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Tangential Acceleration Formula

In the rotational motion of any object, tangential acceleration is the measure of how quickly a tangential velocity changes. Here tangential velocity will work in the direction of a tangent at the point of motion. Therefore it always acts in the perpendicular direction to the centripetal acceleration of a rotating object. It will be equal to the product of angular acceleration and the radius of the rotation. In this topic, we will discuss the Tangential Acceleration Formula with examples. Let us learn it!

Concept of Tangential Acceleration

Tangential acceleration is similar to the linear acceleration, but it is specific to the tangential direction. This is related to circular motion. Therefore, the rate of change of the tangential velocity of a particle in a circular orbit is known as Tangential acceleration. It always directs towards the tangent to the path of the body.

Tangential acceleration will work if an object is moving in a circular path. Tangential acceleration is like linear acceleration, but it’s different from the straight-line linear acceleration. An object is linearly accelerating if it’s traveling in a straight-line path.

For example, a car accelerating around a curve in the road. The car is accelerating tangentially to the curve of its path. Using the car’s motion we will investigate tangential acceleration a bit more.

Tangential Acceleration Formula

Source:en.wikipedia.org

The Formula for Tangential Acceleration

tangential acceleration = (radius of the rotation) × (angular acceleration)

i.e. \(a_{t}=\frac{\Delta v}{\Delta t}\)

Tangential Acceleration Formula In Terms Of Distance

\(a_{t}=\frac{\mathrm{d}^{2} s}{\mathrm{d} t^{2}}\)

or

\(a_{t}=v.\frac{dv}{ds}\)

Where,

\(a_{t}\) Tangential acceleration
\(\Delta v  or dv\) Change in velocity
\(\Delta t or dt\) Change in time
ds Change in distance covered
v linear velocity
t time taken

Tangential acceleration formula is used to compute the tangential acceleration and the parameters related to it. It is expressed in meter per sec square.

The value of the tangential acceleration may have the following possibilities:

  • Greater than zero: When the body has accelerated motion, that is, the magnitude of the velocity vector increases with time
  • Less than zero: When the body has slowed or decelerated motion, that is, the magnitude of the velocity vector decreases with time
  • Equal to zero: When the body has uniform motion, that is, the magnitude of the velocity vector remains constant

Solved Examples for Tangential Acceleration Formula

Q.1: A body accelerates uniformly on a circular path with a speed of \(30 ms^{-1} to 70ms^{-1}\) in 10 s. Calculate its tangential acceleration.

Solution: Given values are:

Initial velocity \(v_i = 30\) m/s,

Final velocity \(v_f = 70\) m/s

So, change in velocity dv = vf – vi = 70 – 30 = 40 m/s

Also, time taken dt = \(t_f – t_i = 10 – 0 = 10s\)

Thus, the tangential acceleration is given by:

\(a_{t} =\frac{\Delta v}{\Delta t}\)

\(a_{t} = \frac{40}{10}\)

\(= 4 ms^{-2}\)

So, tangential acceleration = \(4 ms^{-2}\)

Q.2 : A runner starts from rest and accelerates at a uniform rate \(20ms^{-1}\) in the time interval of 10s, running on a circular track of radius 50m. Find out the tangential acceleration.

Solution: Given values are as

Initial velocity \(v_i = 0\)

Final velocity \(v_f = 20 ms^{-1}\)

Change in velocity \(dv = v_f – v_i = 20 – 0 = 20 ms^{-1}\)

Time taken dt = 10 s

Thus, the tangential acceleration will be,

\(a_t  = \frac{dv}{dt}\)

\(a_t  = \frac{20}{10} = 2 ms^{-2}.\)

So, tangential acceleration = \(2 ms^{-2}\)

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