The document discusses circular motion, including angular velocity, centripetal force, components of circular motion, and motion in horizontal and vertical planes. It defines key terms like angular displacement, angular velocity, tangential velocity, centripetal acceleration, and centrifugal force. Equations are provided for these quantities. Circular motion concepts are applied to examples like stability of vehicles on banked roads and vertical circular motion with a string.

1. 1
CIRCULAR MOTION
A SHORT NOTES
Arun Umrao
https://sites.google.com/view/arunumrao
DRAFT COPY - GPL LICENSING

2. 2 Circular Motion
Contents
1 Circular Motion 3
1.1 Angular velocity . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Angular Velocity & Tangential Velocity . . . . . . . . . 4
1.1.2 Angular Acceleration . . . . . . . . . . . . . . . . . . . 5
1.2 Centripetal Force . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.1 Centripetal Acceleration . . . . . . . . . . . . . . . . . 6
1.2.2 Components Of Circular Motion . . . . . . . . . . . . . 7
1.2.3 Centrifugal Force . . . . . . . . . . . . . . . . . . . . . 8
1.3 Circular Motion in Horizontal Plane . . . . . . . . . . . . . . . 8
1.3.1 Tangential & Radial Acceleration . . . . . . . . . . . . 8
1.3.2 Stability of Vehicle in Banked Road . . . . . . . . . . . 9
1.4 Circular Motion In Vertical Plane . . . . . . . . . . . . . . . . 12
1.4.1 Constraint Motion in Vertical Plane . . . . . . . . . . . 14
1.4.2 Motion of Pendulum . . . . . . . . . . . . . . . . . . . 20
1.5 Coriolis Force . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.6 Motion In Polar Plane . . . . . . . . . . . . . . . . . . . . . . 27
1.6.1 Radial & Transverse Velocity . . . . . . . . . . . . . . 27

3. 1.1. ANGULAR VELOCITY 3
1Circular Motion
When an object moves in a curve path of constant radius then motion is
called circular motion. During circular motion body experiences centrifugal
force. If body is bound with center then its centrifugal force is balanced by
centripetal force.
r θ
~
vT
~
vT
b
O
A
B
Figure 1.1: Circular Motion.
Angular Displacement The distance of an moving object in a circular
path remains constant (r) from the center of the circular path. Therefore,
displacement of the object is measured in angle substanded by the arc of
circular path between initial and final positions of the object at a time t at
center of path. This displacement of the object is called its angular displace-
ment, θ.
1.1 Angular velocity
Angular velocity tells us that how fastly, angular position of the object is
changing with time. Therefore, angular velocity is ratio of angular displace-
ment to the corresponding time.

4. 4 Circular Motion
r δθ
~
vT
~
vT
b
O
A
B
δt
ω
Figure 1.2: Angular Velocity.
So,
~
ω =
d~
θ
dt
(1.1)
The direction of angular velocity is the direction of angular displacement. SI
unit of the angular velocity is radian per second.
1.1.1 Angular Velocity & Tangential Velocity
The instantaneous velocity of the object along the tangent to the circular
path at any given time t, is called its tangential velocity (linear velocity).
An object, moving in the circular path of radius r. It swaps angle θ at center
in time t. During the motion body travels a circumferential length of x. Now
from definition of angle
~
x = ~
r × ~
θ
r δθ
~
vT
~
vT
b
O
A
B
δx
Figure 1.3: Tangential Velocity.
If t → 0 then θ → 0 and, straight path AB and arc path x are of same
length and follow same path (from limits in mathematics). In this case,

5. 1.2. CENTRIPETAL FORCE 5
velocity along arc and tangential velocity becomes same. So, differentiating
it with respect to time
d~
x
dt
= ~
r ×
d~
θ
dt
Substituting the derivative values
~
vT = ~
r × ~
ω (1.2)
Here ω is angular velocity of the object.
1.1.2 Angular Acceleration
An object, moving in circular path experiences tangential and angular accel-
erations. The relation between them is given by derivating tangential velocity
and angular velocity relation ~
vT = ~
r × ~
ω with respect to time.
d~
vT
dt
= ~
r
d~
ω
dt
Or
aT = rα
Here aT and α are tangential acceleration and angular acceleration of the
object respectively.
1.2 Centripetal Force
An object, forced to move in a circular path requires a force that acts along
the center of the circular path to maintain its motion. This force along
the radial axis (line joining to the object and center of the path) is called
centripetal force. The resultant acceleration due to centripetal force is called
centripetal acceleration. Centripetal force is give by
Fc = m ac
Here ac is the centripetal acceleration whose value is v2
T /r. So
Fc =
mv2
T
r
(1.3)
Here vT is tangential velocity of the object.

6. 6 Circular Motion
1.2.1 Centripetal Acceleration
Let an object is moving with constant speed vT and initially it is at point
A on the circumference of the circular path. It is moving with tangential
velocity of vT at this point. After small time dt, velocity of the object is
changed to tangential velocity v′
. During this motion, angle swept at center
of circle is ∆θ. Now from figure 1.4
vT
v
′
∆θ
b b
A
b
B
∆θ
vT
v
′
∆θ
b b
A
b
B
∆θ
v′
cos ∆θ
v′
sin ∆θ
Figure 1.4:
Change in velocity along the radial axis is always equal to
∆vc = v′
sin ∆θ
If ∆θ → 0 then v′
→ vT and sin ∆θ ≈ ∆θ. Therefore,
∆vc = vT ∆θ
Diving both side by ∆t and if instantaneous time is very small then
lim
t→0
∆vc
∆t
= lim
t→0
vT
∆θ
∆t
The equation becomes
dvc
dt
= vT ω
Substituting the value of vT = r ω the result will be
ac = r ω2
(1.4)

7. 1.2. CENTRIPETAL FORCE 7
This is centripetal acceleration of circular motion of an object. Substituting
value of ω = vT /r equation (1.4) becomes
ac =
v2
T
r
(1.5)
Second Way of Above Figure
vT
v
′
∆θ
b
O b
A
b
B
∆θ vT
v
′
∆θ
b
O b
A
b
B
∆θ
∆vc
Figure 1.5:
For uniform circular motion, from the figure 1.5, for ∆θ → 0, △AOB
and triangle between velocity lines vT , ∆v′
, and vc are similar triangles as
v′
= vT for uniform circular motion. Therefore, relation of angle in radian
∆θ =
∆vc
vT
⇒ ∆vc = vT ∆θ
1.2.2 Components Of Circular Motion
~
vT
~
aT
Fcp
ac Fcf
b
O
~
vT
~
aT
Fcp
ac
Fcf
b
O
Figure 1.6: Components of Circular Motion

8. 8 Circular Motion
Figure 1.6 shows the all components of circular motion.
1.2.3 Centrifugal Force
An object moving in a circular path required a centripetal force to maintain
its motion. If centripetal force is removed body starts moving in straight line
away from center. The force which lets body to move outward in a straight
line is called centrifugal force. This force is responsible for separation of
cream from milk or curd. Centrifugal forces are pseudo in nature.
Solved Problem 1.1 A car of mass 500kg is traveling with speed of v in a
circular race track of radius 100m. If the friction between road and tire is 0.5
then find (a) How fast can the car travel before it starts to skid? (b) What
is the angular velocity and (c) Velocity of car at which driver can to pass the
turn by adding extra mass 500kg to increase friction force.
Solution (a) , (b) and (c) v does not depend on mass of car.
1.3 Circular Motion in Horizontal Plane
When an object moves in a circular path over the friction-less surface, motion
is said to be a circular motion in horizontal plane. For example, car is moving
in circular path in horizontal plane. A plane turn 180o
in horizontal plane
in air etc. To describe these types of motion, we should know about some
specific terms explained in following subsections.
1.3.1 Tangential & Radial Acceleration
A particle, moving in a curved path, experiences both tangential and cen-
tripetal accelerations, therefore velocity of the object changes continuously
either in magnitude or in direction or in both. The acceleration of the object
has two components, i.e. (i) tangential acceleration and (ii) radial (cen-
tripetal) acceleration. Net acceleration is
~
a = ~
aT + ~
ac (1.6)
Tangential acceleration is responsible for the change in speed of a particle
along the circular path. Its direction is always parallel to the instantaneous

9. 1.3. CIRCULAR MOTION IN HORIZONTAL PLANE 9
velocity. The magnitude of tangential acceleration is
aT =
d
dt
|~
vT | (1.7)
The radial acceleration arises from the change in direction of the velocity
vector. The absolute magnitude of centripetal or radial force is
ac = ar =
v2
T
r
(1.8)
Here r is the instantaneous radius of the curve path where body is moving
and its direction is always towards the center of the instantaneous curve path.
Radial and tangential accelerations are normal to each other, hence resultant
acceleration is
a =
q
a2
T + a2
r (1.9)
Solved Problem 1.2 A ball of mass m is swung around on a string in a circle
of radius r in a horizontal plane with a constant speed. String makes angle α
with vertical and ball takes t second to make a complete rotation. (a) Make
free body diagram of the ball at any position. (b) What is the velocity of ball
at S and what is its angular velocity ω? (c) What is centripetal acceleration
of ball? (d) What is the direction and magnitude of the net force at a point
of rotation? and (e) What is the value of α.
Solution
1.3.2 Stability of Vehicle in Banked Road
A heavy vehicle takes turn in a bend, it experiences a force that acts outward
of the bend. If velocity of vehicle is high then vehicle becomes unstable. To
overcome this problem outer edge of road at bend is made slightly high over
the inner edge at the bend. To find the slop angle, assume a vehicle of
mass m which takes turn in bend at instantaneous radius r. During the
turn centrifugal force is balanced by inward component of vehicle weight and
friction force. This friction force is µR where R is reaction of the road on
vehicle and it is equal to mg. µ is coefficient of friction.

10. 10 Circular Motion
R
µR
001
b
b
b
b
b
b
mg
Fcf
r
θ
b
b
b
b
b
b
R
µR
m
g
c
o
s
θ
mg sin θ
θ
mg
Fcf
Fcf cos θ
F
c
f
s
i
n
θ
Figure 1.7:
Now illustrating all the components of forces acting in the vehicle are
shown in second part of the above figure. Now, R is balanced by mg cos θ and
Fcp sin θ. Taking motion parallel to road surface, outward centrifugal force
is balanced by inward components of vehicle weight mg sin θ and frictional
force µR. Now balancing the equations
R = mg sin θ + Fcp sin θ (1.10)
Fcp cos θ = mg sin θ + µR (1.11)
Substituting value of R from equation 1.10 into equation 1.11
Fcp cos θ = mg sin θ + µ(mg sin θ + Fcp sin θ)
Substituting the value of Fcp in above equation and simplifying the relation
mv2
r
cos θ − µ
mv2
r
sin θ = (mg sin θ + µ sin θ)g
Solving it from v
v =
s
rg(sin θ + µ cos θ)
cos θ − µ sin θ
(1.12)
This is the maximum velocity for safe passing of curved road. For minimum
velocity
v =
s
rg(sin θ − µ cos θ)
cos θ + µ sin θ
(1.13)

11. 1.3. CIRCULAR MOTION IN HORIZONTAL PLANE 11
If there is not any frictional force on banked road then µ = 0. And
v =
p
rg tan θ (1.14)
Here v is velocity of vehicle at bend and θ is the angle which needs to slope
the road for balanced crossing of the bend.
R
µR
001
b
b
b
b
b
b
mg
Fcf
r
Figure 1.8:
If in any turn there is no slope in the road then frictional force between
tires and road must be equal or greater than the centrifugal force on the
body. From figure 1.8 the required coefficient of friction for balanced passes
of bend is
µR =
mv2
r
µ =
v2
rg
(1.15)
This is the minimum coefficient of friction between road and tires to make
safe crossing of vehicle at the flat road bend.
Solved Problem 1.3 A car is taking turn at a bend of the circular road.
The radius of circular road at the bend is ~
r. Coefficient of friction between
tyres and road is µ. The velocity at bend is v and mass of car is m. Find
the condition for µ if (a) is just to slip and (b) car is just to roll. Distance
between two tires car is b and height of center of gravity of the car is hCG
from the ground.
Solution

12. 12 Circular Motion
1.4 Circular Motion In Vertical Plane
Let an object attached with string is moving in circumference of a circle in
vertical plane. The centripetal force on the body is mv2
/r where r is radius
of the circle. Suppose body is at the top of the circle ie point A and tension
in the string is zero then centripetal force must be equal to the weight of the
body.
mg =
mv2
h
r
vh =
√
rg
T
Fcf
θ
mg
~
vT
bcb
b
A
b
B
b
T
Fcf
θ
mg
~
vT
bcb
b
A
b
B b
Figure 1.9: Vertical Circular Motion.
If body has this velocity at top of the circle then string is not loosing.
To find the minimum velocity at bottom of the circle so that body can reach
atleast highest point of the circle without loosing the string, the kinetic
energy of body at point B must be equal or greater than the kinetic and
potential energy of the body at the point A. At point A velocity of body is
√
rg and height is 2r with respect to B So
1
2
mv2
l = mg(2r) +
1
2
mv2
h
vl =
p
5rg
This is the minimum velocity at point B that the body must have to reach
at the highest point of the circle without loosing the string. The tension in
the string at the highest point of the circle is
Tt = Fcf − mg

13. 1.4. CIRCULAR MOTION IN VERTICAL PLANE 13
Th = 0 (1.16)
Similarly the tension in string when body is at lowest point is T
Tl = Fcf + mg
Tl = 5mg + mg = 6mg (1.17)
This is shown that tension in string is zero when body is at highest point of
circle and is six times of body weight at the lowest point.
Solved Problem 1.4 An object is placed at height of h in a sliding track. The
track makes a perfect circular loop at the ground and then become flat. Ob-
ject is let to move from a certain vertical height in the sliding track. Find the
minimum height of the object in terms of R such that object moves around
the perfect circular loop at ground without falling off the top. Consider there
is no friction.
Solution At the top of the loop, velocity is vT . Net forces at the top of
loop is mg + N that is working vertically downward. If centripetal force on
the object is equal to the net force then object would not fall off but would
make a complete round. So
mg + N =
mv2
T
r
(1.18)
If object is about to fall at the top then N ≡ 0 and
vT =
p
gR (1.19)
From the conservation of the energy
mgh + 0 = mg(2r) +
1
2
mv2
T
On simplification and substituting the value of vT
h =
5
2
r (1.20)

14. 14 Circular Motion
Solved Problem 1.5 An object of mass m is slide in a circular sliding plane
that is a quadrant of a circle. Find the velocity of the object at bottom
assuming there is no friction. Find the work done by friction, if velocity at
bottom is vb and there is friction.
Solution As motion is circular and direction of motion changes contin-
uously, the problem can not solved by using acceleration method. Let the
velocity of body at the bottom is vt when there is no friction then from the
law of conservation of energy
1
2
mv2
t + 0 = 0 + mgr
Where r is the radius of circular loop. On simplification
vt =
p
2gr (1.21)
If there is friction force between track and body then work done by friction
force is equal to the change in total energy (change is kinetic energy + change
in potential energy). Let the velocity of body at the bottom is vb when there
is friction then
W =
1
2
mv2
b −
1
2
mv2
i + mg × 0 − mgr
As initial velocity vi of the object is zero, hence
W =
1
2
mv2
b − mgr (1.22)
1.4.1 Constraint Motion in Vertical Plane
bc
O
A
T
mg
~
vT
b
O
θ
bc
v
FN
Fcf
mg mg
mg cos θ
mg sin θ
Ft
FN
θ
(a) (b) (c)
Figure 1.10: Vertical circular motion.

15. 1.4. CIRCULAR MOTION IN VERTICAL PLANE 15
Assume a particle of mass m is hanging from a fixed point O with an in-
extensible string of length r. In equilibrium tension in string is balanced
by particle weight mg. (See figure 1.10-a). If particle is forced to move in
vertical plane with initial velocity u, it reaches to point P. As particle is
in motion, tangential and normal forces are acted upon it along with body
weight. (See figure 1.10-b). At point P all the forces are resolved. (See figure
1.10-c). Now equating forces
m
d2
s
dt2
= −mg sin θ (1.23)
Negative sign shown that mg sin θ opposes the tangential force. And for
normal forces
m
v2
r
= T − mg cos θ (1.24)
Solving equation 1.23
v
dv
ds
= g sin θ
Substituting s = rθ in above equation
v
dv
r dθ
= g sin θ
On integration Z
v dv = −gr
Z
sin θ dθ
Or
v2
2
= gr cos θ + C
When θ is zero ie particle is at its lowest point then v = u and
C =
u2
2
− gr
On substituting the value of constant C
v2
= u2
+ 2gr cos θ − 2gr (1.25)
From equation 1.25, substituting value of v in equation 1.24
m
2gr cos θ + u2
− 2gr
r
= T − mg cos θ

16. 16 Circular Motion
T =
m
r
u2
+ 3gr cos θ − 2gr
(1.26)
Equation 1.25 and 1.26 represents the motion of particle in vertical plane
curve.
Frictional Vertical Circular Path
A particle of mass m is moving in frictional, vertical circular path of radius R.
At any instant of time, the particle’s position is at point P(x, y), represented
by Pxy. Line, joining the particle and the center of circular path makes angle
θ from the vertical diameter. The coefficient of friction between particle and
surface is µ.
x
y
mg
mg cos θ
mg sin θ
θ
v0
θ
bc
v
FN
Ff Fcf
mg
Pxy
The position of particle is given by
x = R sin θ; y = R(1 − cos θ)
Resolving the forces for particle, normal force on the surface by the particle
is given by
FN − mg cos θ =
mv2
R
Here, v is velocity of the particle at the point P(x, y). The friction force
between particle and the surface is
Ff = µ
mg cos θ +
mv2
R

17. 1.4. CIRCULAR MOTION IN VERTICAL PLANE 17
Take the lowest point of vertical circular path as reference level, i.e. origin
of this motion system. From the law of conservation of energy, we have
mv2
0
2
+ 0 =
mv2
2
+ mgy +
Z lθ
0
Ff ds
Here, v0 is velocity at the lowest point of the vertical path. y is vertical
displacement of the particle when its velocity is v. lθ is arc length of the
path covered by the particle when it makes angle θ from the vertical. We
have ds = R dθ, lθ = Rθ. y is given by y = R(1 − cos θ). The above relation
becomes
mv2
0
2
=
mv2
2
+ mgR(1 − cos θ) +
Z Rθ
0
Ff R dθ (1.27)
This relation can be used to get the velocity of the particle when it move in
vertical frictional circular path.
Just Reach Condition
When it is said that particle is just reach at a point, there are two cases (i)
when velocity is vanished and (ii) When tension is vanished. Velocity can
vanished at any point in the vertical circular path. If velocity vanishes at
a point, particle return back to its lower most point without loosing string
tension. If tension looses at a point and velocity does not vanish then parti-
cle moves forward from the tension vanishing point in trajectory path (like
projectile).
b
O
T
bc
v
Fcf
mg
b
O
T
θ
bc
v = 0
Fcf
mg
b
O
T = 0
≈
bc
v
Fcf
mg
(a) (b) (c)
Figure 1.11: Motion of particle when tension or velocity is vanished.

18. 18 Circular Motion
If tension and velocity vanish at a point, particle felled vertically under
the gravitational acceleration. For a complete circular motion, tension can
be vanish only at top most position of the vertical circular path but velocity
does not vanish at the topmost position.
Motion In Convex Surface
Let a particle is at the top of the convex side of vertical circle and starts slid-
ing with initial speed of u. At any instant, particle is at point P and it makes
angle θ with vertical. At this point, reaction on the particle is R normal to
the surface of circle and weight of circle is acting vertically downward. Now
resolving the components of particle weight along the tangent and normal at
the point P.
bc
v
R
mg
θ
P
u
mg
mg sin θ
mg cos θ
R
θ
(a) (b)
Figure 1.12: Vertical circular motion in convex side of circle.
m
d2
s
dt2
= mg sin θ (1.28)
Here the tangential downward force let the particle slide towards downward.
and
mg cos θ − R = m
v2
r
(1.29)
Substituting s = rθ in equation 1.28 and solving it
r
d2
θ
dt2
= g sin θ
Or
r ω
dω
dθ
= g sin θ

19. 1.4. CIRCULAR MOTION IN VERTICAL PLANE 19
On simplifying and solving it
r
ω2
2
= −g cos θ + c
If v is instantaneous velocity of the particle at the point P then
v2
2r
= −g cos θ + c
When particle is at the top most position of the circle where θ = 0 and
initial velocity is u (ie v = u), where u is the initial tangential velocity of the
particle at highest point. Constant c becomes
u2
2r
+ g = c
Substituting the value of c we get
v2
2r
= −g cos θ +
u2
2r
+ g
On simplification
v2
= 2gr − 2gr cos θ + u2
(1.30)
Substituting this value of v in equation 1.29 we have
mg cos θ − R =
m
r
2gr − 2gr cos θ + u2
On simplifying
R =
m
r
3gr cos θ − 2gr − u2
(1.31)
Equation 1.30 and equation 1.31 are the required relation of particle sliding
in the convex side of the circle.
Particle Leaves the Circular Path Particle leaves the circular path when
R is zero and the angular displacement of the particle with vertical is
cos θ =
2gr + u2
3gr
(1.32)
After leaving the circle, particle moves in parabolic path under the gravita-
tional acceleration of earth.

20. 20 Circular Motion
1.4.2 Motion of Pendulum
A heavy mass body revolves around a fixed axis is called pendulum. Gen-
erally there are two types of pendulum. (i) Simple pendulum and (ii) Com-
pound pendulum.
Simple Pendulum (Lossless)
A simple pendulum is a system in which a heavy bob is hang from any
fulcrum with help of a string. The body moves left and right from its equi-
librium position with a constant period. The maximum displacement from
equilibrium is called amplitude of the pendulum. The time taken by pen-
dulum to complete its one cycle is called time period of the oscillation of
pendulum. Here, air friction is negligible.
2θ
T
mg
~
F
b
b
Figure 1.13: Simple Pendulum.
Let the final position of the pendulum is B which is left to the equilib-
rium. Weight of bob acts vertically downward. Force is acting horizontally
outward and tension along the string length upward. String makes angle θ
with vertical. Now from the figure 1.13, the vertical component of tension is
T cos θ and horizontal component is T sin θ. The weight of body is balanced
by vertical component of tension while force F is balanced by horizontal
component of tension. Now
T cos θ = mg (1.33)
T sin θ = F (1.34)
Now solving equations 1.33 and 1.34 the result is
T =
p
(mg)2 + F2 (1.35)

21. 1.4. CIRCULAR MOTION IN VERTICAL PLANE 21
And the angle θ is
tan θ =
F
mg
(1.36)
Angular Motion of Simple Pendulum
2θ
mg
~
F
Tr
b
b
l
From the above figure, the torque in the pendulum is given by τ = ~
Fl.
The restoring torque on the pendulum is Tr = Iα. Now, from conservation
of energy, we have
Tr = −τ
It gives
ml2
×
d2
θ
dt2
= −mg sin θ × l
Here, m is mass of the pendulum and l is length of the pendulum. ml2
is
inertia of the bob about its fulcrum. Solving above equation, we have
d2
θ
dt2
= −
g
l
sin θ
This is differential form of simple pendulum.
Simple Pendulum (Lossy)
If in the motion of simple pendulum, the friction of air is taken in the con-
sideration, then motion of simple pendulum is lossy type. The air friction is
proportional to the angular velocity, i.e. fr = b dθ/dt.

22. 22 Circular Motion
2θ
mg
~
F
Tr
fr
b
b
l
From the above figure, the torque in the pendulum is given by τ = ~
Fl.
The restoring torque on the pendulum is Tr = Iα. Now, from conservation
of energy, we have
Tr + fr = −τ
It gives
ml2
×
d2
θ
dt2
+ b
dθ
dt
= −mg sin θ × l
Here, m is mass of the pendulum and l is length of the pendulum. ml2
is
inertia of the bob about its fulcrum. Solving above equation, we have
d2
θ
dt2
+
b
ml2
dθ
dt
+
g
l
sin θ = 0
This is differential form of simple lossy pendulum.
Inverted Pendulum
Conical Pendulum
Let a simple pendulum is rotating about its vertical axis in a circle’s circum-
ference of radius r and making an angle θ with vertical. Now from figure 1.14
it is seen that the centripetal force is balancing the outward force on pendu-
lum and bob weight is balanced by vertical component of tension. Now from
figure

23. 1.4. CIRCULAR MOTION IN VERTICAL PLANE 23
T
Fcf
mg
TV
TH
b
b
θ
Figure 1.14: Conic pendulum.
T cos θ = mg (1.37)
T sin θ =
mv2
r
(1.38)
Now dividing the equations (1.37) and (1.38) the angle θ is
tan θ =
v2
rg
(1.39)
The equation (1.39) is same as (1.14). And tension is found by squaring and
adding the equations (1.37) and (1.38).
T =
r
(mg)2 +
m2 v4
r2
(1.40)
The resultant acceleration of pendulum bob would be
a =
q
a2
T + a2
c (1.41)
Solved Problem 1.6 A bead is put in friction-less vertical circular loop of
radius R as shown in the figure given below. The loop is rotating about its
diameter and bead is in equilibrium. Find the angle θ in terms of g, R and ω.
Also, find the minimum value of ω so that bead does not sit at the bottom
of the half circular loop.
Solution

24. 24 Circular Motion
b
mg
Ff
b
R
r
θ
mg
Ff
Ff cos θ
mg sin θ
b
θ b
A bead is put in a friction-less, vertical, rotating circular loop with angular
velocity ω about its diametrical axis as shown in above figure. The bead rises
upward along the surface of the loop due to centrifugal force. At equilibrium,
of the bead, only those forces are effective which are tangential to the circular
loop. Other forces are balanced by action and equal opposite reaction. We
resolute the weight and centrifugal forces along the tangent to the circular
loop at the position of the bead. Now
mg sin θ = Ff cos θ
The centrifugal force on the bead is
Ff =
mv2
r
=
mr2
ω2
r
= mrω2
Substituting it in in-equilibrium equation, we have
mg sin θ = mrω2
cos θ
From the first part of above figure, r = R sin θ. So,
mg sin θ = mR sin θ × ω2
× cos θ
On simplification, we have
cos θ =
g
Rω2
This is first part of the answer. Ans-1.

25. 1.5. CORIOLIS FORCE 25
Now, the bead does not sit at lowest point of the circular loop, i.e. angle
θ is not exact zero but it is larger than 0. It means, θ ≈ 0, i.e. cos θ ≈ 1.
Now, expanding cos θ in geometric series.
1 +
θ2
2!
+ . . . =
g
Rω2
Neglecting, terms having θ as θ ≈ 0, we get
ω2
=
g
R
⇒ ω =
r
g
R
This is the minimum angular frequency, which is required so that bead does
not sit at the bottom of the loop.
1.5 Coriolis Force
In physics, the Coriolis force is an inertial force (also called a fictitious force)
that acts on objects that are in motion relative to a rotating reference frame.
In a reference frame with clockwise rotation, the force acts to the left of the
motion of the object. In one with anticlockwise rotation, the force acts to
the right.
N
b N b S
Figure 1.15: The Earth rotating about its axis of rotation. When rotation
of Earth is viewed from N pole, its direction is in counter clockwise. The
direction of rotation of earth when viewed from south pole is clockwise.

26. 26 Circular Motion
b
bc
b
bc
b
bc
b
bc
b
bc
Figure 1.16:
An object moves in straight line in inertial frame of reference while an
observer standing in a rotating/non inertial frame of reference observes that
object is moving in curve path rather than in straight path. This is due to
the Coriolis and centrifugal forces present in this frame.
Formula The vector formula for the magnitude and direction of the
Coriolis acceleration is
~
aC = −2 ~
ω × ~
vT (1.42)
Where ~
aC is the acceleration of the particle in the rotating system, ~
vT is
the velocity of the particle with respect to the rotating system, and ω is the
angular velocity vector of rotating frame of reference. Based on this relation
Coriolis force:
~
FC = −2 m ~
ω × ~
vT (1.43)
The vector cross product can be evaluated as the determinant of a matrix:
~
ω × ~
vT =