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Course by: Sunil Kumar Singh. E-mail the author

# Uniform circular motion

Module by: Sunil Kumar Singh. E-mail the author

Summary: Uniform circular motion (UCM) is the basic unit of rotational kinematics just like the uniform linear motion is the basic unit of translational kinematics.

Uniform circular motion denotes motion of a particle along a circular arc or a circle with constant speed. This statement, as a matter of fact, can be construed as the definition of uniform circular motion.

## Requirement of uniform circular motion

The uniform circular motion represents the basic form of rotational motion in the same manner as uniform linear motion represents the basic form of translational motion. They, however, are different with respect to the requirement of force to maintain motion.

Uniform linear motion is the reflection of the inherent natural tendency of all natural bodies. This motion by itself is the statement of Newton’s first law of motion : an object keeps moving with its velocity unless there is net external force. Thus, uniform linear motion indicates “absence” of force.

On the other hand, uniform circular motion involves continuous change in the direction of velocity without any change in its magnitude (v). A change in the direction of velocity is a change in velocity (v). It means that an uniform circular motion is associated with an acceleration and hence force. Thus, uniform circular motion indicates “presence” of force.

Let us now investigate the nature of force required to maintain uniform circular motion. We know that a force acting in the direction of motion changes only the magnitude of velocity. A change in the direction of motion, therefore, requires that velocity of the particle and force acting on it should be at an angle. However, such a force, at an angle with the direction of motion, would have a component along the direction of velocity as well and that would change the magnitude of the motion.

In order that there is no change in the magnitude of velocity, the force should have zero component along the direction of velocity. It is possible only if the force be perpendicular to the direction of velocity such that its component in the direction of velocity is zero (Fcos90° = 0). Precisely, this is the requirement for a motion to be uniform circular motion.

In plain words, uniform circular motion (UCM) needs a force, which is always perpendicular to the direction of velocity. Since the direction of velocity is continuously changing, the direction of force, being perpendicular to velocity, should also change continously.

The direction of velocity along the circular trajectory is tangential. The perpendicular direction to the circular trajectory is, therefore, radial direction. It implies that force (and hence acceleration) in uniform direction motion is radial. For this reason, acceleration in UCM is recognized to seek center i.e. centripetal (seeking center).

This fact is also validated by the fact that the difference of velocity vectors, whose time rate gives acceleration, at two instants (Δv) is radial.

The yet another important aspect of the UCM is that the centripetal force is radial and hence does not constitute a torque as the force is passing through the axis of rotation. A torque is force multiplied by the perpendicular distance of the line of action of the force from the axis of rotation. It must be clearly understood that the requirement of centripetal force is essentially additional or different to the force, which is required in non-uniform circular motion to accelerate the particle tangentially. The centripetal force is required to accelerate particle in radial direction and is different to one required to accelerate particle tangentially.

Irrespective of whether circular motion is uniform (constant speed) or non-uniform (varying speed), the circular motion inherently associates a radial acceleration to ensure that the direction of motion is continuously changed – at all instants. We shall learn about the magnitude of radial acceleration soon, but let us be emphatic to differentiate radial acceleration (accounting change in direction that arises from radial force) with tangential acceleration (accounting change in the speed that arises from tangential force or equivalently a torque).

Motion of natural bodies and sub-atomic particles are always under certain force system. Absence of force in the observable neighborhood is rare. Thus, uniform linear motion is rare, while uniform circular motion abounds in nature as there is availability of external force that continuously changes direction of the motion of the bodies or particles. Consider the electrostatic force between nucleus and an electron in an atom. The force keeps changing direction as electron moves. So is the case of gravitational force (indicated by red arrow in the figure), which keeps changing its direction as the planet moves around sun.

It may sometimes be perceived that a force like centripetal force should have caused the particle or body to move towards center ultimately. The important point to understand here is that a force determines initial direction of motion only when the particle is stationary. However, if the particle is already in motion, then force modifies the direction in accordance with initial inclination between velocity and force such that the resulting acceleration (change in vector velocity) is in the direction of force. We shall soon see that this is exactly the case in uniform circular motion.

Let us summarize the discussion of uniform circular motion so far :

• The trajectory of uniform circular motion is circular arc or a circle and hence planar or two dimensional.
• The speed (v) of the particle in UCM is a constant.
• The velocity (v) of the particle in UCM is variable and is tangential or circumferential to the path of motion.
• Centripetal force (F) is required to maintain uniform circular motion against the natural tendency of the bodies to move linearly.
• Centripetal force (F) is variable and is radial in direction.
• Centripetal force (F) is not a torque and does not cause acceleration in tangential direction.
• Centripetal acceleration ( a R a R ) is variable and radial in direction.

## Analysis of uniform circular motion

It has been pointed out that any motion, that changes directions, requires more than one dimension for representation. Circular motion by the geometry of the trajectory is two dimensional motion.

In the case of circular motion, it is a matter of convenience to locate origin of the two dimensional coordinate system at the center of circle. This choice ensures the symmetry of the circular motion about the origin of the reference system.

### Coordinates of the particle

The coordinates of the particle is given by the "x" and "y" coordinate pair as :

x = r cos θ y = r sin θ x = r cos θ y = r sin θ
(1)

The angle "θ" is measured anti-clockwise from x-axis.

### Position of the particle

The position vector of the position of the particle, r, is represented in terms of unit vectors as :

r = x i + y j r = r cos θ i + r sin θ j r = r ( cos θ i + sin θ j ) r = x i + y j r = r cos θ i + r sin θ j r = r ( cos θ i + sin θ j )
(2)

### Velocity of the particle

The magnitude of velocity of the particle (v) is constant by the definition of uniform circular motion. In component form, the velocity (refer to the figure) is :

v = v x i + v y j v = v x i + v y j
(3)

From the geometry as shown in the figure,

v x = - v sin θ v y = v cos θ v = - v sin θ i + v cos θ j v x = - v sin θ v y = v cos θ v = - v sin θ i + v cos θ j
(4)

We may emphasize here that it is always easy to find the sign of the component of a vector like velocity. Decide the sign by the direction of component (projection) with respect to positive direction of reference axis. Note from the figure that component along x-direction is opposite to the positive reference direction and hence negative.

From the ΔOAP,

sin θ = y r cos θ = x r sin θ = y r cos θ = x r

Putting these values in the expression for velocity, we have :

v = - v y r i + v x r j v = - v y r i + v x r j
(5)

### Centripetal acceleration

We are now sufficiently equipped and familiarized with the nature of uniform circular motion and the associated centripetal (center seeking) acceleration. Now, we seek to determine this acceleration.

Knowing that speed, "v" and radius of circle, "r" are constants, we differentiate the expression of velocity with respect to time to obtain expression for centripetal acceleration as :

a = - v r ( đ y đ t i - đ x đ t j ) a = - v r ( đ y đ t i - đ x đ t j )

a = - v r ( v y i - v x j ) a = - v r ( v y i - v x j )

Putting values of component velocities in terms of angle,

a = - v r ( v cos θ i + v sin θ j ) = a x i + a y j a = - v r ( v cos θ i + v sin θ j ) = a x i + a y j
(6)

where

a x = - v 2 r cos θ a y = - v 2 r sin θ a x = - v 2 r cos θ a y = - v 2 r sin θ

It is evident from the equation of acceleration that it varies as the angle with horizontal, "θ" changes. The magnitude of acceleration is :

a = | a | = ( a x 2 + a y 2 ) a = | a | = v r { v 2 ( cos 2 θ + cos 2 θ ) } a = v 2 r a = | a | = ( a x 2 + a y 2 ) a = | a | = v r { v 2 ( cos 2 θ + cos 2 θ ) } a = v 2 r
(7)

The radius of the circle is constant. The magnitude of velocity i.e. speed, "v" is constant for UCM. It, then, follows that though velocity changes with the motion (angle from reference direction), but the speed of the particle in UCM is a constant.

#### Example 1

Problem : A cyclist negotiates the curvature of 20 m with a speed of 20 m/s. What is the magnitude of his acceleration?

Solution : The speed of the cyclist is constant. The acceleration of cyclist, therefore, is the centripetal acceleration required to move the cyclist along a circular path i.e. the acceleration resulting from the change in the direction of motion along the circular path.

Here, v = 20 m/s and r = 20 m

a = v 2 r = 20 2 20 = 20 m / s 2 a = v 2 r = 20 2 20 = 20 m / s 2

This example points to an interesting aspect of circular motion. The centripetal acceleration of the cyclist is actually two (2) times that of acceleration due to gravity (g = 10 m / s 2 m / s 2 ). This fact is used to create large acceleration in small space with appropriate values of “v” and “r” as per the requirement in hand. The large acceleration so produced finds application in particle physics and for equipments designed to segregate material on the basis of difference in density. This is also used to simulate large acceleration in a centrifuge for astronauts, who experience large acceleration at the time of take off or during entry on the return.

Generation of high magnitude of acceleration during uniform circular motion also points to a potential danger to pilots, while maneuvering circular trajectory at high speed. Since a pilot is inclined with the head leaning towards the center of motion, the blood circulation in the brain is low. If his body part, including brain, is subjected to high acceleration (multiple of acceleration due to gravity), then it is likely that the pilot experiences dizziness or sometimes even looses consciousness.

Circular motion has many interesting applications in real world and provides explanations for many natural events. In this module, however, we restrict ourselves till we study the dynamics of the circular motion also in subsequent modules.

### Direction of centripetal acceleration

Here, we set out to evaluate the angle “α” as shown in the figure. Clearly, if this angle “α” is equal to “θ”, then we can conclude that acceleration is directed in radial direction.

Now,

tan α = a y a x = - v 2 r sin θ - v 2 r cos θ = tan θ α = θ tan α = a y a x = - v 2 r sin θ - v 2 r cos θ = tan θ α = θ

This proves that centripetal acceleration is indeed radial (i.e acting along radial direction).

### Time period of uniform circular motion

A particle under UCM covers a constant distance in completing circular trajectory in one revolution, which is equal to the perimeter of the circle.

s = 2 π r s = 2 π r

Further the particle covers the perimeter with constant speed. It means that the particle travels the circular trajectory in a constant time given by its time period as :

T = 2 π r v T = 2 π r v
(8)

#### Exercise 1

An astronaut, executing uniform circular motion in a centrifuge of radius 10 m, is subjected to a radial acceleration of 4g. The time period of the centrifuge (in seconds) is :

(a) π (b) 2π (c) 3π (d) 4π

## Uniform circular motion and projectile motion

Both these motions are two dimensional motions. They are alike in the sense that motion in each case is subjected to continuous change of the direction of motion. At the same time, they are different in other details. The most important difference is that projectile motion has a constant acceleration, whereas uniform circular motion has a constant magnitude of acceleration.

Significantly, a projectile motion completely resembles uniform circular motion at one particular instant. The projectile has only horizontal component of velocity, when it is at the maximum height. At that instant, the force of gravity, which is always directed downward, is perpendicular to the direction of velocity. Thus, projectile at that moment executes an uniform circular motion.

Let radius of curvature of the projectile trajectory at maximum height be “r”, then

a = g = v x 2 r = v 2 cos 2 θ r r = v 2 cos 2 θ g a = g = v x 2 r = v 2 cos 2 θ r r = v 2 cos 2 θ g

### Note:

The radius of curvature is not equal to maximum height – though expression appears to be same. But, they are actually different. The numerator here consists of cosine function, whereas it is sine function in the expression of maximum height. The difference is also visible from the figure.

## Exercise

### Exercise 2

A particle moves along a circle of radius “r” meter with a time period of “T”. The acceleration of the particle is :

(a) 4 π 2 r T 2 4 π 2 r T 2 (b) 4 π 2 r 2 T 2 4 π 2 r 2 T 2 (c) 4 π r T 4 π r T (c) 4 π r T 4 π r T

### Exercise 3

Which of the following quantities are constant in uniform circular motion :

(a) velocity and acceleration

(b) magnitude of acceleration and radius of the circular path

(c) radius of the circular path and speed

(d) speed and acceleration

### Exercise 4

Which of these is/are correct for a particle in uniform circular motion :

(a) the acceleration of the particle is tangential to the path

(b) the acceleration of the particle in the direction of velocity is zero

(c) the time period of the motion is inversely proportional to the radius

(d) for a given acceleration, velocity of the particle is proportional to the radius of circle

### Exercise 5

A particle moves with a speed of 10 m/s along a horizontal circle of radius 10 m in anti-clockwise direction. The x and y coordinates of the particle is 0,r at t = 0. The velocity of the particle (in m/s), when its position makes an angle 135° with x - axis, is :

(a) - 5 2 i - 5 2 j - 5 2 i - 5 2 j (b) 5 2 i - 5 2 j 5 2 i - 5 2 j (c) 5 2 i + 5 2 j 5 2 i + 5 2 j (d) - 5 2 i + 5 2 j - 5 2 i + 5 2 j

### Exercise 6

A car moves along a path as shown in the figure at a constant speed. Let a A a A , a B a B , a C a C and a D a D be the centripetal accelerations at locations A,B,C and D respectively. Then

(a) a A < a C a A < a C (b) a A > a D a A > a D (c) a B < a C a B < a C (d) a B > a D a B > a D

### Exercise 7

A particle executes uniform circular motion with a speed “v” in xy plane, starting from position (r,0), where “r” denotes the radius of the circle. Then

(a) the component of velocity along x - axis is constant

(b) the component of acceleration along x - axis is constant

(c) the angle swept by the line joining center and particle in equal time interval is constant

(d) the velocity is parallel to external force on the particle

### Exercise 8

A particle executes uniform circular motion with a speed “v” in anticlockwise direction, starting from position (r,0), where “r” denotes the radius of the circle. Then, both components of the velocities are positive in :

(a) first quadrant of the motion

(b) second quadrant of the motion

(c) third quadrant of the motion

(d) fourth quadrant of the motion

### Exercise 9

A particle executes uniform circular motion with a constant speed of 1 m/s in xy - plane. The particle starts moving with constant speed from position (r,0), where "r" denotes the radius of the circle. If center of circle is the origin of the coordinate system, then what is the velocity in "m/s" of the particle at the position specified by the coordinates (3m,-4m) ?

(a) 1 5 ( 4 i + 3 j ) 1 5 ( 4 i + 3 j ) (b) 5 ( 4 i - 3 j ) 5 ( 4 i - 3 j ) (c) 1 5 ( 3 i - 4 j ) 1 5 ( 3 i - 4 j ) (d) 1 5 ( 3 i + 4 j ) 1 5 ( 3 i + 4 j )

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