- Rotational Motion
- |
- Angular velocity
- |
- Angular acceleration
- |
- Rotation with constant angular acceleration
- |
- Kinetic energy of Rotation
- |
- Calculation of moment of inertia
- |
- Parallel Axis Theorem|Theorems of Moment of Inertia
- |
- Perpendicular Axis Theorem
- |
- Torque
- |
- work and power in rotational motion
- |
- Angular acceleration
- |
- Relationship between Angular momentum and torque
- |
- Conservation of Angular momentum
- |
- Radius of gyration
- |
- Rolling Motion|Kinetic Energy of rolling bodies
- |
- Rotational Motion problems with solutions

- In any inertial frame of refrance the moment of linear momentum of a particle is known as angular momentum or, angular momentum of a particle is defined as the moment of its linear momentul.

- In rotational motion angular momentum has the same significance as linear momentum have in the linear motion of a particle.

- Value of angular momentum of angular momentum is equal to the product of linear momentum and
**p**(=m**v**) and the position vector**r**of the particle from origin of axis of rotation.

- Angular mmomentum vector is usually represented by
**L**.

- If the linear momentum of any particle is
**p**=m**v**and its position vector from any constant point be**r**then abgular momentum of the particle is given by

**L**=**r**×**p**= m(**r**×**v**) (1)

- Angular momentum is a vector quantity and its direction is perpandicular to the direction of
**r**and**p**and could be found out by right hand screw rule.

- From equation 1 scalar value or magnitude of angular momentum is given as

|**L**|=rpsinθ (2)

where V is the angle between**r**and**p**.

- For a particle moving in a circular path

**v**=**ω**×**r**; (3)

where**ω**is the angular velocity.

Therefore

**L**=m[**r**×(**ω**×**r**)] = m{**ω**(**r**.**r**)-**r**(**r**.**ω**)} = mr^{2}**ω**=I**ω**; (4)

(**r**.**ω**)=0 because in circular motion**r**and**ω**are perpandicular to each other. Here I is the moment of inertia of the particle about the given axis also the direction of**L**and**ω**is same and this is a axial vector.

writing equation 1 in the component form we get

- Writing angular momentum in component form we get

writing equation 5 again we get

Comparing unit vectors on both the sides we get

- Unit of angular momentum in CGS is gm.cm
^{2}/sec and in MKS system it is Kgm.m^{2}/sec or Joule/sec.

- Differentiating equation 1 w.r.t. t we get

- But from Newton's second law of motion we have

Hence rate of change of angular momentum with time is equal to the torque of the force.

a. L

b. L/4

c. L/2

d. 2L

Angular momentum for this is defined as

=mr

First case

L=mr

Second case

L

So, L

a. is zero

b. remains constant

c. goes on increasing

d. goes on decreasing

L=(mv)Xr

or

L=mvrsinθ

Now rsinθ=perpendicular distance from x axis which is constant

So Angular momentum is constant

Class 11 Maths Class 11 Physics Class 11 Chemistry