In this article, we will discuss how to find tangential acceleration from different rotational motions by solving some problems with examples.

**The tangential acceleration is the change in tangential velocity of the object in a circular track which is perpendicular to the centripetal force acting inward.**

**How to find Tangential Acceleration from Angular Velocity?**

On calculating the variation from the alternating angular velocity with time we can find the angular acceleration and thus the tangential acceleration of the object.

**The tangential acceleration can be found out by measuring the difference between the tangential velocity which is equal to the product of the angular velocity and the radius of the circular path covered by the object.**

Consider an object traveling with angular velocity ω in a circular track of radius ‘r’.

Let ‘s’ be displacement of the object in time ‘t’ and ‘θ’ be an angle formed due to the displacement of the object. The following figure represents the same.

By geometry, the arc length ‘s’ of the circle covered by the object in time ‘t’ will be equal to

s = rθ —(1)

Here, in the above equation, there are two variables. The location of the object is displaced in a circular path making an angle ‘θ’ between 0-360^{0}. Hence, we can rewrite the above equation as

Δs=rΔθ

On dividing both the sides by the time variable, we have

Δs/θt=rΔθ / Δt

**The tangential velocity of the object is the displacement of the object between the two intervals of time. That is the same as, v=Δs/Δt}; and the angular displacement of the object with varying time is equal to the angular velocity of the object. Therefore, the previous equation becomes,**

v=rω —(2)

Where v is the tangential velocity

If ω is an angular velocity of the object, then the angular acceleration of the object will be

α =Δω/ Δt—(3)

Where α is the angular acceleration of the object.

The tangential acceleration of the particle is the change in the radial velocity of the object with time varied as the direction of the velocity changes.

The above equation shows the relationship between the tangential acceleration and the angular velocity of the object. The tangential acceleration is equal to the ratio of the change in the angular velocities of the object with time and is directly proportional to the radius of the circular track covered by the object.

**Read more on Angular Acceleration.**

**How to find Tangential Acceleration in Circular Motion?**

The centripetal force keeps the object in a circular motion and the direction of tangential velocity remains perpendicular to this force.

**The tangential acceleration in a circular motion is the change in the velocity caused due to the changing direction of the angular acceleration of the object.**

Consider an object moving in a circular motion experiencing a force equal to the centripetal force.

F=F_{c}

Let ‘r’ be the radius of the circle and v and a be the radial velocity and the radial acceleration respectively.

ma=mv^{2}/r

a=v^{2}/r —(6)

On substituting eq(2) in the above equation, we have

a=rω^{2} —(7)

This is the radial acceleration of the object, and the tangential acceleration of the object will be

a_{t}=dv/dt=rdω/dt=rα

Where ω is the angular velocity and

α is the angular acceleration of the object

**The total acceleration of the object in a circular motion will be the vector sum of the tangential acceleration and the radial acceleration.**

And also,

Therefore,

**How to find Tangential Acceleration without Time?**

The tangential acceleration relies upon the angular acceleration of the object.

**The tangential acceleration is the ratio of variation of the tangential velocity roused due to the changing direction of the path of an object along with time.**

Referring to the above equation No. (5), we can write

a_{t}=rα—(10)

This equation shows the relation between the tangential acceleration and the angular acceleration of the object independent of time.

**How to find Tangential Acceleration of a Pendulum?**

The pendulum oscillates in the harmonic motion making an angle θ along the string length.

**The restoring force acting on the string brings back the pendulum to its original position which is acting tangential to the arc. Based on it we can find the tangential acceleration of a pendulum.**

Consider a pendulum in a SHM. A string of length ‘L’ is attached to a bob of mass ‘m’. Let ‘s’ be the displacement of the bob due to harmonic motion.

**A force equal to mgCosθ is acting along the length of the string which is canceled by the tension across the string. The restoring force acting on the bob is given by**

F=-mgSinθ

ma=-mgSinθ

For small angles

a = gθ

The angle θ can be calculated by measuring the arc length and dividing by the length of the string.

θ = s/L

Therefore the above equation becomes

a=gs/L

**The tangential acceleration of the pendulum is equal to the acceleration due to gravity and displacement of the bob by the length of the string.**

**How to find Tangential Acceleration given Time?**

Tangential velocity will escalate with time if the rate of tangential acceleration is positive.

**The tangential acceleration can be calculated by finding the difference in the radial velocity of the object which is obviously varying as the direction of the object with angular velocity keeps on changing with time.**

This is given by the formula

a_{t}=dv_{t}/dt

Where v_{t} is a radial velocity

**Read more on How To Find Acceleration In Velocity Time Graph: Problems And Examples.**

**Frequently Asked Questions**

**Problem 1: An object is accelerating in a circular motion around a circle of radius 10m. The angular velocity of the object increases to 6m/s from 4m/s between the time intervals of 4 seconds. Calculate the tangential acceleration of the object.**

**Given:** r=10m

ω_{1}=4m/s

ω_{2}=6m/s

Δt=4 sec

Therefore the tangential acceleration is

=10 x (6-4)/4

=10 x 2/4=5 m/s^{2}

Hence the tangential acceleration of the object is 5m/s^{2}.

**Problem 2: Calculate the tangential acceleration and angular acceleration of a ball traveling in a circular path of radius 5 meters with the velocities between 2m/s to 4m/s in 4 seconds.**

**Given:** V_{1}=2m/s

V_{2}=4m/s

T=4s

R=5m

The tangential acceleration is

Since, a_{t}=rα

α =a_{t}/r

α = 0.5m/s^{2}/5m=0.1radians/s^{2}

The angular acceleration of a ball is 0.1radians/s^{2}.