### Chapter : Circular Motion & Rotational Dynamics

__8. Type of Circular Motion__

__8. Type of Circular Motion__

**8. Type of Circular Motion**

**8.1 Uniform circular motion**

**8.2 Non Uniform Circular Motion:**

**8.1 Uniform Circular Motion:**

If m = mass of body,

r = radius of circular orbit,

v = magnitude of velocity

a

_{c}= centripetal acceleration,

a

_{t}= tangential acceleration

In uniform circular motion :

**(i)**|| = || = || = constant i.e. speed is constant

**(ii)**As || is constant so tangential acceleration a

_{t}= 0

**(iii)**Tangential force F

_{t}= 0

**(iv)**Total acceleration

(towards the centre)

**Note:**

**(i)**Because F

_{c}is always perpendicular to velocity or displacement, hence the work done by this force will always be zero.

**(ii)**Circular motion in horizontal plane is usually uniform circular motion.

**(iii)**There is an important difference between the projectile motion and circular motion.

In projectile motion, both the magnitude and the direction of acceleration (g) remain constant, while in circular motion the magnitude remains constant but the direction continuously changes.

Hence equations of motion are not applicable for circular motion.

Remember that equations of motion remain valid only when both the magnitude & direction of acceleration are constant.

**8.1.1 Hint to solve numerical problem :**

**(i)**Write down the required centripetal force

**(ii)**Draw the free body diagram of each component of system.

**(iii)**Resolve the forces acting on the rotating particle along radius and perpendicular to radius

**(iv)**Calculate net radial force acting towards centre of circular path.

**(v)**Make it equal to required centripetal force.

**(vi)**For remaining components see according to question.

**8.1.2 Motion In Horizontal Circle : Conical pendulum**

This is the best example of uniform circular motion

A conical pendulum consists of a body attached to a string, such that it can revolve in a horizontal circle with uniform speed. The string traces out a cone in the space.

**(i)**The force acting on the bob are

(a) Tension T (b) weight mg

**(ii)**The horizontal component T sin θ of the tension T provides the centripetal force and the vertical component T cos θ balances the weight of bob

∴ T sin θ = and T cos θ = mg

From these equations

Also if h = height of conical pendulum tan θ = OP/OS = r/h ....(iii)

From (ii) & (iii),

The time period of revolution

[where os = l]

**8.2 Non-uniform Circular Motion :**

**(i)**In non-uniform circular motion :

|| ≠ constant ω ≠ constant

i.e. speed ≠ constant

i.e. angular velocity ≠ constant

**(ii)**If at any instant

v = magnitude of velocity of particle

r = radius of circular path

w = angular velocity of particle,

then v = rω

**(iii)**Tangential acceleration :

a

_{t}=

where v = and s = arc - length

(iv) Tangential force :

F

_{t}= ma

_{t}

(v) Centripetal force :

F

_{c}= = mω

^{2}r

(vi) Net force on the particle :

= +

⇒ F =

If θ is the angle made by [Note angle between F

_{c}and F

_{t}is 90°] F with F

_{c}, then

tan θ = F

_{t}/F

_{c}

⇒ θ = tan

^{–1}(F

_{t}/F

_{c})

Angle between F & F

_{t}is (90° – θ)

**(vii)**Net acceleration towards the centre = centripetal acceleration

⇒

**(viii)**Net acceleration,

The angle made by 'a' with a

_{c},

**Special Note :**

**(i)**In both uniform & non-uniform circular motion F

_{c}is perpendicular to velocity ; so work done by centripetal force will be zero in both the cases.

**(ii)**In uniform circular motion F

_{t}= 0, as a

_{t}= 0, so work done will be zero by tangential force.

But in non-uniform circular motion F

_{t}≠ 0, thus there will be work done by tangential force in this case.

Rate of work done by net force in non-uniform circular motion = Rate of work done by tangential force

⇒

**Motion in Vertical Circle : Motion of a body suspended by string :**

This is the best example of non-uniform circular motion.

When the body rises from the bottom to the height h apart of its kinetic energy converts into potential energy

Total mechanical energy remains conserved

Total (P.E. + K.E.) at A = Total (P.E. + K.E.) at P

⇒ 0 + mu

^{2}= mgh + mv

^{2}

⇒

[Where l is length of the string]

**Tension at a point P :**

**(i)**At point P

**required centripetal force**=

**(a) Net force towards the centre :**

T – mg cos θ, which provides required centripetal force.

∴ T – mg cos θ =

T = m [ g cos θ +] = [u

^{2}– gl (2 – 3cos θ)]

**(b)**

**Tangential force for the motion**

F

^{t}= mg sin θ

This force retards the motion

**(ii) Results :**

**(a) Tension at the lowest point A :**

**(b) Tension at point B :**

**(c) Tension at point C :**

Thus we conclude that

T

_{A}> T

_{C}> T

_{B}

and also T

_{A}– T

_{B}= 6 mg

T

_{A}– T

_{C}= 3 mg

T

_{C}– T

_{B}= 3 mg

**(iii) Cases :**

**(a)**If u > √5gλ

In this case tension in the string will not be zero at any of the point, which implies that the particle will continue the circular motion.

**(b)**If u = √5gλ

In this case the tension at the top most point (B) will be zero, which implies that the particle will just complete the circular motion.

**(c) Critical Velocity :**The minimum velocity at which the circular motion is possible

The critical velocity at A = √5gλ

The critical velocity at B = √gλ

The critical velocity at C = √3gλ

Also TA = 6 mg, TB = 0, TC = 3 mg

**(d)**If √2gλ < u < √5gλ

In this case particle will not follow circular motion. Tension in string becomes zero somewhere between points C & B whereas velocity remain positive. Particle leaves circular path and follow parabolic trajectory

**(e)**If u = √2gλ

In this case both velocity and tension in the string becomes zero between A and C and particle will oscillate along semi-circular path.

**(f)**If u < √2gλ

The velocity of particle remains zero between A and C but tension will not be zero and the particle will oscillate about the point A.

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