Chapter – 9,Circular Motion

The motion along circular track is called circular motion. Like moving car at turning point, motion of a fan, motion of planet on its orbit etc.

Some important terminology :

1 ) Angular displacement : It is the angle sweeps out by the position vector of a particle in a given interval of time.
2 ) Angular velocity or Angular frequency : It is the time rate of change of angular displacement.
3 ) Time period : It is the time in which an object covers one complete revolution.
4 ) Frequency : It the number of revolutions or rotations covered by an object in one second.It is the inverse of time period.
5 ) Average angular acceleration : It is the ratio of change in angular velocity to the change in time ( in which change in angular velocity is occurred ).
6 ) Instantaneous Angular acceleration : It is the limiting value of the average angular acceleration of the particle in a small interval of time as the time interval approaches to zero. It is the double derivative of Angular Displacement with respect to time whereas First order derivative of angular velocity.
7 ) Centripetal force : It is an external force which deflects the particle from its linear path to make move along a circular path. It is directed radially inward.
8 ) Centripetal acceleration : It is the acceleration acceleration through which a particle deflects from its linear path to a circular path. It is also directed along radially inward.
9 ) Centrifugal force : It is fictitious force not a reaction force. It has a concept only in a rotating  frame of reference.

Some important relations :

1 ) Relation between angular displacement and linear displacement : Let

s = linear displacement

r = radius of the circular track

θ = angular displacement

θ = s / r

2 ) Relation between angular velocity and linear velocity : Let

v = linear velocity
ω = angular velocity
θ = angular displacement
s = linear displacement
t = time
ω = θ / t
= ( s / r ) / t
= ( s / t ) / r
= v / r

ω = v / r

3 ) Relation between angular acceleration and linear acceleration : Let
v = linear velocity
ω = angular velocity
θ = angular displacement
s = linear displacement
t = time
a = linear acceleration
α = angular acceleration
α = ω / t
= ( v / r ) / t
= ( v / t ) / r
= a / r
4 ) Relation between angular velocity and linear displacement : Let
v = linear velocity
ω = angular velocity
θ = angular displacement
s = linear displacement
t = time
ω = θ / t
= ( s / r ) / t
= s / ( r * t )
5 ) Relation between angular velocity and linear acceleration: Let
v = linear velocity
ω = angular velocity
θ = angular displacement
s = linear displacement
t = time
a = linear acceleration
α = angular acceleration
α = ω / t
ω = α * t
= ( a / r ) / t
= a / ( r * t )

Expression for centripetal force : Let

r = radius of circular path

v = constant speed

Let us suppose a particle moves from point A to point B in small interval of time Δt .

v1 = velocity vector at point A

v2 = velocity vector at point B

and magnitude of v1 = magnitude of v2 = v

Here, vis drawn along the tangent at point A and v2 is drawn along the tangent at point B.

θ = angle between two tangents drawn at points A and B = angle between the two corresponding radii = < AOB

Δ= change in velocities due to change in direction in moving form point  A to point B.

Applying triangle law of vectors to the vector triangle PQS, we get

v1 + v2 = Δv

Δv =  v2 – v1

ΔAOB and ΔPQS are the isosceles triangle having the same vertex angle θ , therefore these triangles are similar triangle

∴ PQ / AB = QS / OB

Δv / AB = v / r ————————– ( 1 )

Δt is very small

∴  chord AB = arc AB = vΔt

putting the value of Ab in equation ( 1 ) we get

Δv / vΔt = v / r

Δv / Δt = v2 / r

a = v2 / r

This is the expression for centripetal acceleration.

∴ Centripetal force = mv2 / r

Expression of centripetal force in terms of angular velocity :

Centripetal force =  mv2 / r = m ( ωr )2 / r = mω2r

where

ω = angular velocity = angular frequency = v / r

Expression of centripetal force in terms of time period  :

Centripetal force =  mv2 / r = m ( ωr )2 / r = mω2r = m ( 2π / T )2r  = 4π2mr / T2

where

ω = angular velocity = angular frequency = 2π / T

T = time period

Expression of centripetal force in terms of frequency  :

Centripetal force =  mv2 / r = m ( ωr )2 / r = mω2r = m ( 2π / T )2r  = 4π2mr / T2 = 4π22

where

μ = 1 / T = frequency

T = time period

ω = angular velocity = angular frequency = 2π / T

Analogy Between Linera and Circular motion : Moment of inertia & angular momentum will be discussed later.

Bending of a cyclist : Let

m = sum of the mass of the cycle and cyclist

R = Reaction force on the cycle

g =  acceleration due to gravity

r = radius of the circular track

mg = combine weight of the cyclist and cycle

θ = angle made by the cycle with the vertical

Here, Rcosθ  and Rsinθ  are the rectangular components of R.

Rcosθ balances mg whereas Rsinθ provides centripetal force to the cyclist.

∴  Rcosθ = mg ———————————- ( 1 )

Rsinθ = mv2 / r  ——————————— ( 2 )

( 1 ) ÷ ( 2 ) we get

cosθ / sinθ = mgr / mv2

tanθ = v2 / rg

θ = tan-1 ( v2 / rg )

This is the angle through which a cyclist must bend.

Moving Car On The Level Road : Let

M = mass of the car

F1 and F= the frictional force on the two tyres ( for simplicity two tyres are considered )

R1 and R2 = Normal reactions on the tyres

μ = coefficient of the friction

∴ F1 = μR1   &    F2 = μR2

F = total frictional force = F1 + F2 = μR1 +μR2 = μ ( R1 + R2 ) = μR

where R = total normal reaction on the tyres = Mg

F = μMg ——————- ( 1 )

The required centripetal force should be less than equal to frictional force.

∴ Mv2 / r ≤ μMg

≤ √( μgr )

Where v = constant speed

r = radius of the circular track

maximum velocity = √( μgr )

Moving Car On The Banked Road Let

M = mass of the car

g = acceleration due to gravity

R = normal reaction on the car

θ = angle of banking

v = constant speed

r = radius of the circular path

Following the figure, we can write

Rcosθ = Mg ——————— ( 1 )

Rsinθ provides centripetal force

Rsinθ = Mv2 / r  —————————— ( 2 )

( 2 ) ÷ ( 1 ) we get

tanθ = v2 / rg

θ = tan-1 ( v2 / rg )

v = √( rgtanθ )

Practice Questions

1 ) Derive the expression for centripetal force. Let

r = radius of circular path

v = constant speed

Let us suppose a particle moves from point A to point B in small interval of time Δt .

v1 = velocity vector at point A

v2 = velocity vector at point B

and magnitude of v1 = magnitude of v2 = v

Here, vis drawn along the tangent at point A and v2 is drawn along the tangent at point B.

θ = angle between two tangents drawn at points A and B = angle between the two corresponding radii = < AOB

Δ= change in velocities due to change in direction in moving form point  A to point B.

Applying triangle law of vectors to the vector triangle PQS, we get

v1 + v2 = Δv

Δv =  v2 – v1

ΔAOB and ΔPQS are the isosceles triangle having the same vertex angle θ , therefore these triangles are similar triangle

∴ PQ / AB = QS / OB

Δv / AB = v / r ————————– ( 1 )

Δt is very small

∴  chord AB = arc AB = vΔt

putting the value of Ab in equation ( 1 ) we get

Δv / vΔt = v / r

Δv / Δt = v2 / r

a = v2 / r

This is the expression for centripetal acceleration.

∴ Centripetal force = mv2 / r

2 ) Derive the expression for centripetal force. Let

r = radius of circular path

v = constant speed

Let us suppose a particle moves from point A to point B in small interval of time Δt .

v1 = velocity vector at point A

v2 = velocity vector at point B

and magnitude of v1 = magnitude of v2 = v

Here, vis drawn along the tangent at point A and v2 is drawn along the tangent at point B.

θ = angle between two tangents drawn at points A and B = angle between the two corresponding radii = < AOB

Δ= change in velocities due to change in direction in moving form point  A to point B.

Applying triangle law of vectors to the vector triangle PQS, we get

v1 + v2 = Δv

Δv =  v2 – v1

ΔAOB and ΔPQS are the isosceles triangle having the same vertex angle θ , therefore these triangles are similar triangle

∴ PQ / AB = QS / OB

Δv / AB = v / r ————————– ( 1 )

Δt is very small

∴  chord AB = arc AB = vΔt

putting the value of Ab in equation ( 1 ) we get

Δv / vΔt = v / r

Δv / Δt = v2 / r

a = v2 / r

This is the expression for centripetal acceleration.