**Class – 11 Physics, 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 : **

**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 :**

**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 .

**v _{1} **= velocity vector at point A

**v _{2}** = velocity vector at point B

and magnitude of **v _{1}** = magnitude of

**v**v

_{2}=Here, **v _{1 }**is drawn along the tangent at point A and

**v**is drawn along the tangent at point B.

_{2}θ = angle between two tangents drawn at points A and B = angle between the two corresponding radii = < AOB

Δ**v **= 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

– **v _{1}** +

**v**= Δ

_{2}**v**

Δ**v = v _{2} – v_{1}**

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

**∴ **PQ / AB = QS / OB

Δ**v** / 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 = v^{2} / r

a = v^{2} / r

This is the expression for centripetal acceleration.

**∴ Centripetal force = mv ^{2} / r**

**Expression of centripetal force in terms of angular velocity : **

Centripetal force = mv^{2} / r = m ( ωr )^{2} / r = mω^{2}r

where

ω = angular velocity = angular frequency = v / r

**Expression of centripetal force in terms of time period : **

Centripetal force = mv^{2} / r = m ( ωr )^{2} / r = mω^{2}r = m ( 2π / T )^{2}r = 4π^{2}mr / T^{2}

where

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

T = time period

**Expression of centripetal force in terms of frequency : **

Centripetal force = mv^{2} / r = m ( ωr )^{2} / r = mω^{2}r = m ( 2π / T )^{2}r = 4π^{2}mr / T^{2} = 4π^{2}mμ^{2}

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 = mv^{2} / r ——————————— ( 2 )

( 1 ) **÷** ( 2 ) we get

cosθ / sinθ = mgr / mv^{2}

**tanθ = v ^{2} / rg**

**θ = tan ^{-1} ( v^{2} / rg ) **

This is the angle through which a cyclist must bend.

**Moving Car On The Level Road :**

Let

M = mass of the car

F_{1} and F_{2 }= the frictional force on the two tyres ( for simplicity two tyres are considered )

R_{1} and R_{2} = Normal reactions on the tyres

μ = coefficient of the friction

**∴ **F_{1} = μR_{1} & F_{2} = μR_{2}

F = total frictional force = F_{1} + F_{2} = μR_{1} +μR_{2} = μ ( R_{1} + R_{2} ) = 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.

**∴ **Mv^{2} / r ≤ μMg

v ≤ √( 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 = Mv^{2} / r —————————— ( 2 )

( 2 ) ÷ ( 1 ) we get

tanθ = v^{2} / rg

**θ = tan ^{-1} ( v^{2} / rg ) **

**v = √( rgtanθ )**

**Practice Questions **

**Long answer type questions : **

**1 ) Derive the expression for centripetal force. **

**Answer : **

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 .

**v _{1} **= velocity vector at point A

**v _{2}** = velocity vector at point B

and magnitude of **v _{1}** = magnitude of

**v**v

_{2}=Here, **v _{1 }**is drawn along the tangent at point A and

**v**is drawn along the tangent at point B.

_{2}θ = angle between two tangents drawn at points A and B = angle between the two corresponding radii = < AOB

Δ**v **= 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

– **v _{1}** +

**v**= Δ

_{2}**v**

Δ**v = v _{2} – v_{1}**

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

**∴ **PQ / AB = QS / OB

Δ**v** / 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 = v^{2} / r

a = v^{2} / r

This is the expression for centripetal acceleration.

**∴ Centripetal force = mv ^{2} / r**

**2 ) Derive the expression for centripetal force. **

**Answer : **

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 .

**v _{1} **= velocity vector at point A

**v _{2}** = velocity vector at point B

and magnitude of **v _{1}** = magnitude of

**v**v

_{2}=Here, **v _{1 }**is drawn along the tangent at point A and

**v**is drawn along the tangent at point B.

_{2}θ = angle between two tangents drawn at points A and B = angle between the two corresponding radii = < AOB

Δ**v **= 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

– **v _{1}** +

**v**= Δ

_{2}**v**

Δ**v = v _{2} – v_{1}**

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

**∴ **PQ / AB = QS / OB

Δ**v** / 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 = v^{2} / r

**a = v ^{2} / r**

This is the expression for centripetal acceleration.