 # Derivation of Centripetal Acceleration  View Notes

Acceleration that acts on the object in a circular motion is called the centripetal acceleration. It is a property of the motion of the body traversing a circular path. It acts radially towards the center of the circle.

Consider a particle of mass, ‘m’ moving with a constant speed, ‘v’’ having a uniform angular velocity, ‘ω,’  around a circular path of radius, ‘r’ with center O.

Let on time t, the particle be at point P where OP = r1 and on time  t + ∆t, the particle is at Q i.e., OQ = r2.

Where ∠POQ equals to  ∆Ө (∆Ө  = arc traced by the particle from P to Q or simply angular displacement)

Since, |r1| = |r2| = r

Angular velocity ω =  ∆Ө/∆t…(1)

Let v1 and v2 be the velocity vectors at P and Q respectively.

So, velocity along the tangent to the circular path  at a location is represented by the tangents:

|AL| = |AM| = |v|

Now the change in velocity with time from t to t + ∆t is represented by,

|O’a| = v1 and |O’b| =v2

Clearly, ∠aO’b = Ө

Applying  ∆ law of vectors:

O’a + ab  = O’b or ab = O’b - O’a  = v2 - v1 =  ∆v

At  ∆t -> 0,  a lies close to b.

Now taking arc ab = radius r.

Then O’a = |v|......(2)

Therefore,   ∆Ө = ab/ O’a = |∆v| / |v|

From eq(1),

ω ∆t =  |∆v| / |v|

|∆v|/∆t  = ω|v|...(3)

Since, v = rω putting in eq(3)

= (rω)r = ωr² …(4)

When  ∆t->0,

|∆v|/∆t = magnitude of centripetal acceleration, |a| at P given by,

|a| =   |∆v|/∆t = ωr² = $\frac{v}{r^{2}/r}$

Thus

 |a| = $\frac{v^{2}}{r}$

### Centripetal Acceleration Formula Proof

Consider a particle traversing a circular path of radius r with center O.

Initially particle is at P with linear velocity v and angular velocity ω.

Since,   v = r x ω …(a)

Differentiating both the sides w.r.t:

dv/dt  = ω dr/dt  +  r dω/dt…(b)

Here, dv/dt = a (resultant acceleration of particle at P)....(c)

dω/dt  =  α (angular acceleration at P)....(d)

Where, angular acceleration is the time rate of change in the angular velocity of an object traversing a circular path.

dr/dt = v (linear velocity at P)....(e)

Putting values of (c), (d), (e) in (b)

 a  = ω x v + r x α…(f)

Here, we can see the resultant acceleration has two components:

(i) ω x v  and (ii) r x α

|a(c)| = |v x ω|

a(c) =  radial or centripetal acceleration.

Both are perpendicular to each other.

|a(c)| = |v x ω| = v ωSin 90° =vω

Putting v = rω, we get,

 Centripetal acceleration, a(c) = ω²r = v²/r

Derive an expression for centripetal acceleration in uniform circular motion

As we know that resultant acceleration of the particle at P is given by,

a  = ω x v + r x α

Where the component, a(t)  = r x α

r x α,

When r and α are perpendicular to each other , then,

 a(t) =  r x α =  r α Sin 90° = r α

a(t) is called the tangential acceleration acting along the tangent to the circular path at point P.

Since, In case of uniform circular motion, the object moves with a constant speed (v),

Therefore,  α  = dω/dt = 0

So a(t) = 0

While a(c) ≠ 0

Thus in a circular motion, only centripetal acceleration acts on the body which is given by,

 a(c) = ω x v

This expression can be zeroed when ω = 0.

This is possible only when a particle moves in a straight line.

### Centrifugal Acceleration

In Newtonian mechanics, a kind of fictitious acceleration (appears to) acts in a body having a circular motion. It is always directed away from the center around which the body moves.

### Centripetal Force Derivation

The force that acts on a body moving in a circular path and is directed towards the centre around which the body is moving is called the centripetal force.

The circle represents the orbit of any satellite of radius R moving from point A to B with speed v in time t.

Now, draw vector AP to represent the initial velocity of the satellite at A, which is along a tangent at A, and second vector, BQ, to represent new velocity at B.

Redraw the initial and new vectors, both starting from the same point D.

They both have a magnitude equal to v.

FG represents the change in velocity, and must be added to the old velocity v to generate a new velocity, having the same magnitude i.e., v.

From Fig.3,

∆AOB and ∆FDG

AO = DF, OB = DG and  ∠AOB = ∠FDG = X

Hence,  ∆AOB ～  ∆FDG

So, change in velocity/v = AB/ R

Acceleration, a = change in velocity/ time taken A to B

= AB x v = R x time to A to B = v²/r

Using the relation, F = ma, we get,

 Fc = mv²/r = mrω

This expression is for the centripetal force.

Q1: What Causes Centripetal Acceleration?

Ans: Centripetal force causes centripetal acceleration. This force causes the circular motion of the Earth around the Sun. Therefore, it results in gravitational attraction between these two bodies.

Q2: Is Gravity a Centripetal Force?

Ans: Yes, gravity is a centripetal force, and this force is directed towards the center. Earth has gravity. It attracts all bodies towards its center.

Q3: Why is There no Work Done in Circular Motion?

Ans:  In a circular motion, the direction of velocity varies with time and force acts orthogonal to the direction of motion. Here, the body moves with a constant speed. Therefore, no work is done and the energy remains constant.

Q4: How is Centripetal Force Used in Everyday Life?

Ans: There are certain everyday applications of centripetal force as discussed here below:

1. The normal force and the static friction both acting on the tires of the car keeps the car in a circular motion.

2. Standing still: We are in constant motion with the rotation of the earth.

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