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XII SCIENCE
[PHYSICS – I]
CIRCULAR MOTION
XII SCIENCE NOTES
Prof. R. S. Gade.
March 10, 2013
THEORY QUESTIONS
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Define angular displacement. State right hand rule and right handed screw rule
which gives direction of angular displacement.
Define angular displacement, angular velocity and angular acceleration.
Prove the relation, v = ω x r, where symbols have their usual meanings.
Define uniform circular motion. Show that linear speed of particle performing
circular motion is the product of radius of circle and angular speed of particle.
Define period and frequency of particle performing uniform circular motion.
State their SI units.
Obtain the relation between magnitude of linear acceleration and angular
acceleration in circular motion.
Obtain an expression for acceleration of particle performing uniform circular
motion by using calculus method.
Obtain an expression for acceleration of particle performing uniform circular
motion by using geometrical method.
Define centripetal force. Give its any four examples.
Define centrifugal force. Give its any four examples.
Distinguish between centripetal and centrifugal force.
Derive expression for maximum safety speed with which vehicle should move
along a curved horizontal road. State the significance of it.
What is banking of roads? Why is it necessary?
Obtain the expression for maximum safety speed with which vehicle can be
safely driven along banked curved road.
Show that angle of banking is independent of mass of vehicle.
Define conical pendulum. Obtain an expression for the angel made by string
with the vertical. Hence deduce the expression for linear speed of bob of the
conical pendulum.
Define period of conical pendulum and obtain the expression for its period.
What is vertical circular motion? Show that motion of the object revolving in
vertical circle is non-uniform circular motion.
What is the difference between uniform circular motion and non-uniform
circular motion?
Derive expressions for linear velocity at lower point, midway and top position
for a particle revolving in a vertical circle if it has to just complete circular
motion without slackening the string at top.
Derive an expression for difference in tensions at highest and lowest point for a
particle performing vertical circular motion.
Obtain expressions for tension at highest position, midway position and bottom
position for an object revolving in a vertical circle.
Obtain expressions of energy at different positions in the vertical circular
motion.
Write the kinematical equations for circular motion in analogy with linear
motion.
* * * * *
[RSG/XII/PHY-I/CIRCULAR MOTION]
Page 2
CIRCULAR MOTION
CIRCULAR MOTION: The motion of the particle along the circumference of a
circle is called circular motion.
Examples: i) Motion of tip of blades of fan.
ii) Motion of merry-go-round.
iii) Motion of satellite around the planet.
There are two types of circular motions,1) Non-uniform circular motion: The motion of particle along the circumference of
a circle with variable speed. e.g. Motion of the particle in a vertical circle
2) Uniform circular motion: The motion of a particle along the circumference of a
circle with constant speed.
Examples: i) Motion of tip of the hands of a clock.
ii) Motion of an electron around the nucleus.
iii) Motion of earth around the sun.
* In uniform circular motion as the speed of the particle is constant, in U.C.M.
revolutions are repeated after equal interval of time. Hence UCM is a periodic
motion.
* In any type of circular motion the linear velocity of the particle is always directed
along the tangent to circular path. Hence in UCM even the speed remains
constant, the direction of linear velocity goes on changing continuously. On
account of this change in direction the linear velocity of particle varies. Hence
UCM is an accelerated motion.
* A vector drawn from the centre of the circular path to the position of the particle
is called the radius vector of the particle. The magnitude of the radius vector is
constant but its direction changes continuously, which is always directed away
from the centre of the circular path.
ANGULAR DISPLACEMENT:
“The angle traced by radius vector of the particle in given time is called its
angular displacement”
* Angular displacement is measured in radian or rad
* For smaller magnitude it is a vector quantity. Its direction is along the axis of
rotation, perpendicular to plane of circular motion, as given by Right Hand Rule.
Arc length
Its magnitude is given by: Angular displacement =
∴ δθ = δs / r
Radius
In vector form we write: δs = δθ x r
* Finite angular displacement is not a vector because it does not obey the
commutative and associative laws of vector addition.
* Right hand rule: “Imagine the axis of rotation to be held in our right hand with
fingers curled round the axis and the thumb stretched along the axis. If the curled
fingers indicate the sense of rotation then the thumb indicates the direction of the
vector (i.e. of angular displacement).”
* Right handed screw rule: “Hold the right handed screw with its head in the
plane of circular motion. Rotate the screw head in the direction of motion then the
direction in which the screw advances gives the direction of angular
displacement.”
[RSG/XII/PHY-I/CIRCULAR MOTION]
Page 3
ANGULAR VELOCITY:
“The rate of change of angular displacement with respect to time is called angular
velocity.”
If the particle has an infinitesimal angular displacement δθ in a short time δt,
δθ
dθ
then its instantaneous angular velocity is given by: ω = lim = .
δt→0 δt
dt
* The SI unit of angular Velocity is rad/s and its dimensions are [M0L0T-1]
* The angular Velocity is a vector quantity having direction same as that of angular
displacement i.e. along the axis perpendicular to the plane of rotation as given by
right hand rule.
ANGULAR ACCELERATION:
“The rate of change of angular velocity with respect to time is called angular
acceleration.”
δω
dω
* The instantaneous angular acceleration is given by: α = lim = .
δt→0 δt
dt
* If the magnitude of the angular velocity changes from ω1 to ω2 in time t, the
magnitude of the average angular acceleration is given by:
ω − ω1
2π(n 2− n 1 )
αav = 2
=
.
t
t
2
* The SI unit of angular acceleration is rad/s and its dimensions are [M0L0T-2]
* The direction of angular acceleration is similar to the change in angular velocity.
i.e. when angular velocity is increasing then dω is positive and direction of
angular acceleration is same as that of angular velocity. While
when angular
velocity is decreasing then dω is negative and the direction of angular acceleration
is apposite to that of angular velocity.
RELATION BETWEEN LINEAR VELOCITY & ANGULAR VELOCITY:
Consider a particle moving along a circular path of
B
radius, r in anticlockwise direction with uniform angular
δs
speed ω and linear speed v. Suppose that the particle
moves from the point A to the point B in a very short
δθ
A
O r
time δt. During this time the distance covered by the
particle is the arc length δs and its angular displacement
is δθ. As δt is very small, δs is also so small that it can
be considered as a straight line distance i.e linear
displacement of the particle in time δt. Therefore the linear speed of the particle
δS
can be given by: V = lim -------- (i) But, δs = r δθ .
δt→0 δt
n
∴ Eq (i) becomes: V = lim
r δθ
= r lim
δt→0 δt
δθ
δt→0 δt
= r dθ/dt ∴
v = rω ------- (ii)
∴ Linear velocity = Radius × Angular velocity.
This gives the relation between the linear speed and angular speed.
In vector form,
V = ω × r ------ (iii)
RELATION BETWEEN LINEAR ACCELERATION AND ANGULAR
ACCELERATION: Consider a particle performing non uniform circular motion
along a circular path of radius r in anticlockwise direction.
Then we have: V = ω × r
Differentiating this equation w.r.t. time, t we get the total linear acceleration as:
a=ω ×
[RSG/XII/PHY-I/CIRCULAR MOTION]
dr
dt
+
dω
dt
× r = (ω × V) + (α × r)
Page 4
*
*
*
*
This equation shows that in non-uniform circular motion the acceleration of the
particle has two components. The component, (ω × V) is called radial component
and is found to be directed along the radius towards the centre. While the
component (α × r) is called tangential component and is directed along the
tangent to circular path.
Therefore the acceleration of particle performing non-uniform circular motion is
given by:
a = aT + aR
Therefore magnitude of acceleration in non-uniform circular motion is:
a = a2T + a2R
PERIOD AND FREQUENCY OF U.C.M.:
“The time taken by a particle performing UCM to complete one revolution is
called period of UCM.”
2πr
2π
T=
=
v
ω
“The number of revolutions performed per unit time by a particle performing
UCM is called its frequency (n).”
1
v
ω
n= =
= .
T
2πr
2π
The SI unit for periodic time is second (s) while that of frequency is hertz (Hz)
Dimensions of period is [M0L0T1] and that of frequency are [M0L0T-1]
Radial acceleration / centripetal acceleration / acceleration of particle in
UCM:
a) Geometrical method:
Consider a particle performing UCM along the circular path of radius r
in anticlockwise direction with constant linear
R
speed v and constant angular speed ω. let in a very
𝜹𝑽
short time δt, particle moves from the point A to the
Q
𝑽𝟏
point B.
P
𝑽𝟐
Let V1 and V2 be the linear velocities of the
δθ
particle when it is at the point A and B resp. as
B
represented by vector AP and vector BQ resp. in
𝑽𝟏
fig. Draw a vector BR equal and parallel to vector
δs
AP; and join RQ.
δθ
By triangle law of vector addition, we have
A
O
r
BR + RQ = BQ. ⇒ RQ = BQ – BR
= V2 - V1 = δV
Thus, δV is the change in velocity of the particle in
time δt. Therefore by definition: a =
δV
δt
= lim
δV
δt→0 δt
Now, when radius vector rotates through an angle δθ, the tangent also rotates
through an angle δθ. When δt → 0, δθ becomes very small.
∴ δθ = RQ / BR = δv / v ⇒ δv = v δθ
∴ The magnitude of the acceleration is given by:
δV
Vδθ
δθ
a = lim = lim
= v lim
δt→0 δt
δt→0
[RSG/XII/PHY-I/CIRCULAR MOTION]
δt
δt→0 δt
Page 5
V2
∴
a = v ω = r ω2 =
------- (i)
r
This is the required equation for magnitude of acceleration of the particle
performing UCM. The direction of this acceleration is same as that of RQ, and in
the limiting case of δt → 0 the direction of RQ is found to be perpendicular to the
tangent at A i.e. it is along the radius directed towards the centre. Hence it is
known as the centripetal acceleration or radial acceleration.
* In vector form: a = - ω2 r.
b) Calculus method:
Consider a particle performing UCM along the circular path of radius r and centre
at the origin of x-y co-ordinate system. Let the
Y
particle moves from the point A to the point B(x, y)
in a time, t. Let the angular displacement of the
B(x,y)
particle is, θ and angular velocity is, ω. Then, θ = ω
N
t. At time t radius vector of the particle is given by,
y 𝒓
X
θ
X’
r = rcosωt i + rsinωt j ----------- (i)
x M A
Differentiating this equation w.r.t. „t‟ we get
O
velocity of particle as:
V = −rωsinωt i + rωcosωt j ------ (ii)
Differentiating this equation w.r.t.„t‟ we get
Y’
acceleration of particle as:
a = −rω2 cosωt i − rω2 sinωt j
= −ω2 (rcosωt i + rsinωt j ) = −ω2 r ---- (iii)
This is the required equation for acceleration of the particle in UCM, which
clearly represents that the magnitude of the acceleration is rω2 and – ve sign in
this equation represents that the direction of acceleration is opposite to the radius
vector i.e. it is directed towards the centre. Therefore this acceleration is known as
the centripetal acceleration. As it is along the radius of the circular path it is also
known as radial acceleration.
Centripetal and Centrifugal Forces:
Centripetal force: “The force which acts on the particle along the radius towards the
centre of the circle and which keeps the particle moving along circular path is
called centripetal force.”
For the body of mass m moving in a circular path of radius r with the
speed v or ω the magnitude of the centripetal force is given by:
F = mv2/r = mrω2 = 4π2n2mr = 4π2mr/T2.
Examples:
i) For the circular motion of satellite around the planet the centripetal force is
the gravitational force exerted by the planet on the satellite.
ii) For the circular motion of the electron around the nucleus the centripetal force
is the force of attraction between + ve charge on nucleus & –ve charge on
electron.
iii) For circular motion of an object tied to a string & revolved in horizontal circle
around the other end the centripetal force is the tension produced in the string.
iv) In case of train negotiating the curve, the necessary centripetal force is
provided by push due to rails on the wheels of a train.
[RSG/XII/PHY-I/CIRCULAR MOTION]
Page 6
v) For a car travelling round a circular horizontal road with uniform speed, the
necessary centripetal force for negotiating the curve is provided by the force
of friction between tyres of vehicle and road surface.
* From these examples it is clear that the centripetal force is due to any one of the
known interactions, hence it is a real force.
* It exist in inertial as well as non-inertial frame of reference, and it has
independent existence
* The work done by this force on the body is zero.
Centrifugal Force:
“A force experienced by a particle in circular motion, along the radius away from
the centre of the circular path is called centrifugal force.”
* Magnitude of the centrifugal force is equal to the centripetal force.
Examples:
i) If vehicle is moving along a curved path a passenger in the vehicle
experiences a push in the outward direction, which is nothing but the
centrifugal force.
ii) A coin placed slightly away from the centre of a rotating gramophone disc
slips towards the edge of the disc. This is due to centrifugal force acting on
the coin.
iii) Centrifugal force is used for drying cloths in washing machine.
iv) The bulging of earth at equator and flattening at poles is due to centrifugal
force acting on it.
v) A bucket full of water is rotated in a vertical circle at a particular speed, so
that water does not fall. This is because weight of water is balanced by
centrifugal force acting on it.
* As centrifugal force is not due to any of the known interaction, it is not a real
force but it is an imaginary force. It arises due to the acceleration of frame of
reference; hence it is called as pseudo force.
* It exists in non-inertial frame of reference only, and it has no independent
existence.
Distinguishing between centripetal force and centrifugal force:
Centripetal Force
Centrifugal force
This force is directed towards the
This force is directed away from
1.
1.
centre of circular path.
the centre of circular path.
2. This force is real force.
2. This force is imaginary force
It exist due to interaction between
It exist due to acceleration of
3.
3.
the two bodies.
frame of reference.
It exist in both inertial and nonIt exist in only non-inertial frame
4.
4.
inertial frame of references.
of reference.
5. It exist independently of motion.
5. It exists only during the motion.
2
6. Vector form: F = - m ω r
6. Vector form: F = + m ω2 r
Maximum speed of vehicle along a horizontal curved road:Consider a vehicle of mass m is moving along a horizontal curved road of
radius r with constant speed v. When vehicle moves along the curved horizontal
road, it performs the circular motion. For this motion necessary centripetal force
[RSG/XII/PHY-I/CIRCULAR MOTION]
Page 7
is provided by the frictional force between road and vehicle. Therefore, if μ is the
coefficient of friction between road and vehicle, for the safety motion of vehicle,Centripetal force = Frictional force ∴ mv2/ r = μmg ⟹ vmax = μrg
This is the maximum speed of vehicle with which it can be safely driven along
a horizontal curved road without skidding.
Banking of roads:
Necessity: We know that when a vehicle goes along a horizontal curved road,
for its circular motion centripetal force is provided by the force of friction. But
this force of friction is not reliable, because it changes as the condition changes.
Also for the fast moving vehicles it will not sufficient to provide the centripetal
force. In that case we can increase the frictional force by making road surface
rougher. But it will result in wear and tear of the tyres. To avoid all these
difficulties banking of roads is necessary.
Meaning: “Banking of road is the process of constructing the road surface at the
curve, in which the outer edge of the road is kept at higher level than the inner
edge, so that the road gets the slop towards the centre of curvature of road.”
Angle of banking: “when the road is banked, the angle between the inclined
surface of the road and the horizontal is called angle of banking.”
Maximum speed of the vehicle along banked curved road:
Consider a vehicle of mass m is moving along a curve road of radius of
curvature, r and banked at an angle of, θ. Let f be the frictional force between the
tyres of the vehicle and road surface. Let the maximum safety speed of the vehicle
is Vmax. Then the different forces acting on the vehicle are: i) Its weight, mg
acting vertically downward, ii) Normal reaction, N exerted by road surface on
vehicle. iii) Frictional force between tyres and road surface.
The
frictional
Ncosθ
N
force can be resolved into
N
Ncosθ
two components f cosθ and
f sinθ as shown in figure.
θ
θ
The normal reaction, N
f cosθ
•
• C. G.
can also be resolved into
Nsinθ
θ
Nsinθ
two components- Ncosθ
f
f sinθ
and Nsinθ as shown in
f sinθ
θ
f
figure. The component
mg
Ncosθ is balanced by the
mg
weight, mg and component
f sinθ. ∴ Ncosθ = mg + f sinθ i.e. mg = Ncosθ - fsinθ ----- (i)
While the component Nsinθ along with component f cosθ is in the horizontal
direction directed towards the centre of the circular path and provides the
necessary centripetal fore for circular motion of vehicle.
∴ we write:
mv 2
r
= Nsinθ + f cosθ ----- (ii)
Dividing equation (ii) by (i) we get:
v2
rg
=
Nsin θ +fcos θ
Ncos θ−fsin θ
- - - - (iii)
The magnitude of frictional force depends upon speed of vehicle for given road.
Let vmax is maximum speed of vehicle, then frictional force is f = μ s N where μs is
the coefficient of friction. With these equation (iii) becomes:
[RSG/XII/PHY-I/CIRCULAR MOTION]
Page 8
v 2max
rg
=
Nsin θ +fcos θ
∴ vmax =
Ncos θ−fsin θ
μs + tan θ
rg
1 − μs tan θ
- - - - - (iv)
This is the maximum possible speed with which vehicle can be driven
without slipping off the road.
* Special cases: To derive the vehicle without wear and tear of tyres, μs = 0 and this
speed is called optimum speed (vo). Hence vo = r g tanθ .
* From above equation, θ = tan-1
v 2o
Rg
. This equation of angle of banking does not
involve mass of vehicle m. hence angle of banking is independent of mass of
vehicle.
* Elevation of the road is the vertical distance through which outer edge of the road
is raised above the inner edge, and is given by: h = ℓsin θ =
ℓv 2
, where ℓ is the
rg
width of the road or distance between the two rails.
Conical Pendulum: “Conical pendulum is a simple pendulum, which is given
such a motion that bob describes a horizontal circle and the string describes a
cone.”
Expression for period: Consider a conical pendulum consisting a bob of mass, m
and length, ℓ. Let bob revolves along the
S
horizontal circle of radius, r so that string
describes a cone of height, h and semi-vertical
θ
angle, θ as shown in fig. When bob is in position,
ℓ
A, the forces acting on the bob are: i) it‟s weight,
mg & ii) tension in the string, T´. From these
T’
h
T’ Cos θ
forces tension, T´ can be resolved into two
components: T´cosθ & T´sinθ. The component
θ
T´cosθ is balanced by mg. ∴ T´cosθ = mg ----- (i)
r
O
A
While the component T´sinθ provides the
T’Sin θ
centripetal force for the circular motion of the
mg
bob, ∴ T´sinθ =
mv 2
r
------- (ii)
2
Dividing equation (ii) by (i) we get: tan θ = V / rg ∴ θ = tan
V2
-1
rg
----- (iii) This
is the expression for angle made by string with the vertical. Also, from equations
(i) & (ii) we get: Speed of the bob, v = r g tanθ ----- (iv)
Now v = r ω and tanθ = r / h
∴ω=v/r=
g
h
Now, period of conical pendulum is: T =
2πr
v
But from fig. sinθ = r/ℓ ⇒ r = ℓsinθ ∴ T = 2π
=
2πr
rgtan θ
ℓsin θ
gtan θ
=
r
2π
gtan θ
ℓcos θ
g
---- (v)
Again from fig, cosθ = h/ℓ i.e. h = ℓcosθ ∴ T = 2π h/g ------------- (vi)
* Thus the period of conical pendulum is same as that of simple pendulum of length
h, where h is the axial height of cone.
[RSG/XII/PHY-I/CIRCULAR MOTION]
Page 9
* Also from these equations it is clear that the period of conical pendulum depends
upon: (i) length of pendulum, (ii) angle of inclination to vertical, (iii) the
acceleration due to gravity at that place, but (iv) it is independent of mass of the
pendulum bob and for smaller θ it is independent of θ also because from equation
(v) if θ is small then cosθ ≌ θ ∴ T = 2π
ℓ
g
2
2
* Now squaring & adding equations (i) & (ii) we get: T´ = (mg) +
2
mv 2
r
2
... (vii)
Again tanθ = v / rg = r / h ∴ v2 = r2 g / h. using this in equation (vii) we get:
Tension in the string as: T´ = mg
1 +
r 2
h
Distinguishing between Non-uniform and uniform circular motion:
(i) In the non-uniform circular motion velocity as well as speed of the particle
changes continuously while in uniform circular motion speed of the particle is
constant only velocity changes continuously. (ii) In the non-uniform circular
motion linear acceleration of the particle have two components i.e. radial
component and tangential component while in uniform circular motion linear
acceleration of particle has only one component i.e. radial component. (iii) In
non-uniform circular motion some power is delivered by force acting on the
particle while in case of uniform circular motion power delivered is zero.
Examples: The motion of the object attached to one end of string and whirled in
vertical circle about the other end is non-uniform circular motion, while if the
same object is whirled in horizontal circle with constant speed then it is a uniform
circular motion.
Vertical circular motion:
When a body tied to one end of a string and is revolved around the other end in
vertical plane then the motion of
V1 = rg
body is called vertical circular
𝑇1 = 0
motion. In this motion when the
A
particle moves from bottom to top
T1
position its motion is opposed by
mg
the acceleration due to gravity and
V3 = 3rg
V3 = 3rg
O
D
C
hence speed goes on decreasing
r T
T3 = 3mg
T3 = 3mg
θ
r-h
continuously. Similarly when
C
P
particle moves from top to bottom
θ
h T2
its motion is assisted by the gravity
mgcosθ
V2 = 5rg B
and hence its speed goes on
T2 = 6mg mgsinθ
increasing continuously. In this
mg
mg
way the speed of the particle in
vertical circular motion varies
continuously. Hence vertical circular motion is non-uniform circular motion
because; speed of body goes on changing continuously due to effect of
acceleration due to gravity.
Let at any time, t, the body is at position, P, where the string makes an
angle θ with the vertical as shown in fig. When the body is at position P, the
forces acting on the body are: i) Its weight, mg and ii) tension in the string, T,
[RSG/XII/PHY-I/CIRCULAR MOTION]
Page 10
which is directed towards the centre of the circular path, O. From this weight, mg
can be resolved into two components: a) mg cosθ & b) mg sinθ as shown in
figure.
Equation for tension in string at any point, P: From fig the net force acting on
body, directed towards centre is (T - mg cosθ), which provides the necessary
centripetal force.
∴
mv 2p
r
= T − mg cos θ , where vp is velocity of body at position P.
mv 2p
∴ Tension in the string: Tp =
+ mgcosθ ---------- (i)
r
This is the general expression for tension in string.
* At the highest point A, θ = 1800 ∴ cosθ = -1. Hence tension at the highest point is:
mv 2
T1 = 1 – m g - - - - - (ii) This is minimum value of tension and to continue
r
the circular motion this minimum tension must be zero.
* At the lowest point B, θ = 00 ∴ cos θ = 1. Hence tension at the lowest point is:
mv 2
T2 = 2 + m g - - - - - - (iii)
r
* At the midpoint such as C, θ = 900 ∴ cos θ = 0. Hence tension at the middle point
is:
T3 = (mv32 / r) + 0 = mv32 / r - - - - - (iv)
* Expression for difference in tensions at lowest and highest points:
At the lowest point B, T2 =
At highest point A, T1 =
m v 22
m v 22
m v 21
r
mv 21
r
+ mg - - - - - (i)
– m g - - - - (ii)
m
∴ T2 – T1 =
+ 2 mg = (v22 – v12) + 2mg - - - - - (iii)
r
r
r
Now according to principle of conservation of energy:
Total energy at B = Total energy at A
∴
(P. E. + K. E.)at B = (P. E. + K. E.)at A
i.e.
0 + ½ m v22 = mg.2r + ½ m v12 ⇒ v22 – v12 = 4 gr.
m
Using this in equation (iii) we get: T2 – T1 = x 4 gr + 2mg = 6 mg - - - - - (iv)
r
Equation of velocity at any position P:
To find velocity at point P, applying law of conservation energy at point B & P,
we get: Total energy at B = Total energy at P
i.e. K.E.at B + P.E. at B = K.E at P + P.E.at P
∴
½ mv22 + 0
= ½ mvp2 + m g h ⇒ Vp2 = V22 − 2gh -------- (i)
Now in ∆OCP, cosθ =
Therefore, from (i),
* Special cases:
r −h
r
∴ r cosθ = r – h
𝑜r
h = r (1 – cosθ)
Vp2 = V22 − 2g r (1 – cosθ) ----- (ii)
I) At the highest position: From the figure we write, T1 + m g =
mV 21
mV 21
r
∴ T1 =
− mg ------------ (iii)
r
i.e. tension in this position is minimum. For looping the loop this tension must
be greater than zero. i.e. minimum tension is zero.
∴ 0=
mV 21
r
− mg ∴
[RSG/XII/PHY-I/CIRCULAR MOTION]
mV 21
r
= mg ∴ v1 =
rg
------------ (iv)
Page 11
II) At the lowest position: When the particle moves from the top position to lowest
position, its potential energy is converted into kinetic energy.
∴ Decrease in potential energy = Increase in kinetic energy
∴
mg (2r) – 0 = ½ m v22 – ½ m v12
But, V1 = rg ∴ v22 = rg + 4 rg = 5 rg ⇒ v2 = 5rg
---------- (v)
III) At the horizontal position (i.e. at point C or D): According principle of
conservation of energy: Total energy at C = Total energy at B
∴ mg r + ½ m v32 = 0 + ½ m v22 where v2 = 5rg
∴
V3 = 3rg ------ (vi)
Equations for energy at different positions of vertical circular motion:
I) At any point P: Relative height of point P is h from point B & vp2 = v22 - 2 g h
∴ Potential energy of body, P.E. = mgh. and Kinetic energy, K.E. = ½ mvp2
5
∴ K.E. = ½ m (v22 – 2gh) = mgr − mgh (∵ V2 = 5rg )
2
5
∴ Total energy, E = P. E. + K.E. = mgr
2
II) At the lowest position: h = 0 & V2 =
5rg
5
∴ The potential energy, P.E. = 0 & Kinetic energy, K.E. = ½ mv22 = mgr
∴ Total energy,
5
E=
2
2
mgr
III) At the highest position: h = 2r & v1 = rg
1
∴ The potential energy, P.E. = 2 mgr & Kinetic energy, K. E. = ½ mv12 = mgr
∴ Total energy,
E=
5
2
2
mgr
IV) At the horizontal position: h = r & V3 =
∴ The potential energy, P.E. = mgr &
3rg
3
Kinetic energy, K.E. = ½ mv32 = mgr
5
2
∴ Total energy, E = mgr
2
* From these cases it is clear that in vertical circular motion the total energy of the
particle remains same i.e. conserved.
Kinematical equations for circular motion:
Consider a particle performing circular motion with uniform angular
acceleration α changes its angular velocity from ω0 to ω in time t so that angular
displacement of the particle in time t is θ, then by using analogy between the
uniform linear motion and circular motion we can write:
ω = ω0 + α t - - - - (i); θ = ω0 t + ½ α t2 - - - - (ii); ω2 = ω20 + 2 α . θ - - - - (iii)
U. C. M.
Non – U. C. M.
1 Circular motion with constant
1 Circular motion with variable
speed.
angular speed.
2 For this motion α = 0
2 For this motion, α ≠ 0
3 Work done by tangential force is
3 Work done by tangential force is
zero
not zero.
4 Motion of Earth around Sun.
4 Motion of body in vertical circle.
* * * * * *
[RSG/XII/PHY-I/CIRCULAR MOTION]
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IMPORTANT FORMULAE
dx
1.
dθ =
2.
ω=
3.
4.
α=
=
dt
V = rω
5.
15. For vertical circular motion,At general position velocity:
V = 3rg + 2rgcosθ
* VH = rg ; VL = 5 rg ; VM = 3rg
At general position tension:
dr
dθ
dt
dω
ω 2 −ω 1
t
aT = rα ; aR =
2πr
6. V =
T
v2
7.
a=
8.
F=
;ω=
2π
T
=
V2
r
2π(n 2 −n 1 )
t
T=
2
= rω = Vω
= rω2 = Vω , for U.C.M.
r
mV 2
r
= mrω2 = mVω = 4π2n2mr
9. V = μrg , on horizontal road
10. Optimum velocity to avoid the wear
and tear of tyres: Vo = rg tan θ
11. On banked road the maximum
velocity to avoid skidding is:
vmax =
Rg
+ mg cosθ
r
mv 2H
m v2
TH =
- mg; TL = L + mg
r
r
At any position total energy:
5
E = mgr = EL = EH
2
16. Kinematical equations for C.M.
ω = ω0 + αt
1
θ = ω0 t + αt 2
2
2
2
ω = ω0 + 2αθ
*
; ω = 2πn.
m V2
μs + tan θ
1 − μs tan θ
12. h = ℓ sinθ = elevation
13. θ = tan−1
v 2o
rg
14. For conical pendulum, Period,
T = 2π
r
gtan θ
h
= 2π
m V2
g
*
Tension, T´ =
*
Angular speed, ω =
r sin θ
= 2π
ℓcos θ
1 +
= mg
g
r 2
h
IMPORTANT CONVERSIONS
1 radian = (360/2π)0 = 57.30
1 km/hr = 5/18 m/s
1 N = 105 dyne
1 rpm = 1/60 Hz
g tan θ
r
[RSG/XII/PHY-I/CIRCULAR MOTION]
Page 13
PROBLEMS FOR PRACTICE
[FROM BOARD TEXT BOOK]
1.
Soln:
2.
Soln:
3.
Soln:
4.
Soln:
5.
Soln:
6.
Soln:
7.
Soln:
Calculate the angular velocity and linear velocity of a tip of minute hand of length
10 cm. (g = 9.8 m/s2)
For minute hand, T = 60 min = 60 x 60 s.
ω = 2π / T = 6.28 / 60 x 60 = 1.745 x 10-3 rad/s
V = r ω = 10-1 x 1.745 x 10-3 = 1.745 x 10-4 m/s
Propeller blades in aeroplane are 2 m long.
(a) When propeller is rotating at 1800 rpm, compute the tangential velocity of tip of
the blade.
(b) What is the tangential velocity at a point on blade midway between tip and axis?
V = r ω = r x 2πn = 2 x 2 x 3.14 x (1800/60) = 376.8 m/s
For midpoint, r = 1 m
∴ V = r ω = 1 x 2 x 3.14 x (1800/60) = 188.4 m/s
A car of mass 2000 kg rounds a curve of radius 250 m at 90 km/hr. Compute its, (a)
angular speed, (b) centripetal acceleration, (c) centripetal force.
V = 90 x (5/18) m/s = 25 m/s
(a) ω = V / r = 25 / 250 = 0.1 rad/s
(b) aC = V ω = 25 x 0.1 = 2.5 m/s2
(c) F = m aC = 2000 x 2.5 = 5000 N
A bucket containing water is whirled in a vertical circle at arm‟s length. Find the
minimum speed at top to ensure that no water spills out. Also find corresponding
angular speed. (Assume, r = 0.75 m)
The water will not spills out at the top if,
V = r g = 0.75 x 9.8 = 7.35 = 2.711 m/s
ω = V / r = 2.711 / 0.75 = 3.615 rad/s.
A motor cyclist at a speed of 5 m/s is describes a circle of radius 25 m. Find his
inclination with vertical. What is the value of coefficient of friction between the
tyres and ground?
r = 25 m; V = 5 m/s ; θ = ?; μ = ?
tan θ = V2 / r g = 25 / 25 x 9.8 = 0.1020 ∴ θ = tan-1 (0.1020) = 5051'
μ = V2 / r g = tanθ = 0.1020
A stone weighing 1 kg is whirled in a vertical circle attached at the end of a rope of
length 0.5 m. Find the tension at – (i) lowest position; (ii) mid position; (iii) highest
position
(i) At lowest position, TL = 6 mg = 6 x 1 x 9.8 = 58.8N
(ii) At midpoint, TM = 3 mg = 3 x 1 x 9.8 = 29.9 N
(iii) At highest position, TH = 0
An object of mass 0.5 kg attached to a rod of length 0.5 m is whirled in a circle at
constant angular speed. If the maximum tension in the sting is 5 kg wt, calculate: (i)
speed of stone; (ii) maximum number of revolutions it completes in a minute.
(i) Max. Tension = mg + mV2 / r ∴ 5 x 9.8 = 0.5 x 9.8 + (0.5 x V2 / 0.5)
∴ V2 = 4.5 x 9.8 ⇒ V = 6.641 m/s
(ii) n = ω / 2π = V / 2πr
∴ N = V t / 2πr = 6.641 x 60 / 6.28 x 0.5 = 126.9 rpm
[RSG/XII/PHY-I/CIRCULAR MOYION]
Page 14
8.
A stone weighing 4400 kg rounds a level curve of radius 200 m on unbanked road
at 60 km/hr. What should be minimum value of coefficient of friction to prevent
skidding? At what angle the road should be banked for this velocity?
n
Sol : m = 4400 kg; r = 200 m; V = (60 x 5 / 18) m/s; μ = ?
μ = V2 / r g = 50 x 50 / 200 x 9 x 9.8 = 0.1417
θ = tan-1(V2 / rg) = tan-1(0.1417) = 804'
9.
A string of length 9.924 m carries a bob of mass 100 g revolving with a period 2π s
in horizontal plane. Calculate angle of inclination of string with vertical and tension
in the string.
Soln: For conical pendulum, period T = 2π
ℓ cos θ
g
⇒ cos θ = T2 g / 4π2 ℓ
∴ cosθ = 9.8 / 9.924 = 0.9875 ∴ θ = cos-1(0.9875) = 905'
Tension = m g / cos θ = 0.1 x 9.8 / 0.9875 = 0.9924 N
Pilot
10. A pilot of mass 50 kg in a jet aircraft while executing a
loop-the-loop with constant speed of 250 m/s. If the radius
N
of circle is 5 km, compute the force exerted by seat on the
mg
pilot at: (a) the top of loop; (b) the bottom of loop.
N
Soln: The force exerted by seat means normal reaction.
2
(a) At top position: N1 + mg = mV / r
Pilot
∴ N1 = mV2 / r – mg = m (V2 / r – g)
mg
= 50 [(250 x 250 / 5000) – 9.8] = 135.0 N
(b) At the bottom: N2 – mg = mV2 / r
∴ N2 = mV2 / r + mg = m (V2 / r + g)
= 50 [(250 x 250 / 5000) + 9.8] = 1115 N
11. A ball is released from height along the slop and
move along a circular track of radius R without
A
falling vertically downwards, as shown in figure.
Then show that: h = 5 R / 2.
R
n
Sol : When the ball is height, h it has the potential energy h
equal to (m g h). When this ball is released, this
potential energy is converted into kinetic energy
B
when coming to bottom of the track.
∴ m g h = ½ m V2 . But to complete the loop of radius R, the velocity at the bottom
must be 5 R g ∴ m g h = ½ m (5 R g) ⇒ h = 5 R / 2
12. A block of mass 1 kg is released from the point
P on a frictionless track which ends in quarter P
Q
circular track of radius 2 m at the bottom as
shown in figure. What is the magnitude of H = 6
r = 2m
radial acceleration and total acceleration of the m
block when it arrives at Q?
n
Sol : Height lost by the body in the motion from P to Q is, h = 6 – 2 = 4 m
From equation, v2 = u2 + 2gh ⇒ v2 = 0 + (2 x 9.8 x 4) = 78.4
∴ Radial acceleration, aR = v2 / r = 78.4 / 2 = 39.2 m/s2 & aT = g = 9.8 m/s2
∴ Total acceleration, a = (39.2)2 + (9.8)2 = 1537 + 96.04 = 1633.04
∴ a = 40.4 m/s2
1
2
[RSG/XII/PHY-I/CIRCULAR MOYION]
Page 15
A circular race course track has radius of 500 m and is banked at 100. If the
coefficient of friction between the tyres of vehicle and the road surface is 0.25.
Compute: (a) the maximum speed to avoid slipping.
(b) the optimum speed to avoid wear and tear of tyres. (g = 9.8 m/s2).
Soln: R = 500 m; θ = 100; μ = 0.25; Vmax = ?; VO = ?
(a) To avoid the slipping the maximum speed is:
13.
Vmax =
=
Rg
μs + tan θ
1 − μs tan θ
50 x 98
=
500 x 9.8
0.25+0.1763
1−0.25x0.1763
0.25 + tan ⁡10 0
1 − 0.25 tan 10 0
= 46.75 m/s
(b)
Optimum speed,
VO = R g tan θ = 500 x 9.8 x 0.1763 = 29.39 m/s
14. The length of hour hand of wrist watch is 1.5 cm. Find magnitude of: (a) angular
velocity; (b) linear velocity; (c) angular acceleration; (d) radial acceleration; (e)
tangential acceleration; (f) linear acceleration of a particle on tip of hour hand.
n
Sol : (a) ω = 2π / T = 2 x 3.14 / 12 x 60 x 60 = 1.454 x 10-4
(b) V = r ω = 1.5 x 10-2 x 1.454 x 10-4 = 2.182 x 10-6
(c) α = 0 for UCM
(d) Radial accln, aR = V ω = 1.454 x 2.182 x 10-10
= 3.175 x 10-10
(e) Tangential acceleration = r α = 0
(f) For UCM, a = aR = 3.175 x 10-10 m/s2
* * * * *
[RSG/XII/PHY-I/CIRCULAR MOYION]
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