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August 18, 202239 Insightful Publications

**Circular motion** is defined as the movement of an object along a circular route while spinning. Lets us look at some circular motion examples.** **Have you ever wondered how the giant wheel moves or why we feel a force in the outward direction while riding it? The type of motion we undergo while riding the giant wheel is circular motion.

We can analyse this motion just as we analyse the linear motion, and also, we can apply the equation of motion in the parameters of the circular motion.

The motion of a body such that its distance from a particular point remains constant is known as circular motion. The speed may or may not be constant while the body is moving, but the velocity is always changing as the direction of the motion is always changing. Therefore the body will always experience some acceleration, and hence there will always be some force on the body. If the speed of the body remains constant, then it is known as uniform circular motion.

To analyse the circular motion, we have a set of physical quantities that describe the motion. Example: angular displacement, angular velocity, angular acceleration etc.

For circular motion, we have a different set of angular variables.

Angular displacement: It is the angle displaced by the body with respect to a reference line. It is denoted by \(\theta .\)

Angular velocity: It is the rate of change of angular displacement with respect to time. It is denoted by \(\omega .\)

Angular acceleration: It is the rate of change of angular velocity with respect to time.

It is denoted by \(\alpha .\)

The equation of motion is the same as it is for the linear variables; just linear parameters are changed with the corresponding analogous angular parameters.

\(\overrightarrow s \to \vec \theta \)

\(\overrightarrow u \to \overrightarrow {{\omega _i}} \)

\(\overrightarrow v \to \overrightarrow {{\omega _f}} \)

\(\overrightarrow a \to \overrightarrow \alpha \)

This gives us the equation of motion for angular variables to be as follows.

The first equation of motion in angular variables is given by,

\(\overrightarrow {{\omega _f}} \to \overrightarrow {{\omega _i}} + \overrightarrow \alpha t\)

The second equation of motion in angular variables is given by,

\(\overrightarrow \theta = \overrightarrow {{\omega _i}} t + \frac{1}{2}\overrightarrow \alpha {t^2}\)

The third equation of motion in angular variables is given by,

\({\overrightarrow {{\omega _f}} ^2} – {\overrightarrow {{\omega _i}} ^2} = 2\overrightarrow \alpha .\overrightarrow \theta \)

It is important to note that the direction of the angular displacement and angular velocity is perpendicular to the plane of motion given by the direction of the thumb of the right hand when fingers are curled according to the motion of the body.

If the angular velocity is increasing, then the direction of angular acceleration is in the same direction as angular velocity, but if the angular velocity is decreasing, then the angular acceleration is in the opposite direction to that of angular velocity.

When the angular acceleration is zero, that is, the speed of the particle executing the circular motion is constant, then the motion is said to be a uniform circular motion. The magnitude of the force experienced by the particle remains constant.

The magnitude of the force is given by,

\(F = \frac{{M{v^2}}}{R}\)

Thus, the normal acceleration is,

\({a_n} = \frac{{{v^2}}}{R}\)

Where,

\(M\) is the mass of the particle performing the circular motion.

\(v\) is the magnitude of the velocity of the particle.

\(R\) is the radius of the path of the particle.

Using the property of the circle, we can derive the relation between the linear and angular variables.

The arc length of the circle and the angle subtended by it at the centre gives us the relation between the angular displacement and the linear displacement.

\(s = R\theta \)

Where,

\(R\) is the radius of the path.

On differentiating the above equation with respect to time we get,

\(\frac{{ds}}{{dt}} = R\frac{{d\theta }}{{dt}}\)

\( \Rightarrow v = R\omega \)

On differentiating the above equation with respect to time we get,

\( \Rightarrow {a_t} = R\alpha \)

Any type of motion can be considered to be a part of a circle at any instance. For example, a straight-line motion can be considered part of a circle whose radius is infinitely large.

\(R = \frac{{{{\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)}^{\frac{3}{2}}}}}{{\left| {\frac{{{d^2}y}}{{d{x^2}}}} \right|}}\)

So, this implies that every curve will have an instantaneous radius of curvature that may vary with time and position but will always satisfy the above equation.

The centripetal force is the force that is required to keep a particle moving in a curved path that is directed towards the centre of rotation or towards the centre, whereas the centrifugal force is the apparent force that is experienced by the particle which is moving in a curved path that acts outwardly away from the centre of rotation.

Centripetal force is an actual force, whereas centrifugal force is a pseudo force.

The giant wheel in an amusement park is an example of circular motion.

Satellites are moving in a circular motion around the earth.

** Q.1. Two cars having masses \({{\rm{m}}_1}\) and \({{\rm{m}}_2}\) move in circles of radii \({r_1}\) and \({r_2}\) respectively. If they complete the circle in equal time, then what will be the ratio of their angular speeds \(\frac{{{\omega _1}}}{{{\omega _2}}}?\)**:

Sol

The time taken to complete one rotation on a circle is called time period, i.e., in a time period \(T,\) the particle suffers an angular displacement of \(2\pi .\) Thus, angular speed,

\(\omega = \frac{{2\pi }}{T}\)

This relationship of angular speed and time period is independent of mass of the moving particle or radius of the circle. So,

\(\frac{{{\omega _1}}}{{{\omega _2}}} = \frac{{{T_2}}}{{{T_1}}}\)

*Q.2. A particle is kept fixed on a turntable rotating uniformly. As seen from the ground the particle goes in a circle, its speed is \(20\,{\rm{cm}}\,{{\rm{s}}^{ – 1}}\) and acceleration is \(20\,{\rm{cm}}\,{{\rm{s}}^{ – 2}}.\) The particle is now shifted to a new position to make the radius half of the original value. The new values of the speed and acceleration will be.Sol:*

For a uniformly rotating turntable, the angular speed is constant. So, tangential speed and centripetal acceleration are given by,

\({v_t} = \omega r\)

\({a_c} = {\omega ^2}r\)

for \(\omega = \) constant,

\({v_t}\propto r\)

\({a_c}\propto r\)

So, on reducing the radius to half, both tangential speed and radial acceleration also reduces to half, i.e., \(10\,{\rm{cm}}\,{{\rm{s}}^{ – 1}}\) and \(10\,{\rm{cm}}\,{{\rm{s}}^{ – 2}},\) respectively.

Circular motion is the type of periodic motion in which the body follows a circular path that is the distance from a fixed point remains constant.

The angular variables associated with the kinematics of circular motion are similar to the linear variable. The equations of motion are also the same as the linear motion, with analogous terms used in place of the linear variable.

The direction of the angular vectors is perpendicular to the plane. The direction can be determined by the right-hand thumb rule. That is when we curl the finger of our right hand, then the thumb points towards the direction of the vector.

Centripetal and centrifugal forces are two types of forces associated with circular motion.

It is important to note that the centripetal force is a real force, whereas centrifugal force is an apparent force.

The instantaneous radius of curvature is given by,

\(R = \frac{{{{\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}.\)

** Q.1. What is circular motion?**Circular motion is a type of periodic motion in which the body follows a circular path that is the distance from a fixed point remains constant.

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** Q.2. Does Uniform circular motion have acceleration?** Yes, uniform circular motion does have an acceleration that is perpendicular to the direction of velocity; though the magnitude of the velocity remains constant, the direction of the velocity keeps on changing. Thus, it has some acceleration.

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** Q.3. What is the expression for the instantaneous radius of curvature?**If the equation of the path is given then,

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The instantaneous radius of curvature is given by,

\(R = \frac{{{{\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}.\)

** Q.4. What is the magnitude and direction of the acceleration in uniform circular motion?** The acceleration in uniform circular motion is towards the centre, and the magnitude of the acceleration is given by,

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\({a_n} = \frac{{{v^2}}}{R}\)

Where,

\(v\) is the magnitude of the velocity of the particle.

\(R\) is the radius of the path of the particle.

** Q.5. Does acceleration always towards the centre in a circular motion?**No, the direction of the acceleration is not always towards the centre of the path. If the motion is uniform circular motion, then the net acceleration will be towards the centre, but if the circular motion is not uniform, then there can be two components of the acceleration one will be in the tangential direction, and another will be in the normal direction.

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