HC Verma Summary: Circular Motion | Physics Class 11 - NEET PDF Download

 Page 1


Circular Motion
July 5, 2025
Contents
1 Angular V ariables 2
1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Unit V ectors Along the Radius and T angent 3
3 A cceler ation in Circular Motion 3
3.1 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.2 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.3 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
4 Dynamics of Circular Motion 6
4.1 Example 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
4.2 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
5 Circular Turnings and Banking of Roads 6
5.1 Example 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
5.2 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
6 Centrifugal F orce 7
6.1 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
6.2 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
7 Effect of Earth’ s Rotation on Apparent W eight 8
8 W ork ed Out Examples 8
8.1 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.2 Example 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.3 Example 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.4 Example 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.5 Example 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.6 Example 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.7 Example 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.8 Example 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.9 Example 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.10 Example 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.11 Example 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.12 Example 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.13 Example 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.14 Example 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1
Page 2


Circular Motion
July 5, 2025
Contents
1 Angular V ariables 2
1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Unit V ectors Along the Radius and T angent 3
3 A cceler ation in Circular Motion 3
3.1 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.2 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.3 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
4 Dynamics of Circular Motion 6
4.1 Example 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
4.2 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
5 Circular Turnings and Banking of Roads 6
5.1 Example 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
5.2 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
6 Centrifugal F orce 7
6.1 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
6.2 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
7 Effect of Earth’ s Rotation on Apparent W eight 8
8 W ork ed Out Examples 8
8.1 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.2 Example 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.3 Example 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.4 Example 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.5 Example 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.6 Example 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.7 Example 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.8 Example 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.9 Example 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.10 Example 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.11 Example 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.12 Example 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.13 Example 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.14 Example 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1
1 Angular V ariables
When an object moves in a circle, lik e a car on a roundabout or a stone tied to a string being
swung around, we can describe its motion using angles instead of str aight-line distances. Imag-
ine a particleP moving on a circle of r adiusr , with the center at pointO (lik e the hub of a wheel).
W e setO as the origin and dr aw a lineOX as theX -axis. The position ofP is given b y the angle
? between the lineOP (from the center to the particle) and theX -axis. This angle? is called the
angular position, measured in r adians (r ad).
As the particle moves along the circle, ? changes. If it moves to a nearb y point P
'
in a tin y
time?t , the angle increases b y?? . The r ate at which the angle changes is called angular velocity ,
denoted b y? :
? = lim
?t?0
??
?t
=
d?
dt
.
Angular velocity tells us how fast the particle is spinning around the circle, in r adians per
second (r ad/s).
The r ate at which angular velocity changes is called angular acceler ation, denoted b ya :
a =
d?
dt
=
d
2
?
dt
2
.
Angular acceler ation, measured in r ad/s/s, tells us how quickly the spinning is speeding up or
slowing down.
Ifa is constant, we can use these equations, similar to linear motion:
? = ?
0
t+
1
2
at
2
,
? = ?
0
+at,
?
2
= ?
2
0
+2at,
where?
0
is the angular velocity att = 0 ,? is the angular velocity at timet , and? is the angular
position at timet .
The line ar distance tr aveled along the circle (arc length) is related to the angle:
?s = r??.
Dividing b y t ime:
?s
?t
=
??
?t
=? v = r?,
where v is the linear speed (in m/s). This shows that linear speed depends on both how fast
the angle changes (? ) and how far the particle is from the center (r ).
The r ate o f change of linear speed is the tangential acceler ation:
a
t
=
dv
dt
=
d?
dt
=
ra
r
= a.
This acceler ation is along the tangent to the circle (the direction of motion) and changes the
speed, not the direction. Figure 7.1 shows a particleP on a circle with r adiusr , angle? , and the
arc fromP toP
'
.
1.1 Example 1
A particle moves in a circle of r adius 20 cm with a linear speed of 10 m/s. Find the angular
velocity .
Solution: Convert r adius to meters: r = 20 cm = 0.20 m. Use:
? =
v
r
=
10 m/s
0.20 m
= 50 r ad/s.
2
Page 3


Circular Motion
July 5, 2025
Contents
1 Angular V ariables 2
1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Unit V ectors Along the Radius and T angent 3
3 A cceler ation in Circular Motion 3
3.1 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.2 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.3 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
4 Dynamics of Circular Motion 6
4.1 Example 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
4.2 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
5 Circular Turnings and Banking of Roads 6
5.1 Example 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
5.2 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
6 Centrifugal F orce 7
6.1 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
6.2 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
7 Effect of Earth’ s Rotation on Apparent W eight 8
8 W ork ed Out Examples 8
8.1 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.2 Example 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.3 Example 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.4 Example 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.5 Example 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.6 Example 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.7 Example 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.8 Example 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.9 Example 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.10 Example 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.11 Example 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.12 Example 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.13 Example 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.14 Example 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1
1 Angular V ariables
When an object moves in a circle, lik e a car on a roundabout or a stone tied to a string being
swung around, we can describe its motion using angles instead of str aight-line distances. Imag-
ine a particleP moving on a circle of r adiusr , with the center at pointO (lik e the hub of a wheel).
W e setO as the origin and dr aw a lineOX as theX -axis. The position ofP is given b y the angle
? between the lineOP (from the center to the particle) and theX -axis. This angle? is called the
angular position, measured in r adians (r ad).
As the particle moves along the circle, ? changes. If it moves to a nearb y point P
'
in a tin y
time?t , the angle increases b y?? . The r ate at which the angle changes is called angular velocity ,
denoted b y? :
? = lim
?t?0
??
?t
=
d?
dt
.
Angular velocity tells us how fast the particle is spinning around the circle, in r adians per
second (r ad/s).
The r ate at which angular velocity changes is called angular acceler ation, denoted b ya :
a =
d?
dt
=
d
2
?
dt
2
.
Angular acceler ation, measured in r ad/s/s, tells us how quickly the spinning is speeding up or
slowing down.
Ifa is constant, we can use these equations, similar to linear motion:
? = ?
0
t+
1
2
at
2
,
? = ?
0
+at,
?
2
= ?
2
0
+2at,
where?
0
is the angular velocity att = 0 ,? is the angular velocity at timet , and? is the angular
position at timet .
The line ar distance tr aveled along the circle (arc length) is related to the angle:
?s = r??.
Dividing b y t ime:
?s
?t
=
??
?t
=? v = r?,
where v is the linear speed (in m/s). This shows that linear speed depends on both how fast
the angle changes (? ) and how far the particle is from the center (r ).
The r ate o f change of linear speed is the tangential acceler ation:
a
t
=
dv
dt
=
d?
dt
=
ra
r
= a.
This acceler ation is along the tangent to the circle (the direction of motion) and changes the
speed, not the direction. Figure 7.1 shows a particleP on a circle with r adiusr , angle? , and the
arc fromP toP
'
.
1.1 Example 1
A particle moves in a circle of r adius 20 cm with a linear speed of 10 m/s. Find the angular
velocity .
Solution: Convert r adius to meters: r = 20 cm = 0.20 m. Use:
? =
v
r
=
10 m/s
0.20 m
= 50 r ad/s.
2
1.2 Example 2
A particle tr avels in a circle of r adius 20 cm at a speed that uniformly increases. If the speed
changes from 5.0 m/s to 6.0 m/s in 2.0 s, find the angular acceler ation.
Solution: T angential acceler ation:
a
t
=
?v
?t
=
6.0-5.0
2.0
= 0.5 m/s
2
.
Convert r adius: r = 20 cm = 0.20 m. Angular acceler ation:
a =
a
t
r
=
0.5
0.20
= 2.5 r ad/s
2
.
1.3 Example 3
A wheel of r adius 50 cm rotates with an angular velocity of 20 r ad/s. What is the linear speed of
a point on its rim?
Solution: Convert r adius: r = 50 cm = 0.50 m.
v = r? = 0.50×20 = 10 m/s.
2 Un it V ectors Along the Radius and T angent
T o describe circular motion mathematically , we use special directions: r adial (from the center to
the particle) and tangential (along the circle’ s path). Imagine a particleP on a circle with centerO
as the origin;OX as theX -axis, andOY as theY -axis (Figure 7.2). The particle’ s angular position
is? .
W e defin e two unit vectors (vectors of length 1):
- ˆ e
r
: The r adial unit vector , pointing outwar d from O to P . - ˆ e
t
: The tangential unit vector ,
along the tangent atP , in the direction of increasing? (counterclockwise).
T o find these vectors, dr aw lines from P par allel to the axes. The r adial unit vector ˆ e
r
along
OP has components:
ˆ e
r
= cos?ˆ i+ sin?ˆ ?,
whereˆ i and ˆ ? are unit vectors alongX andY . The tangential unit vector ˆ e
t
perpendicular to
ˆ e
r
is:
ˆ e
t
=- sin?ˆ i+ cos?ˆ ?.
These ve ctors help us describe position, velocity , and acceler ation in circular motion.
3 A cceler ation in Circular Motion
A particle moving in a circle is alwa ys acceler ating because its direction changes, even if its speed
is constant. Let’ s derive the acceler ation using the position vector , following the steps from the
original document. Consider a particle P on a circle of r adius r , with center O as the origin
(Figure 7.2). The position vector of the particle is:
? r = O
?
P = r? e
r
= r( cos?
?
i+ sin?
?
j).
T o find t he velocity , differentiate the position vector with respect to time:
? v =
d? r
dt
=
d
dt
[r( cos?
?
i+ sin?
?
j)].
Sincer is constant (the particle sta ys on the circle), we differentiate the components:
? v = r
[(
- sin?
d?
dt
)
?
i+
(
cos?
d?
dt
)
?
j
]
= r
d?
dt
[
- sin?
?
i+ cos?
?
j
]
.
3
Page 4


Circular Motion
July 5, 2025
Contents
1 Angular V ariables 2
1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Unit V ectors Along the Radius and T angent 3
3 A cceler ation in Circular Motion 3
3.1 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.2 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.3 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
4 Dynamics of Circular Motion 6
4.1 Example 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
4.2 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
5 Circular Turnings and Banking of Roads 6
5.1 Example 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
5.2 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
6 Centrifugal F orce 7
6.1 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
6.2 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
7 Effect of Earth’ s Rotation on Apparent W eight 8
8 W ork ed Out Examples 8
8.1 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.2 Example 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.3 Example 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.4 Example 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.5 Example 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.6 Example 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.7 Example 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.8 Example 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.9 Example 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.10 Example 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.11 Example 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.12 Example 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.13 Example 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.14 Example 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1
1 Angular V ariables
When an object moves in a circle, lik e a car on a roundabout or a stone tied to a string being
swung around, we can describe its motion using angles instead of str aight-line distances. Imag-
ine a particleP moving on a circle of r adiusr , with the center at pointO (lik e the hub of a wheel).
W e setO as the origin and dr aw a lineOX as theX -axis. The position ofP is given b y the angle
? between the lineOP (from the center to the particle) and theX -axis. This angle? is called the
angular position, measured in r adians (r ad).
As the particle moves along the circle, ? changes. If it moves to a nearb y point P
'
in a tin y
time?t , the angle increases b y?? . The r ate at which the angle changes is called angular velocity ,
denoted b y? :
? = lim
?t?0
??
?t
=
d?
dt
.
Angular velocity tells us how fast the particle is spinning around the circle, in r adians per
second (r ad/s).
The r ate at which angular velocity changes is called angular acceler ation, denoted b ya :
a =
d?
dt
=
d
2
?
dt
2
.
Angular acceler ation, measured in r ad/s/s, tells us how quickly the spinning is speeding up or
slowing down.
Ifa is constant, we can use these equations, similar to linear motion:
? = ?
0
t+
1
2
at
2
,
? = ?
0
+at,
?
2
= ?
2
0
+2at,
where?
0
is the angular velocity att = 0 ,? is the angular velocity at timet , and? is the angular
position at timet .
The line ar distance tr aveled along the circle (arc length) is related to the angle:
?s = r??.
Dividing b y t ime:
?s
?t
=
??
?t
=? v = r?,
where v is the linear speed (in m/s). This shows that linear speed depends on both how fast
the angle changes (? ) and how far the particle is from the center (r ).
The r ate o f change of linear speed is the tangential acceler ation:
a
t
=
dv
dt
=
d?
dt
=
ra
r
= a.
This acceler ation is along the tangent to the circle (the direction of motion) and changes the
speed, not the direction. Figure 7.1 shows a particleP on a circle with r adiusr , angle? , and the
arc fromP toP
'
.
1.1 Example 1
A particle moves in a circle of r adius 20 cm with a linear speed of 10 m/s. Find the angular
velocity .
Solution: Convert r adius to meters: r = 20 cm = 0.20 m. Use:
? =
v
r
=
10 m/s
0.20 m
= 50 r ad/s.
2
1.2 Example 2
A particle tr avels in a circle of r adius 20 cm at a speed that uniformly increases. If the speed
changes from 5.0 m/s to 6.0 m/s in 2.0 s, find the angular acceler ation.
Solution: T angential acceler ation:
a
t
=
?v
?t
=
6.0-5.0
2.0
= 0.5 m/s
2
.
Convert r adius: r = 20 cm = 0.20 m. Angular acceler ation:
a =
a
t
r
=
0.5
0.20
= 2.5 r ad/s
2
.
1.3 Example 3
A wheel of r adius 50 cm rotates with an angular velocity of 20 r ad/s. What is the linear speed of
a point on its rim?
Solution: Convert r adius: r = 50 cm = 0.50 m.
v = r? = 0.50×20 = 10 m/s.
2 Un it V ectors Along the Radius and T angent
T o describe circular motion mathematically , we use special directions: r adial (from the center to
the particle) and tangential (along the circle’ s path). Imagine a particleP on a circle with centerO
as the origin;OX as theX -axis, andOY as theY -axis (Figure 7.2). The particle’ s angular position
is? .
W e defin e two unit vectors (vectors of length 1):
- ˆ e
r
: The r adial unit vector , pointing outwar d from O to P . - ˆ e
t
: The tangential unit vector ,
along the tangent atP , in the direction of increasing? (counterclockwise).
T o find these vectors, dr aw lines from P par allel to the axes. The r adial unit vector ˆ e
r
along
OP has components:
ˆ e
r
= cos?ˆ i+ sin?ˆ ?,
whereˆ i and ˆ ? are unit vectors alongX andY . The tangential unit vector ˆ e
t
perpendicular to
ˆ e
r
is:
ˆ e
t
=- sin?ˆ i+ cos?ˆ ?.
These ve ctors help us describe position, velocity , and acceler ation in circular motion.
3 A cceler ation in Circular Motion
A particle moving in a circle is alwa ys acceler ating because its direction changes, even if its speed
is constant. Let’ s derive the acceler ation using the position vector , following the steps from the
original document. Consider a particle P on a circle of r adius r , with center O as the origin
(Figure 7.2). The position vector of the particle is:
? r = O
?
P = r? e
r
= r( cos?
?
i+ sin?
?
j).
T o find t he velocity , differentiate the position vector with respect to time:
? v =
d? r
dt
=
d
dt
[r( cos?
?
i+ sin?
?
j)].
Sincer is constant (the particle sta ys on the circle), we differentiate the components:
? v = r
[(
- sin?
d?
dt
)
?
i+
(
cos?
d?
dt
)
?
j
]
= r
d?
dt
[
- sin?
?
i+ cos?
?
j
]
.
3
Since
d?
dt
= ? (angular velocity), and recognizing that- sin?
?
i+ cos?
?
j =? e
t
(the tangential unit
vector), we get:
? v = r?? e
t
.
The velocity is along the tangent to the circle, and its magnitude isv = r? , consistent with our
earlier result.
Now , t o find the acceler ation, differentiate the velocity vector:
? a =
d? v
dt
=
d
dt
[r?(- sin?
?
i+ cos?
?
j)].
Apply t he product rule, noting thatr is constant, but? and? ma y vary with time:
? a = r
[
?
d
dt
[- sin?
?
i+ cos?
?
j]+
d?
dt
(- sin?
?
i+ cos?
?
j)
]
.
Compute the derivative of the tangential unit vector:
d
dt
[- sin?
?
i+ cos?
?
j] =
(
- cos?
d?
dt
)
?
i+
(
- sin?
d?
dt
)
?
j =
d?
dt
(- cos?
?
i- sin?
?
j).
Since
d?
dt
= ? , this becomes:
?(- cos?
?
i- sin?
?
j).
Notice t hat cos?
?
i+ sin?
?
j =? e
r
, so:
- cos?
?
i- sin?
?
j =-? e
r
.
Thus:
?
d
dt
(- sin?
?
i+ cos?
?
j) = ?·?(-? e
r
) =-?
2
? e
r
.
The seco nd term involves
d?
dt
= a , and- sin?
?
i+ cos?
?
j =? e
t
:
d?
dt
(- sin?
?
i+ cos?
?
j) = a? e
t
.
Combine:
? a = r
[
-?
2
? e
r
+a? e
t
]
=-?
2
r? e
r
+ra? e
t
.
Sincera =
dv
dt
(fromv = r? ), the acceler ation is:
? a =-?
2
r? e
r
+
dv
dt
? e
t
.
The a cceler ation has two components:
- Radial acceler ation: a
r
=-?
2
r =-
v
2
r
, directed toward the center (along-? e
r
). - T angential
acceler ation: a
t
=
dv
dt
, alo ng the tangent, changing the speed.
Uniform Circular Motion: If the speed is constant (
dv
dt
= 0 ):
? a =-?
2
r? e
r
, a
c
=
v
2
r
,
directed toward the center , called centripetal acceler ation. The magnitude is:
a
c
= ?
2
r =
v
2
r
,
sincev = r? .
Nonuniform Circular Motion: If the speed changes, both r adial and tangential components
exist. The r adial acceler ation is:
a
r
=-?
2
r =-
v
2
r
,
4
Page 5


Circular Motion
July 5, 2025
Contents
1 Angular V ariables 2
1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Unit V ectors Along the Radius and T angent 3
3 A cceler ation in Circular Motion 3
3.1 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.2 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.3 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
4 Dynamics of Circular Motion 6
4.1 Example 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
4.2 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
5 Circular Turnings and Banking of Roads 6
5.1 Example 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
5.2 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
6 Centrifugal F orce 7
6.1 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
6.2 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
7 Effect of Earth’ s Rotation on Apparent W eight 8
8 W ork ed Out Examples 8
8.1 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.2 Example 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.3 Example 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.4 Example 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.5 Example 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.6 Example 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.7 Example 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.8 Example 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.9 Example 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.10 Example 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
8.11 Example 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.12 Example 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.13 Example 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8.14 Example 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1
1 Angular V ariables
When an object moves in a circle, lik e a car on a roundabout or a stone tied to a string being
swung around, we can describe its motion using angles instead of str aight-line distances. Imag-
ine a particleP moving on a circle of r adiusr , with the center at pointO (lik e the hub of a wheel).
W e setO as the origin and dr aw a lineOX as theX -axis. The position ofP is given b y the angle
? between the lineOP (from the center to the particle) and theX -axis. This angle? is called the
angular position, measured in r adians (r ad).
As the particle moves along the circle, ? changes. If it moves to a nearb y point P
'
in a tin y
time?t , the angle increases b y?? . The r ate at which the angle changes is called angular velocity ,
denoted b y? :
? = lim
?t?0
??
?t
=
d?
dt
.
Angular velocity tells us how fast the particle is spinning around the circle, in r adians per
second (r ad/s).
The r ate at which angular velocity changes is called angular acceler ation, denoted b ya :
a =
d?
dt
=
d
2
?
dt
2
.
Angular acceler ation, measured in r ad/s/s, tells us how quickly the spinning is speeding up or
slowing down.
Ifa is constant, we can use these equations, similar to linear motion:
? = ?
0
t+
1
2
at
2
,
? = ?
0
+at,
?
2
= ?
2
0
+2at,
where?
0
is the angular velocity att = 0 ,? is the angular velocity at timet , and? is the angular
position at timet .
The line ar distance tr aveled along the circle (arc length) is related to the angle:
?s = r??.
Dividing b y t ime:
?s
?t
=
??
?t
=? v = r?,
where v is the linear speed (in m/s). This shows that linear speed depends on both how fast
the angle changes (? ) and how far the particle is from the center (r ).
The r ate o f change of linear speed is the tangential acceler ation:
a
t
=
dv
dt
=
d?
dt
=
ra
r
= a.
This acceler ation is along the tangent to the circle (the direction of motion) and changes the
speed, not the direction. Figure 7.1 shows a particleP on a circle with r adiusr , angle? , and the
arc fromP toP
'
.
1.1 Example 1
A particle moves in a circle of r adius 20 cm with a linear speed of 10 m/s. Find the angular
velocity .
Solution: Convert r adius to meters: r = 20 cm = 0.20 m. Use:
? =
v
r
=
10 m/s
0.20 m
= 50 r ad/s.
2
1.2 Example 2
A particle tr avels in a circle of r adius 20 cm at a speed that uniformly increases. If the speed
changes from 5.0 m/s to 6.0 m/s in 2.0 s, find the angular acceler ation.
Solution: T angential acceler ation:
a
t
=
?v
?t
=
6.0-5.0
2.0
= 0.5 m/s
2
.
Convert r adius: r = 20 cm = 0.20 m. Angular acceler ation:
a =
a
t
r
=
0.5
0.20
= 2.5 r ad/s
2
.
1.3 Example 3
A wheel of r adius 50 cm rotates with an angular velocity of 20 r ad/s. What is the linear speed of
a point on its rim?
Solution: Convert r adius: r = 50 cm = 0.50 m.
v = r? = 0.50×20 = 10 m/s.
2 Un it V ectors Along the Radius and T angent
T o describe circular motion mathematically , we use special directions: r adial (from the center to
the particle) and tangential (along the circle’ s path). Imagine a particleP on a circle with centerO
as the origin;OX as theX -axis, andOY as theY -axis (Figure 7.2). The particle’ s angular position
is? .
W e defin e two unit vectors (vectors of length 1):
- ˆ e
r
: The r adial unit vector , pointing outwar d from O to P . - ˆ e
t
: The tangential unit vector ,
along the tangent atP , in the direction of increasing? (counterclockwise).
T o find these vectors, dr aw lines from P par allel to the axes. The r adial unit vector ˆ e
r
along
OP has components:
ˆ e
r
= cos?ˆ i+ sin?ˆ ?,
whereˆ i and ˆ ? are unit vectors alongX andY . The tangential unit vector ˆ e
t
perpendicular to
ˆ e
r
is:
ˆ e
t
=- sin?ˆ i+ cos?ˆ ?.
These ve ctors help us describe position, velocity , and acceler ation in circular motion.
3 A cceler ation in Circular Motion
A particle moving in a circle is alwa ys acceler ating because its direction changes, even if its speed
is constant. Let’ s derive the acceler ation using the position vector , following the steps from the
original document. Consider a particle P on a circle of r adius r , with center O as the origin
(Figure 7.2). The position vector of the particle is:
? r = O
?
P = r? e
r
= r( cos?
?
i+ sin?
?
j).
T o find t he velocity , differentiate the position vector with respect to time:
? v =
d? r
dt
=
d
dt
[r( cos?
?
i+ sin?
?
j)].
Sincer is constant (the particle sta ys on the circle), we differentiate the components:
? v = r
[(
- sin?
d?
dt
)
?
i+
(
cos?
d?
dt
)
?
j
]
= r
d?
dt
[
- sin?
?
i+ cos?
?
j
]
.
3
Since
d?
dt
= ? (angular velocity), and recognizing that- sin?
?
i+ cos?
?
j =? e
t
(the tangential unit
vector), we get:
? v = r?? e
t
.
The velocity is along the tangent to the circle, and its magnitude isv = r? , consistent with our
earlier result.
Now , t o find the acceler ation, differentiate the velocity vector:
? a =
d? v
dt
=
d
dt
[r?(- sin?
?
i+ cos?
?
j)].
Apply t he product rule, noting thatr is constant, but? and? ma y vary with time:
? a = r
[
?
d
dt
[- sin?
?
i+ cos?
?
j]+
d?
dt
(- sin?
?
i+ cos?
?
j)
]
.
Compute the derivative of the tangential unit vector:
d
dt
[- sin?
?
i+ cos?
?
j] =
(
- cos?
d?
dt
)
?
i+
(
- sin?
d?
dt
)
?
j =
d?
dt
(- cos?
?
i- sin?
?
j).
Since
d?
dt
= ? , this becomes:
?(- cos?
?
i- sin?
?
j).
Notice t hat cos?
?
i+ sin?
?
j =? e
r
, so:
- cos?
?
i- sin?
?
j =-? e
r
.
Thus:
?
d
dt
(- sin?
?
i+ cos?
?
j) = ?·?(-? e
r
) =-?
2
? e
r
.
The seco nd term involves
d?
dt
= a , and- sin?
?
i+ cos?
?
j =? e
t
:
d?
dt
(- sin?
?
i+ cos?
?
j) = a? e
t
.
Combine:
? a = r
[
-?
2
? e
r
+a? e
t
]
=-?
2
r? e
r
+ra? e
t
.
Sincera =
dv
dt
(fromv = r? ), the acceler ation is:
? a =-?
2
r? e
r
+
dv
dt
? e
t
.
The a cceler ation has two components:
- Radial acceler ation: a
r
=-?
2
r =-
v
2
r
, directed toward the center (along-? e
r
). - T angential
acceler ation: a
t
=
dv
dt
, alo ng the tangent, changing the speed.
Uniform Circular Motion: If the speed is constant (
dv
dt
= 0 ):
? a =-?
2
r? e
r
, a
c
=
v
2
r
,
directed toward the center , called centripetal acceler ation. The magnitude is:
a
c
= ?
2
r =
v
2
r
,
sincev = r? .
Nonuniform Circular Motion: If the speed changes, both r adial and tangential components
exist. The r adial acceler ation is:
a
r
=-?
2
r =-
v
2
r
,
4
and the tangential acceler ation is:
a
t
=
dv
dt
.
The total acceler ation’ s magnitude is:
a =
v
a
2
r
+a
2
t
=
v
(
v
2
r
)
2
+
(
dv
dt
)
2
.
The anglea that the resultant acceler ation mak es with the r adius (Figure 7.3) is:
tana =
a
t
a
r
=
dv
dt
v
2
r
.
3.1 Example 4
Find the magnitude of the linear acceler ation of a particle moving in a circle of r adius 10 cm with
uniform speed, completing the circle in 4 s.
Solution: Convert r adius: r = 10 cm = 0.10 m. Circumference: 2pr = 2p× 0.10 = 0.2p m.
Linear speed:
v =
2pr
t
=
0.2p
4
= 0.05p m/s.
Centripetal acceler ation:
a =
v
2
r
=
(0.05p)
2
0.10
=
0.0025p
2
0.10
= 0.025p
2
˜ 0.247 m/s
2
.
3.2 Example 5
A particle moves in a circle of r adius 20 cm. Its linear speed is v = 2t m/s, where t is in seconds.
Find the r adial and tangential acceler ation att = 3 s.
Solution: Convert r adius: r = 0.20 m. Att = 3 s:
v = 2×3 = 6 m/s.
Radial a cceler ation:
a
r
=
v
2
r
=
6
2
0.20
=
36
0.20
= 180 m/s
2
.
T angential acceler ation:
a
t
=
dv
dt
=
d
dt
(2t) = 2 m/s
2
.
3.3 Example 6
A particle moves in a circle of r adius 0.5 m with constant speed 4 m/s. Find the centripetal accel-
er ation.
Solution:
a
c
=
v
2
r
=
4
2
0.5
=
16
0.5
= 32 m/s
2
.
5