## Friday, February 8, 2008

### IIT JEE Physics Formula Revision 7. Circular motion

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 = ω0t + ½ αt²

ω = ω0 + αt

ω² = ω0² + 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

at = rate of change of speed of the particle in circular motion

at = 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)

er = unit vector along the radius at a point on the circle

er = i cos θ+j sin θ

et = unit vector along the tangent at a point on the circle

et = - 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 er + dv/dt et

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 er

Acceleration of the particle is in the direction of - er , i.e, towards the centre of the circle. The magnitude of the acceleration is

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 er + dv/dt et

the radial component is ar = - ω²r = -v²/r
the tangential component is at = dv/dt

The magnitude of the acceleration

a = √ (ar² + at²)

= √[( 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

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 = fs/M

Or M v²/r ≤ fs

As fs≤ μsMg

M v²/r ≤ μsMg
That means v²/gr ≤ μs
v≤√( μsgr)

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+ ω4R²sin² 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