Centripetal Acceleration Derivation Explained Step-by-Step

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 centre of the circle.

Centripetal Acceleration

The property of a moving body traversing a circular path is called centripetal acceleration. This acceleration is directed radially in the circular path's centre. The magnitude of centripetal acceleration is directly proportional to the square of the body's speed along the curve divided by the distance between the moving body and the circle's centre. Centripetal force is the force that creates centripetal acceleration and it is directed towards the centre of this circular route.

When an object follows a circular course, its trajectory alters at every point along the way. An object might feel centripetal acceleration even when tracing an arc or a circle at a constant velocity.

There are two perpendicular accelerations at each point of the circular path the object is moving: centripetal acceleration directed towards the body or inward acceleration and tangential acceleration directed directly along the tangent of the circular path.

According to Newton's law, a body in motion has its acceleration in the same direction as the force exerted to move it.

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 centre O. 

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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 \[\omega = \frac{\Delta \theta}{\Delta t} \]…(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

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,  \[\Delta \theta = \frac{ab}{O’a} = \frac{| \Delta v |}{| v |}  \]  

From eq(1), 

\[\omega \Delta t = \frac{| \Delta v |}{| v |} \]   

\[\frac{\Delta v}{\Delta t} = \omega | v | \]......(3)

Since, v = rω putting in eq(3)

= (rω)r = ωr² …(4)

When  ∆t -> 0, 

\[\frac{\Delta v}{\Delta t} = \text{magnitude of centripetal acceleration,} \] |a| at P given by,

\[|a| = \frac{|\Delta v|}{\Delta t} = ωr² = \frac{v}{\frac{r^2}{r}} \]

Thus

\[|a| = \frac{v}{r^2} \]               

Centripetal Acceleration Formula Proof

Consider a particle traversing a circular path of radius r with centre C. 

The Initial 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:

\[\frac{dv}{dt} = \omega \frac{dr}{dt} + r \frac{d \omega}{dt} \]......(b)

Here,\[\frac{dv}{dt} = a\] (resultant acceleration of particle at P)....(c) 

\[\frac{d \omega}{dt}  =  \alpha \] (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.

\[\frac{dr}{dt} = v\] (linear velocity at P)....(e)

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

\[\overline{a} = \overline{\omega} \times \overline{v} + \overline{\alpha} \times \overline{r} \].....(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 ω Sin 90° =vω 

Putting v = rω, we get,

Centripetal acceleration, \[a(c) = \omega ^2 r = \frac{v^2}{r} \]

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 

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,  \[α  = \frac{dω}{dt} = 0\]

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 centre 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.

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∆AOB and ∆FDG

AO = DF, OB = DG and  ∠AOB = ∠FDG = X

Hence,  ∆AOB ～  ∆FDG 

So, \[\frac{\text{change in velocity}}{v} = \frac{AB}{R}\]

Acceleration, \[a = \frac{\text{change in velocity}}{\text{time taken A to B}}\]

= AB x v = R x time to A to B = \[\frac{v²}{r}\]

Using the relation, F = ma, we get,

Fc = \[\frac{mv^2}{r} = mr\omega ^2\]

This expression is for the centripetal force.

Conclusion:

Haven’t we already established that it is important for the Earth to revolve around the Sun? Well, being able to understand what causes that movement is equally important. This article introduces you to Centripetal Acceleration and its derivation. Go through it thoroughly for a better understanding.