large number of extremely small elements. If dm is the mass of the element and it is at a distance x and y from the origin of chosen coordinate system, then x and y coordinates of the centre of mass are given by Finding Centre of Mass of a System when a Part of its Mass is Removed

Consider of system of mass M. If a mass m is removed, the remaining mass = M – m which may be written as M + (–m). Then the x and y coordinates of the centre of mass of the remaining portion are given by where x and y are the coordinates of the centre of mass of the complete part and x’ and y’ are the coordinates of the centre of mass of the removed part

Velocity and Acceleration of Centre of Mass of a System of Particles

If r1, r2, … rn are the position vectors of masses m1, m2, … mn, respectively, the position vector of the centre of mass of the system of particles is given by if no net external force acts on a system, its centre of mass will remain at rest or will move with a constant velocity.

Momentum Conservation and Centre of Mass Motion

We have seen that where P = MvCM is the total linear momentum of the system of particles which is equal to the product of the total mass of the system and the velocity of the centre of mass. Thus, if no net external force acts on a system, the total linear momentum of the system remains constant; the total linear momentum being the vector sum of linear momentum of individual particles,

i.e. P = P1 + P2 + … + Pn

## Torque

If a force F acts on a particle P whose position vector with respect to the origin of an inertial reference frame is r, the torque τ acting on the particle with respect to the origin is defined as [Fig. ] where q is the angle between vectors r and F. Torque t is a vector quantity. Its magnitude is given by τ = rF sin θ; its direction is normal to the plane containing vectors r and F and can be determined by the right–hand screw rule. Unit of Torque

Torque has the same dimensions as those of work (both being force times distance) viz. ML2T–2. The two are, however, very different quantities. Work is a scalar, torque is a vector. To distinguish between the two we express work in joules and torque in newton–metre (N m). Couple Two equal antiparallel forces having different lines of action constitute a couple. The moment of couple or torque = Fr ⊥  (Fig. 5.12) = magnitude of either force ¥ perpendicular distance between the two antiparallel forces Work done by torque If a force F acts on a rigid body at perpendicular r⊥ from the axis of rotation, the work done by the force in rotating the body through an angle △θ is given by where w is the angular velocity.

Angular Momentum

The angular momentum L of a particle P with respect to the origin of an inertial reference frame is defined as [Fig. ]

L = r x p

where r is the position vector of the particle and p its linear momentum. In terms of magnitudes,

L = rp sin θ = r⊥ x p

where q is the angle between vectors r and p. The dimensions of angular momentum are (ML2 T–1) and its SI unit is kg m2 s 1. The direction of L is perpendicular to the plane containing the vectors r and p and its sense is given by the right–hand rule. ### Law of Conservation of Angular Momentum

If no external torque acts, the total angular momentum of a body or a system of particles is conserved. We have seen that the rate of change of angular momentum of a particle is equal to the torque produced by the total force. If, in a certain situation, the torque itself vanishes, then it follows that the angular momentum of the particle will remain constant. This is the law of conservation of the angular momentum of a particle. One trivial situation is when the force vanishes. Then the torque vanishes too. The particle then moves freely in a straight line in accordance with Newton’s first law in which case both linear and angular momenta are conserved. A general situation is when the torque vanishes without the force itself vanishing. The torque τ will vanish if the component F⊥ (the angular component) of F vanishes but the radial component FII does not. The radial component FII is the component of F along the radius (or position) vector r. Hence, if the force acting on the particle is purely radial (i.e. if it is directed along or against its position vector) then the torque acting on the particle vanishes and its angular momentum is conserved and so is its areal velocity.

Moment of Inertia

The moment of inertia of a rigid body about a particular axis may be defined as the sum of the products of the masses of all the particles constituting the body and the squares of their respective distances from the axis of rotation, i.e. Its value depends upon the particular axis about which the body rotates and the way the mass is distributed in the body with respect to the axis of rotation. In the case of a body which does not consist of separate, discrete particles but has a continuous and homogeneous distribution of matter in it, the summation is replaced by integration, so that where dm is the mass of an infinitesimally small element of the body at a distance r from the axis of rotation. Moment of inertia is a scalar quantity. Its SI unit is kg m2 and its dimensions are (ML2).

The radius of gyration of a body about its axis of rotation may be defined as the distance from the axis of rotation at which, if the entire mass of the body were concentrated, its

moment of inertia about the given axis would be the same as with its actual distribution of mass. It is usually denoted by the letter K. If M is the mass of the body, its moment of inertia I in terms of its radius of gyration K can be written I = MK2

Moment of Inertia and Rotational Kinetic Energy

Kinetic energy of a rotating body is related to moment of inertia as where w is the angular velocity (or frequency) of the body.

Moment of Inertia and Torque
The magnitude of torque is given by

τ = Iα
where α is the angular acceleration of the body.

Moment of Inertia and Angular Momentum

The magnitude of angular momentum of a rotating body is given by

L = Iw

Parallel Axes Theorem

If M is the total mass of a body and h the distance between two parallel axes, then according to parallel axes theorem #### Perpendicular Axes Theorem

The theorem of perpendicular axes for a body of plane lamina states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its own plane and intersecting each other at the point where the perpendicular axis passes through it. If Ix and Iy are the moments of inertia of a plane lamina about the perpendicular axes x and y respectively which lie in the plane of the lamina and intersect each other at O, then the moment of inertia I of the lamina about an axis passing through O and perpendicular to its plane is given by Expressions for moment of inertia of bodies of regular shapes about particular axes of rotation :

Kinematics of Rotational Motion with Constant Angular Acceleration

Consider a body rotating with an initial angular velocity w0. It is given a constant angular acceleration α (by applying a constant torque) for a time t. As a result it acquires a final angular velocity w and suffers an angular displacement θ in time t. The equations of rotational motion are Rolling Motion without Slipping

(1) Total Kinetic Energy The total kinetic energy of a body which is moving as well as rotating is equal to the sum of translational K.E. and rotational K.E., i.e. where m = mass of the body, v = linear velocity of its centre of mass, I = moment of inertia of the body about an axis passing through its centre of mass and w = angular velocity of rotation.

Instantaneous Velocity of a Point on a Rolling Body

Consider a wheel of radius R(= AC = BC) rolling without slipping on a horizontal rough surface (Fig. ). Every point on the wheel has instantaneous velocity. For a point P at a distance r from the centre of mass C, the instantaneous velocity is the vector sum of velocity v of the centre of mass and tangential velocity vt of point P relative to the centre of mass, i.e.  If a body rolls on a surface without slipping, the instantaneous velocity of the point of contact with the surface is zero

A Body Rolling without Slipping on a Rough Horizontal Surface

Horizontal force F is applied at the centre of mass of a body (disc, ring, cylinder or sphere) of mass M and radius R (Fig. 5.27) on a rough horizontal surface. If f is the frictional force, and the body rolls without slipping αCM = αR From (i) and (ii), acceleration of centre of mass is If the force is applied tangentially to the body as shown in Fig. , then  A Body Rolling without slipping on a Rough Inclined Plane

A body (ring, disc, cylinder or sphere) of mass M and radius R is rolling (without slipping) down a rough inclined plane of inclination θ (Fig. below). For linear motion parallel to the plane

Mg sin θ – f = Ma                      (i)

where a = linear acceleration of the centre of mass. For rotational motion about the axis through the centre of mass I is the moment of inertia about the centre of mass. Using (i) and (ii), we get  Condition for rolling without slipping

To prevent slipping, f ≤ µN, where µ is the coefficient of static friction between the body and the plane and N = Mg cos θ is the normal reaction. Hence to avoid slipping, Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12