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
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
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
In this case dv/dt = 0
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.
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 = fs/M
Or M v²/r ≤ fs
As fs≤ μsMg
M v²/r ≤ μsMg
That means v²/gr ≤ μs
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