## Unit – II B |

## Centripetal Acceleration:

### Expression for Centripetal Acceleration (Geometric Method):

- The linear velocity of the particle performing uniform circular motion is a vector quantity being always tangential to the circular path and changes constantly in the direction. Therefore the particle in uniform circular motion has linear acceleration.
- Let us consider a particle performing uniform circular motion with linear velocity and magnitude of angular velocity along a circle of radius r with centre O in an anticlockwise sense (moving from initial position A to final position B)as shown in the diagram.

- Let us draw velocity triangle QBR for the velocities of the particles at A and B.

- Now by definition,

- The triangles AOB and RBQ are similar. Hence ∠ QBR = δθ

For smaller angular displacement δθ,

- Substituting in equation (1), the magnitude of acceleration is given by

- Substituting in equation (2) we get

- This is the expression for acceleration of particle performing the uniform circular motion. This acceleration is directed towards the centre of circular path along the radius. This acceleration is called as radial acceleration or centripetal acceleration.
- In vector form centripetal acceleration can be given as .

- The negative sign indicates that centripetal acceleration is oppositely directed to that of radius vector i.e. directed towards the centre of the circle along the radius.

### Expression for Centripetal Acceleration (Cartesian Method or Calculus Method):

- Consider particle revolving with uniform angular velocity w along a circle of radius ‘r’. Let us consider a rectangular Cartesian system of axes in the plane of the circular path of particle and the origin is at the centre of the circular path. 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

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

- 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 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.

### Relation between linear velocity (v) and angular velocity (ω) by calculus method:

- From the equation (3) we get the instantaneous velocity as

- Thus the linear velocity of a particle performing U.C.M. is radius times its angular velocity.

### Angle between linear velocity (v) and radius vector by calculus method:

- From the equations (2) and (3) we have

- Thus the scalar or dot product of velocity of particle performing U.C.M. and the radius vector is zero. Hence the angle between the velocity of particle performing U.C.M. and the radius vector.

### Acceleration of a Body Performing Circular Motion has Two Components:

- The relation between linear velocity and angular velocity In vector form is written as

- Thus the acceleration of the body performing circular motion has two components. one along the radius of the circular path towards the centre and is called centripetal acceleration and other tangential component.

Is it the Maharashtra textbook? If u are making this than pls make it b coz there is no e books for the Maharashtra textbook which helps me to much

Yes Sandeep, it is of Maharashtra textbook. We are also going to make a different section soon.