- Introduction
- Angular velocity and angular acceleration
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- Rotation with constant angular acceleration
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- Kinetic energy of Rotation
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- Calculation of moment of inertia
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- Theorems of Moment of Inertia
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- Torque
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- work and power in rotational motion
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- Torque and angular acceleration
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- Angular momentum and torque as vector product
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- Angular momentum and torque of the system of particles
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- Angular momentum of the system of particles with respect to the center of mass of the system
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- Radius of gyration
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- Kinetic Energy of rolling bodies (rotation and translation combined)
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- Solved examples

- While discussing and defining torque or moment of force ,we found that necessary condition for a body not to rotate is that resultant torque about any point should be zero

- However this condition is necessary but not sufficient for a rigid body to be static for example in absence of resultant torque a body once set in rotation will continue to rotate with constant angular velocity

- Analogous to translation motion when torque acts on a rigid body rotating about a point with constant angular velocity then angular velocity of the body does not remain constant but changes with angular acceleration α which is proportional to the externally applied torque

- Consider a force F
_{i}acting on the ith particle of mass m_{i}of the rigid body pivoted about an axis through point O as shown below in the figure

- This force F
_{i}as discussed earlier has two components one parallel to the radius vector**r**and one perpendicular to the_{i}**r**_{i}

- Component of force parallel to radius vector does not have any effect on the rotation of the body

- Component of force F
_{i}perpendicular does affect the rotation of the body and produces torque about point O through which the body is pivoted which is given by

τ_{i}=F_{i⊥}r_{i}---(21) - if F
_{i⊥}is the resultant force acting on the ith particle ,then from Newton’s second law of motion

F_{i⊥}=m_{i}a_{i⊥}= m_{i}r_{i}α ----(22)

where a_{i⊥}is the tangential acceleration of the body

- From equation (21) and (22)

τ_{i}=m_{i}r_{i}^{2}α

And taking sum over all the particles in the body we have

∑τ_{i}=∑(m_{i}r_{i}^{2}α)=α∑(m_{i}r_{i}^{2}) ---(23)

as angular acceleration is same for all the particles of the body

- we know that

∑(m_{i}r_{i}^{2}) =I

where I is the moment of inertia of the rigid body .Hence in terms of moment of inertia equation 23 becomes

∑τ=Iα ---(24)

we have denoted resultant torque acting on the body ∑τsub>i as ∑τ - Both the torque and angular acceleration are vector quantities so in vector form

∑**τ**=I**α**---(25)

- Alternatively equation (24) which is rotational analogue of Newton second law of motion ( ∑
**F**=m**a**) can be written as

∑**τ**=I**α**= I(d**ω**/dt)=d(I**ω**)/dt ---(26)

which is similar to the equation

**F**=d(m**v**/dt=d**p**/dt

where**p**is the linear momentum - The quantity I
**ω**is defined as the angular momentum of the system of particles

Angular momentum =I**ω**

**L**=I**ω** - From equation 26 we see that resultant torque acting on a system of particles equal to the rate of change of the angular momentum

∑**τ**=d**L**/dt

- 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.

- The turning effect of the force about the axis of rotation is called the moment of force or torque..

- In rotational motion torque has same importance as that of force in the linear motion.

- Torque due to a force
**F**is measured as a vector product of force**F**and position vector**r**of line of action of force from the axis of rotation.

- We already know that orque is denoted by letter
**τ**.

- If
**F**is the force acting on the particle and**r**is the position vector of particle with respect to constant point then the torque acting on the particle is given by

**τ**=**r**×**F**(8)

- FRom equation 8 magnitude or resultant of torque is given by

|**τ**|=rfsinθ (9)

where θ is the angle between**r**and**F**.

- From equation 9 if θ=90
^{0}this menas**r**is perpandicular to**F**then,

- FRom equation 8 magnitude or resultant of torque is given by

|**τ**|=rF

and if θ=0^{0}this menas**r**is parallel to**F**then,

|**τ**|=0

- Unit of torque is Dyne-cm or Newton-m

- 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.

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