Moment of Force : TorqueThe rotational analogue (effect) of force is said to be moment of force or torque.

Torque on a single Particle

The moment of the force acting on the particle with respect to the origin O is defined as the vector product.

Torque is perpendicular to the plane containing r and F, and its direction is given by the right handed screw rule.

= r x F

Dimensions M L2 T -2. Its dimensions are the same as those of work or energy. Moment of a force is a vector, while work is a scalar. The SI unit of moment of force is Newton-metre (Nm).

= rsinθ.F = r⊥F = r.Fsinθ = rF⊥

Another definition of torque

It is defined as product of force and perpendicular distance from the axis of rotation.

r F

r sinθ r cosθ

where r is the magnitude of the position vector r, i.e. the length OP, F is the magnitude of force F and θ is the angle between r and F as shown.

The moment of force (or torque) is a vector quantity.

The magnitude of torque is = rF sin θ

Note that τ = 0 if r = 0, F = 0 or θ = 00 or 1800 . Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin.

F

r

τ

θ

Torque on a rigid body

F1

r2

r1

F2

Fn

rn

It is sum of all individual torques acting on small elementary particles consisting inside a rigid body.

If forces acting on particles is Fi and radius of rotation is ri .Then torque on them is

= ri x Fii

And net external torque on body is

ext=∑ = ∑ ri x Fii

Where I = 1,2,3,……….,n

Angular momentumThe rotational analogue (effect) of Linear momentum is said to be Angular momentum.

Angular Momentum a single Particle

The moment of the linear momentum of particle with respect to the origin O is defined as the vector product.

Angular momentum is perpendicular to the plane containing r and F, and its direction is given by the right handed screw rule.

L = r x P

Dimensions M L2 T -1. The SI unit of moment of force is Newton-metre (Kg m2 s-1).

L = rsinθ.P = r⊥P = r.Psinθ = rP⊥

Another definition of Angular momentum

It is defined as product of Linear momentum and perpendicular distance from the axis of rotation.

r P

r sinθ r cosθ

If θ = 900 ,Then sin θ= 1L= r P= mvr

Where r is the magnitude of the position vector r, i.e. P is the magnitude of force and θ is the angle between r and P as shown.

The Angular momentum is a vector quantity

The magnitude of torque is L= rP sin θ

Note that L = 0 if r = 0, P = 0 or θ = 00 or 1800 . Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin.

P

r

L

θ

Angular Momentum of a rigid body

F1

r2

r1

F2

Fn

rn

It is sum of all individual angular momentums acting on small elementary particles consisting inside a rigid body.

If momentums of particles is Pi and radius of rotation is ri .Then torque on them is

L = ri x Pii

And net external torque on body is

net=∑ = ∑ ri x Pii

L L

Where I = 1,2,3,……….,n

Relation between Torque and angular momentum

Differentiate both side w r t time

dL = d(r x P)dtdt

dL = r x dPdtdt

Pxdr dt+

dL = r x Fdt

Px V + Px V = 0

dtdr =V

dL = dt ext

Conservation of Angular Momentum

When ext= 0 then dL

dt= 0 or, L= constant

Thus, if the total external torque on a system is zero then its angular momentum is conserved that is constant.

Equilibrium of BodiesFor a body to be in equilibrium there would be no translational motion as well as no rotational motion occur.

In that case

extF = ∑ Fi =0

ext= ∑

Both are vectors so they have their components i= 0

∑ Fix =0 ∑ Fiy =0 ∑ Fiz =0

∑ix= 0 ∑

iy= 0 ∑

iz= 0

Couple of Forces

When two equal but opposite forces are acting on a body simultaneously then body is in translational equilibrium but not in rotational equilibrium means that resultant force is zero but resultant torque is non zero.

F

-F

r1 r2

extF =0

ext= +

2 1 = Fr1 + Fr2

Centre of gravity

It is that point where total gravitational torque is zero.

If ri be the position of ith particle wrt its centre of gravity then total torque about the CG

CG= ∑

i= ∑ mi g = 0

Determining the centre of gravity of a body of irregular shape. The centre of gravity G lies on the vertical AA1 through the point of suspension of the body A.

Centre of mass and centre of gravity are two different physical points , however in small uniform bodies they coincides to each others.

Balancing a cardboard on the tip of a pencil. The point of support, CG, is the centre of gravity.

Moment of inertia

Rotational analogue of inertia is said to be Moment of Inertia.

The tendency of bodies to resist any change in state regarding rotational motion .

If a body is in rest then it don’t want to rotates.

If a body is rotating in a certain direction then it don’t want to change its direction of rotational motion.

It means if a body is rotating then it don’t wants to stop.

Moment of Inertia of a Point Body

It is defined as the product of mass and square of perpendicular distance from the axis of rotation. It is scalar quantity

I = Mr2

SI unit = Kg-m2

r

Dimension = M L2 T 0

Moment of Inertia of a Rigid Body

I1

r2

r1

I2

In

rn

It is scalar sum of all individual moment of inertia of particles consisting inside a rigid body.

m2

m1

mn

Let the particles have masses mi radius of rotation ri and moment of inertia Ii

Then total Moment of Inertia

I = ∑ I i = ∑ mi ri

Radius of Gyration

The radius of gyration of a body about an axis is defined as the distance from the axis of a mass point whose mass is equal to the mass of the whole body and whose moment of inertia is equal to the moment of inertia of the body about the axis.

r2

r1

rnm2

m1

mn

MK

MK2

= I = ∑ I i = ∑ mi ri

2

K = ∑ mi ri

2

MSI unit- metre Dimension - L

The moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

Theorem of perpendicular axes

Iz= Ix + Iy

Theorem of perpendicular axes applicable to a planar body; x and y axes are two perpendicular axes in the plane and the z-axis is perpendicular to the plane.

Theorem of parallel axes

The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

Iz′ = Iz + Ma2

z and z′ are two parallel axes separated by a distance a; O is the centre of mass of the body, OO’ = a.

ω = ω0 + αt θ = θ0 + ω0 t + ½ αt 2

ω2 = ω02 +2α (θ –θ0 )

KINEMATICS OF ROTATIONAL MOTIONABOUT A FIXED AXIS

The kinematical quantities in rotational motion, angular displacement (θ), angular velocity (ω) and angular acceleration (α) respectively correspond to kinematic quantities in linear motion, displacement (x), velocity (v) and acceleration (a).

kinematic equations for rotational motion with uniform angular acceleration are

Where θ0= initial angular displacement of the rotating body, and ω0 = initial angular velocity of the body.

DYNAMICS OF ROTATIONAL MOTIONABOUT A FIXED AXIS

Work done by a torque

x'

y'

θ

dθ

ds 1

α1

Ф1

r1

F1

C

P1

P’1

φ1 + α1 = 90°

Displacement of the particle ds1 = r1dθ & direction tangential at P1 to the circular path. Here dθ is the angulardisplacement of the particle, dθ = P∠ 1C P′1.

Let a rigid body rotating about Z-axis so its one of the particle move from P1 to P’1 due to force F1 on that Particle. The particle at P1 describes a circular path of radius r1 with centre C.

dW1 = F1ds1cos Ф1 = F1(r1 dθ)sinα1

The effective torque due to F1 is τ1= CP × F1= r1F1 sinα1 ,

Therefore, dW1 = τ 1dθIf there are more than one forces acting on the body, torques due to the different forces as τ 1, τ 2,… dW = (τ 1+ τ 2+...)dθ (τ = τ 1 + τ 2 + .....)

dW = τ dθ

Power

Kinetic Energy of Rotating body

Relation between Torque & angular accelerationThe rate of change of kinetic energy is

Equate rate of work done and of change in kinetic energy

ANGULAR MOMENTUM IN CASE OF ROTATION ABOUT A FIXED AXIS

As we know the relation between torque and moment of inertia about a fixed axis. Now we use that relation to interrelates angular momentum & moment of inertia.

Taking integration both sides

Conservation of Angular Momentum

If the external torque is zero then

It means if moment of inertia decreased then angular speed increased & vice versa.

Some Examples of Conservation of Angular Momentum

A circus acrobat and a diver take advantage of this principle.

Skaters and classical, Indian or western, dancers performing a pirouette on the toes of one foot display ‘mastery' over this principle.

ROLLING MOTION

All wheels used in transportation have rolling motion. It is combination of translational & rotational motion.

Rolling without Slipping

This means that at any instant of time the bottom of the disc P0 which is in contact with the surface is at rest on the surface.

ω

The velocity of any point of the disc, like P0, P1 or P2, consists of two parts, one is the translational velocity vcm and

the other is the linear velocity vr= rω, on account of rotation.

vr is directed perpendicular to the radius vector of the given point with respect to C.

The centre of mass of the disc moves with velocity, vcm. The disc rotates with angular velocity ω about its axis which passes through C; vcm = Rω, where R is the radius of the disc.

vcm is the velocity of C. It is parallel to the level surface.

The velocity of point at the top of the disc (v1) has a magnitude vcm+ rω or 2 vcm and is directed parallel to the level surface.

Kinetic Energy of Rolling Motion

General result for a system of particles, according to which the kinetic energy of a system of particles (K) can be separated into the kinetic energy of motion of the centre of mass (translation) (MV2/2) and kinetic energy of rotational motion about the centre of mass of the system of particles (K′ ).