# Physics – Circular Motion: Additional Questions

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### Characteristics:

#### Characteristics of Circular Motion:

• In a circular motion, the particle moves along the circumference of a circle.
• It is a translational motion along a curved path.
• The magnitude of radius vector or position vector is constant and equals to the radius of the circular path.
• The direction of radius vector or position vector changes continuously.
• If the magnitude of the velocity (speed) of the particle performing constant, the particle is said to perform the uniform circular motion.
• If the magnitude of the velocity (speed) of the particle performing changes continuously, the particle is said to perform the non- uniform circular motion.

#### Characteristics of Radius or Position Vector:

• It is always directed along the radius of the circular path.
• Its direction is from the centre of the circular path to the position of the particle at that instant i.e. is directed radially outward..
• In a circular motion, the magnitude of radius vector or position vector is constant and equals to the radius of the circular path.
• The direction of radius vector or position vector in circular motion changes continuously.
• Its direction is opposite to that of centripetal acceleration and the centripetal force.
• It is denoted by r. Its dimensions are  [MºL1Tº]. Its S.I. unit is metre (m) and c.g.s. unit is centimetre (cm).

#### Characteristics of Instantaneous Velocity or Tangential Velocity:

• It is always directed along the tangent to the circular path at given point on the circular path at that instant.
• It is always perpendicular to the direction of the radius vector at the point represented by the radius vector on the circular path.
• For uniform circular motion the magnitude of instantaneous velocity is always constant but direction changes continuously.
• For non-uniform circular motion, the magnitude and direction of the instantaneous velocity change continuously.
• It is denoted by v. Its dimensions are  [MºL1T-1]. Its S.I. unit is metre per second  (m s-1) and c.g.s. unit is centimetre per second (cm s-1).

#### Characteristics of Angular Displacement:

• It is the angle, traced by the radius vector at the centre of the circular path in a given time is called the angular displacement of the particle at that time.
• For smaller magnitude (infinitesimal) angular displacement is a vector quantity and its direction is given by the right-hand thumb rule.
• Finite angular displacement is a vector quantity.
• Instantaneous angular velocity is a vector quantity.
• The direction of angular displacement in an anticlockwise sense is considered as positive, while the direction of angular displacement in a clockwise sense is considered as negative.
• The angular displacement of the particle performing uniform circular motion in equal time is equal.
• It is denoted by ‘θ’. Its S.I. unit is radian (rad). It is dimensionless quantity. [MºLºTº].

#### Characteristics of Angular Velocity:

• The rate of change of angular displacement with respect to time is called as the angular velocity of the particle.
• Its direction is given by the right-hand thumb rule.
• The direction of angular velocity is the same as that of angular displacement.
• For uniform circular motion, the magnitude of angular velocity is constant.
• The magnitude of angular velocity (ω) is related with the magnitude of linear velocity (v) by the relation v = rω.
• It is denoted by letter ‘ω’. Its S.I. unit is radians per second (rad s-1). Its dimensions are [MºLºT -1].

#### Characteristics of Angular Acceleration:

• The average angular acceleration is defined as the time rate of change of angular velocity.
• If the angular velocity is increasing angular acceleration is positive (e.g. the angular acceleration of the tip of a fan just switched on). If the angular velocity is decreasing angular acceleration is negative (e.g. the angular acceleration of the tip of a fan just switched off)
• If the angular velocity is increasing then the angular acceleration has the same direction as that of the angular velocity. If the angular velocity is decreasing then the angular acceleration has the opposite direction as that of the angular velocity.
• For uniform circular motion angular acceleration is zero.
• The magnitude of angular acceleration (α) is related with the magnitude of linear acceleration (a) by the relation a = rα.
• It is denoted by letter ‘α’. Its S.I. unit is radians per second square (rad /s2). Its dimensions are [MºLºT -2].

#### Characteristics of Uniform Circular Motion:

• The magnitude of velocity (speed) of the particle performing U.C.M. is constant.
• The magnitude of the instantaneous velocity of the particle performing U.C.M.  remains constant but its direction changes continuously. Hence U.C.M. is accelerated motion.
• It is a periodic motion with definite period and frequency.
• in U.C.M., the magnitude of the centripetal force acting on the body is constant
• in U.C.M., the linear speed, angular speed, radial (centripetal) acceleration, kinetic energy, angular momentum and magnitude of the linear momentum of the body remain constant.
• in U.C.M., the angular acceleration, tangential acceleration is zero.

#### Characteristics of Time Period of U.C.M.:

• The time taken by a particle performing uniform circular motion to complete one revolution is called as the period of revolution or periodic time or simply period (T).
• For U.C.M. it is constant.
• It is denoted by ‘T’. The S. I. Unit of the period is second (s). Its dimensions are[MºLºT 1].

#### Characteristics of Frequency of U.C.M.:

• The number of revolutions by the particle performing uniform circular motion in unit time is called as frequency (n) of revolution.
• For U.C.M. it is constant.
• The frequency is denoted by letter ‘n’ or ‘f’. The S. I. Unit of frequency is hertz (Hz). Its dimensions are [MºLºT-1].
• In time T the particles complete one revolution. Thus the particle completes 1/T revolutions in unit time. Thus n = 1/T.

#### Characteristics of Centripetal Acceleration:

• It is the acceleration of a particle performing the circular motion, which is directed towards the centre of circular path along the radius.
• It is always directed towards the centre of the circular path along the radius.
• The direction of centripetal acceleration changes continuously.
• For U.C.M. the magnitude of the centripetal acceleration is constant.
• It is denoted by the letter ‘a’. Its S.I. unit is metre per square second square (m s-2). Its dimensions are [MºL1T -2].
• The magnitude of centripetal acceleration is given by

#### Characteristics of Centripetal Force:

• It is a real force because it is provided by gravitational, electromagnetic or nuclear interaction.
• It arises in an inertial frame of reference. The inertial reference frame is that frame of reference which is moving with uniform velocity w.r.t. earth.
• It is always directed towards the centre of the circular path.
• Without it, the circular motion is not possible.
• In U.C.M. the magnitude of the centripetal force is constant.
• The work done by the centripetal force is zero because the displacement of the particle (tangential) is perpendicular to the direction of the centripetal force (radial) i.e. there is no displacement in the direction of the centripetal force.
• The torque produced by the centripetal force at the centre of the circular path is zero.
• The direction of centripetal force does not depend on the sense of rotation of the body.
• It is denoted by the letter ‘F’. Its S.I. unit is newton and c.g.s. unit is dyne. Its dimensions are [M1L1T -2].
• The magnitude of centripetal acceleration is given by

#### Characteristics of Centrifugal Force:

• It is an imaginary force or pseudo force.
• It is experienced in non – inertial frame of reference.
• The magnitude of the centripetal force is equal to that of centripetal force.
• It is always directed away from the centre of the circle along the radius
• it is directed opposite to the centripetal force.
• Centrifugal force doesn’t have an independent existence.

#### Characteristics of Angle of Banking:

• The angle of banking is given by

• The angle of banking depends on the speed of the vehicle, the radius of the curved road and the acceleration due to gravity g at that place.
• The angle of banking is independent of mass ‘m’ of the vehicle. Thus the angle of banking is the same for heavy and light vehicles.
• the angle of banking is directly proportional to the square of the velocity (v) of the vehicle, inversely proportional to the radius (r) of curvature of the path and inversely proportional to the acceleration due to gravity (g) at that place.
• The acceleration due to gravity is more on the pole than that at the equator. Thus for the given radius of the curve and speed of the vehicle angle of banking is less at the pole than that at the equator.
• The elevation outer edge of a banked road above the inner edge is given by

h = l sinθ

#### Characteristics of Safe Velocity on Banked Road:

• The safe velocity on the banked road is given by

• The safe velocity of a vehicle on the banked road depends on the radius of the curved road, the angle of banking, and the acceleration due to gravity g at that place.
• Safe velocity is independent of mass ‘m’ of the vehicle. Thus it is the same for heavy and light vehicles.
• For the given radius of the curve and angle of banking, the safe velocity is more at the pole than that at the equator.

#### Characteristics of Semi-vertical Angle of Conical Pendulum:

• The semi-vertical angle is the angle made by the string of conical pendulum with the vertical.
• It is given by the expression

• It is independent of the mass of the bob of the conical pendulum.
• The semi-vertical angle θ depends on the length and period of the conical pendulum. If time period decreases, then θ increases.
• θ can never be 90° because for this period T = 0.

### Factors Affecting:

#### Factors Affecting Angle of Banking:

• The angle of banking is given by

• The angle of banking is independent of mass ‘m’ of the vehicle. Thus the angle of banking is the same for heavy and light vehicles.
• the angle of banking is directly proportional to the square of the velocity (v) of the vehicle, inversely proportional to the radius (r) of curvature of the path and inversely proportional to the acceleration due to gravity (g) at that place.

#### Factors Affecting Time Period of Conical Pendulum:

• The time period of a conical pendulum is given by

• The time period of a conical pendulum is directly proportional to the square root of its length.
• The time period of a conical pendulum is directly proportional to the square root of the cosine ratio of the semi-vertical angle that is the angle made by the string of conical pendulum with the vertical.
• The time period of conical pendulum increases with the increase in the value of the semi-vertical angle.
• The time period of a conical pendulum is inversely proportional to the square root of the acceleration due to gravity at that place.
• The time period of a conical pendulum is independent of the mass of the bob of the conical pendulum.

### Scientific Reasons:

#### Centrifugal Force is a Pseudo Force:

• Centrifugal force is not real force since it does not arise due to either gravitational or electrostatic or nuclear interaction.
• Centrifugal force has no independent existence. It comes to play with the action of the centripetal force. It does not arise due to either gravitational or electrostatic or nuclear interaction.
• Centrifugal force acts in the non-inertial reference frame
Hence centripetal force is a pseudo force.

#### Centripetal Force and Centrifugal Force Do Not Constitute Action-Reaction Pair:

• A statement ” A particle moving uniformly along a circle experiences a force directed towards the centre (centripetal force) and an equal and opposite force directed away from the centre (centrifugal force). The two forces keep the particle in equilibrium” is completely wrong.
• Because the particle in circular motion is not in equilibrium because a net force (centripetal force) acts towards the centre which is required to change the direction of particle continuously. Centripetal force is a real force.
• The centrifugal force is an imaginary force is required by an observer moving with the particle. The observer is a non-inertial reference frame and for the observer, the particle is at rest.
• Centripetal force is required for circular motion. Centrifugal force does not have an independent existence. It comes into play when centripetal force starts acting. Thus the centripetal force and centrifugal force do not constitute the action-reaction pair.

#### Derivation for Expression of Centripetal Acceleration by Calculus or Cartesian Method:

• Let us consider a particle performing uniform circular motion with a linear velocity of magnitude v and angular velocity of magnitude ω along a circle of radius r with centre O in an anticlockwise sense. Let the particle moves from A to P in time ‘t’ Such that ∠ POA = θ.

But for uniform Circular motion θ = ω t. Thus, ∠ POA = ω t

Let us draw seg PM perpendicular to seg OA. Thus, ∠ POM = ω t

The radius vector at time t at P is given by

Substituting these values in equation (1)

The linear velocity of a particle can be obtained by differentiating equation (2) w.r.t. time t.

The linear acceleration of a particle can be obtained by differentiating equation (3) w.r.t. time t.

From equations (1) and (4) we have

• Thus the magnitude of the acceleration is v2/r and its direction is along the radius and the negative sign indicates that it is opposite to the radius vector i.e. the acceleration is directed towards the centre of the circular path. This acceleration is called the centripetal acceleration.

#### Derivation for Expression for Tension in the String of a Conical Pendulum:

• Let us consider a conical pendulum consists of a bob of mass ‘m’ revolving in a horizontal circle with constant speed ‘v’ at the end of a string of length ‘l’. Let the string makes a constant angle ‘θ’ with the vertical. let ‘h’ be the depth of the bob below the support.
• The tension ‘F’ in the string can be resolved into two components. Horizontal ‘Fsin θ’ and vertical ‘Fcos θ’.

The vertical component (F cos θ) balances the weight mg of the vehicle.

F cosθ  = mg  ………….. (1)

The horizontal component (F sin θ) provides the necessary centripetal force.

F sin θ = mv2/r  ………… (2)

Dividing equation (2) by (1) we get,

Squaring equations (1) and (2) and adding

Substituting equation (5)

This is an expression for the tension in the string of a conical pendulum.

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