**Angular variables**

θ = angular position of the particle

ω = angular velocity = d θ/dt = lim∆t→0 ∆θ/∆t

α = angular acceleration = d ω/dt = d²θ/dt²

If the angular acceleration is constant, formulas similar in form to linear formulas can be used to find the angular variables:

θ = ω

_{0}t + ½ αt²

ω = ω

_{0}+ αt

ω² = ω

_{0}² + 2 α θ

**Linear variables of circular motion**

s = Linear distance traveled by the particle in circular motion

∆s = Linear distance traveled by the particle in circular motion in time ∆t

∆s = r∆θ

Where

r = radius of the circle over which the particle is moving

∆θ = angular displacement in time ∆t

∆s/∆t = r∆θ/∆t

v = r ω

where

v = linear speed of the particle

a

_{t}= rate of change of speed of the particle in circular motion

a

_{t}= dv/dt = rdω/dt = r α

**Unit Vectors along the Radius and Tangent on a point on the circle on which the particle is moving.**

If x-axis is horizontal and y axis is vertical (normal representation)

**e**= unit vector along the radius at a point on the circle

_{r}**e**=

_{r}**i**cos θ+

**j**sin θ

**e**= unit vector along the tangent at a point on the circle

_{t}**e**= -

_{t}**i**sin θ+

**j**cos θ

Position vector of the point P

If the angular position is θ, x coordinate is r cos θ, and y coordinate is r sin θ.

Hence position vector is

**r**= r(

**i**cos θ +

**j**sin θ)

Differentiating position vector with respect to time we get velocity

**v**= r ω(-

**i**sin θ +

**j**cos θ)

Differentiating v with respect to time

**a**= dv/dt = - ω²r

**e**+ dv/dt

_{r}**e**

_{t}**Uniform Circular Motion**

If the particle moves in the circle with a uniform speed (v = constant), it is uniform

circular motion.

In this case dv/dt = 0

Hence

**a**= dv/dt = - ω²r

**e**

_{r}Acceleration of the particle is in the direction of -

**e**, i.e, towards the centre of the circle. The magnitude of the acceleration is

_{r}a = ω²r = v²/r (as v = rω)

**Nonuniform circular motion**

In nonuniform circular motion, speed is not constant and the acceleration of the particle has both radial and tangential components.

**a**= - ω²r

**e**+ dv/dt

_{r}**e**

_{t}the radial component is a

_{r}= - ω²r = -v²/r

the tangential component is a

_{t}= dv/dt

The magnitude of the acceleration

a = √ (a

_{r}² + a

_{t}²)

= √[( ω²r) ² + (dv/dt) ²]

The direction of this acceleration makes an angle α with the radius connecting the centre of the circle with the point at that instant.

Tan α = (dv/dt)/ (ω²r)

**Forces in Circular Motion**

If a particle of mass m is moving along a circle with uniform speed, the force acting on it has to be

F = ma = mv²/r = m ω²r

This force is called centripetal force.

**Centrifugal Force**

It is equal to the centripetal force

F = mv²/r = m ω²r

**Circular Turnings on the Road**

When a turn is made, the speed has to be less so that friction can make the vehicle turn in stead of skid

For a safe turn we need to have

v²/r = f

_{s}/M

Or M v²/r ≤ f

_{s}

As f

_{s}≤ μ

_{s}Mg

M v²/r ≤ μ

_{s}Mg

That means v²/gr ≤ μ

_{s}

v≤√( μ

_{s}gr)

Instead of relying on friction, road are given an angle to provide centripetal force.

Tan θ = v²/gr

θ is the angle to be given to the road based v. Road are designed for assumed normal speed.

**Effect of Earth’s Rotation on Apparent Weight**

Axis of rotation of earth = The line joining the north and south pole = SN

Every point P on earth moves in circle.

Draw a perpendicular PC to SN from P.

Centre of the earth = O.

The distance between a point on the equator and centre of earth O = R radius of earth.

The angle between OP and PC is θ

θ is called the colatitude of the place where P is there.

If heavy particle of mass m is suspended through a string at P, the forces on it will be

1. gravitational attraction mg towards earth along PO

2. Centrifugal force = m ω² r

3. Tension in the string

The resultant of mg and m ω² r =

m√(g²+ ω

^{4}R²sin² θ-2g ω²R sin² θ) = mg’

The direction of this resultant force with OP is given by

Tan α = ω²Rsin θ cos θ/(g - ω²Rsin² θ)

A plumbline stays along g’

g’ = √(g²- ω²Rsin² θ(2g- ω²R ))

As 2g> ω²R; g’

At equator θ = 90° and

G’ = g- ω²R

At poles θ = 0° and hence

g’ = g

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