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 :
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, v1 is 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
Δv = 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 = 4π2mμ2
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 F2 = 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
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 = Mv2 / r —————————— ( 2 )
( 2 ) ÷ ( 1 ) we get
tanθ = v2 / rg
θ = tan-1 ( v2 / 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 .
v1 = velocity vector at point A
v2 = velocity vector at point B
and magnitude of v1 = magnitude of v2 = v
Here, v1 is 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
Δv = 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.
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 .
v1 = velocity vector at point A
v2 = velocity vector at point B
and magnitude of v1 = magnitude of v2 = v
Here, v1 is 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
Δv = 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.
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