Tangential Acceleration Formula: Overview, Formula, Direction

Tangential Acceleration Formula: In a circular motion, a particle may speed up or slow down or move with constant speed. When the particle is in a circular motion, it will always have an acceleration toward the centre called centripetal acceleration (even if moving with constant speed). But in the case of changing speed, tangential acceleration in the direction or opposite of the direction of the velocity will act.

For example, a car is accelerating around a curve path on the road, then it experiences both tangential and centripetal acceleration. Do you know what tangential acceleration is in a circular motion and how it works? In this article, we will find the expression of the tangential acceleration, learn formula for acceleration, and more. We will also find the net acceleration in a non-uniform circular motion with the help of centripetal and tangential acceleration. Scroll down to find more!

What is the Formula of Acceleration? An Overview

In a uniform circular motion, the net force acting on the object is in the direction perpendicular to the motion of the object. Hence, this causes a change in the direction continually, but the magnitude of velocity remains constant. Therefore, the object is said to be accelerating in a direction that points towards the centre of the circular pathway. But what happens when the net force acting on the object is not perpendicular? In this case, there will be two-component force vectors that will point along the perpendicular and parallel to the velocity vector.

The perpendicular force component will cause the object to move along a circular pathway as it creates a centripetal acceleration, and the parallel force component will cause the object to accelerate along the tangent as it creates tangential acceleration. Hence, the object will undergo non-uniform circular motion as both the direction and magnitude of the velocity of the object will change.

Formula for Tangential Acceleration

Tangential acceleration is the measure of how quickly the speed of a body changes when an object moves in a circular motion. Let us consider a particle \((P)\) that is moving in a circle of radius \((r)\) and centre \(O,\) as shown in the figure below. The position of the particle \(P\) at a given instant may be described by the angle \(\theta\) between \(OP\) and \(OX.\) This angle \(\theta\) is called the angular position of the particle with respect to \(OX,\) and it changes as the particle moves on the circle. Let us assume the point rotates an angle \(∆θ\) in the time interval of \(∆t.\) The rate of change of angular position is known as the angular velocity \(\left( \omega \right).\)

Thus, we can write,

\(\omega = \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta \theta }}{{\Delta t}} = \frac{{{\text{d}}\theta }}{{{\text{d}}t}}\)

Here \(\omega \) is the angular speed or magnitude of angular velocity. Angular velocity is a vector quantity. The direction of \(\omega \) is perpendicular to the plane of the circle, and it can be given by the screw law or right-hand thumb rule.

We can also write linear velocity as:

\(\vec v = \frac{{d\vec s}}{{dt}}\)  or \(\frac{{d\vec r}}{{dt}}\)

And, the magnitude of linear velocity is called linear speed \((v).\) Thus, we can write

\(v = |\vec v| = \mid \frac{{d\vec s}}{{dt}}\;{\rm{or}}\;\frac{{d \vec r}}{{dt}}\mid \)

Relation Between Linear Speed and Angular Speed

From the above figure, linear distance \(PP’\) travelled by the particle in time \(∆t\) is

\(\Delta s = r\Delta \theta \)

\( \Rightarrow \frac{{\Delta s}}{{\Delta t}} = r\frac{{\Delta \theta }}{{\Delta t}}\)

\( \Rightarrow \frac{{ds}}{{dt}} = r\frac{{d\theta }}{{dt}}\)

\( \Rightarrow v = r\omega ……..\left( 1 \right)\)

Like the velocity, a particle in circular motion has two accelerations that are angular and linear acceleration. Where angular acceleration \(\left( \alpha \right)\) is the rate of change of angular velocity. Thus, we can write that

\(\alpha = \frac{{d\omega }}{{dt}} = \frac{{{d^2}\theta }}{{d{t^2}}}\)

Angular acceleration \(\left( \alpha \right)\) is also a vector quantity. The direction of \(\left( \alpha \right)\) is also perpendicular to the plane of the circle, either parallel or antiparallel to \(\omega .\) If the angular speed of the particle is increasing, then \(\left( \alpha \right)\) is parallel \(\omega \) to and if angular speed is decreasing, then \(\left( \alpha \right)\) is antiparallel to \(\omega .\) If angular speed (or angular velocity) is constant then, angular acceleration will be zero.

Relation between the tangential component of the linear acceleration and angular acceleration can be obtained on differentiating equation \((1)\) with respect to time. The rate of change of speed is

\({a_t} = \frac{{d|\vec v|}}{{dt}}\)

\({a_t} = \frac{{d(r\omega )}}{{dt}} = r\frac{{d\omega }}{{dt}}\)

\({a_t} = r\alpha ……..\left( 2 \right)\)

Where, 

\(r=\)The radius of the circle,

\(\alpha = \)Angular acceleration.

This component of acceleration in the tangential direction is called tangential acceleration \(\left( {{a_t}} \right).\) This component is responsible for the change in the linear speed. If the speed of the particle is constant, then \({a_t}\) is zero. If speed is increasing, then this is positive and in the direction of linear velocity. This component will be negative and in the opposite direction of linear velocity if the speed is decreasing.

Centripetal acceleration: The component of acceleration in the radial direction (towards the centre) is called radial or centripetal acceleration. This component changes the direction of the linear velocity. As the direction continuously keeps on changing. So, this component can never be zero, and the value of this component is given as:

\({a_r} = \frac{{{v^2}}}{r} = r{\omega ^2}\)

Net acceleration: Tangential acceleration is in the direction of the tangent to the circle, whereas centripetal acceleration is in the radial direction of the circle pointing inwards to the centre. These two components are mutually perpendicular, as shown in the figure below. Thus, a particle in a circular motion having centripetal acceleration as well as tangential acceleration has a net acceleration equal to their vector sum, which is given as:

\(\overrightarrow {{a_n}} = \overrightarrow {{a_c}} + \overrightarrow {{a_t}} \)

\( \Rightarrow a = \sqrt {a_t^2 + a_r^2} \)

\( \Rightarrow a = \sqrt {{{\left( {\frac{{dv}}{{dt}}} \right)}^2} + {{\left( {\frac{{{v^2}}}{r}} \right)}^2}} \)

\( \Rightarrow a = \sqrt {{{(r\alpha )}^2} + {{\left( {r{\omega ^2}} \right)}^2}} \)

And,

\( \tan \theta = \frac{{{a_r}}}{{{a_t}}}\)

Or,

\( \Rightarrow \theta = \tan^{-1} \left( {\frac{{{a_r}}}{{{a_t}}}} \right)\)

Solved Examples on Tangential Acceleration Formula

Q.1. Find the angular acceleration of the particle if a particle travels in a circular path of radius \({\rm{20}}\,{\rm{cm}}\) at a speed that uniformly goes increases? If the speed changes from \({\rm{5}}{\rm{.0}}\,{\rm{m}}\,{{\rm{s}}^{ – 1}}\) to \(6.0\;{\rm{m}}\,{{\rm{s}}^{ – 1}}\) in \({\rm{2}}{\rm{.0}}\,{\rm{s}}{\rm{.}}\)
Ans: The magnitude of the tangential acceleration is given by
\({a_t} = \frac{\alpha }{{dt}} = \frac{{{v_2} – {v_1}}}{{{t_2} – {t_1}}}\)
\( \Rightarrow {a_t} = \frac{{6 – 5}}{2} = 0.5\;{\rm{m}}\,{{\rm{s}}^{ – 2}}\)
The angular acceleration formula is given by,
\(\alpha = \frac{{{a_t}}}{r}\)
\( \Rightarrow \alpha {\rm{ = }}\frac{{{\rm{0}}{\rm{.5}}\,{\rm{m}}\,{{\rm{s}}^{{\rm{ – 2}}}}}}{{{20\,\rm{cm}}}}\)
\( \Rightarrow \alpha = 2.5\,{\rm{rad}}\,{{\rm{s}}^{ – 2}}.\)

Q.2. A particle moves in a circular path of radius \({\rm{2}}\,{\rm{m}}{\rm{.}}\) It is moving with its linear speed that is given by \(v = {t^2},\) where \(t\) in second and \(v\) in \({\rm{m}}\,{{\rm{s}}^{{\rm{ – 1}}}}.\) What is the radial, tangential and net acceleration of a particle at \(t{\rm{ = 2}}\,{\rm{s}}\)?
Ans: The linear speed of the particle at \(t{\rm{ = 2}}\,{\rm{s}}\) is,
\(v = {t^2} = {2^2}{\rm{m}}\,{{\rm{s}}^{{\rm{ – 1}}}}\)
\( \Rightarrow v = 4\,{\rm{m}}\,{{\rm{s}}^{{\rm{ – 1}}}}\)
Then, the radial acceleration at \(t = 2\,{\rm{s}}\) is,
\({a_r} = \frac{{{\nu ^2}}}{r} = \frac{{{4^2}}}{2}{\rm{m}}\,{{\rm{s}}^{{\rm{ – 2}}}}\)
\( \Rightarrow {a_r} = 8\,{\rm{m}}\,{{\rm{s}}^{{\rm{ – 2}}}}\)
The tangential acceleration at \(t = 2\,{\rm{s}}\) is,
\({a_t} = \frac{{dv}}{{dt}} = \frac{{d\left( {{t^2}} \right)}}{{dt}} = 2t\)
\( \Rightarrow {a_t} = 2 \times 2\;{\rm{m}}\,{{\rm{s}}^{ – 2}}\)
\( \Rightarrow {a_t} = 4\,{\rm{m}}\,{{\rm{s}}^{{\rm{ – 2}}}}\)
Now, the net acceleration of the particle at \(t = 2\,{\rm{s}}\) is
\(a = \sqrt {a_t^2 + a_r^2} \)
\( \Rightarrow a = \sqrt {{4^2} + {8^2}} \)
\( \Rightarrow a = 8.944\;{\rm{m}}\,{{\rm{s}}^{ – 2}}.\)

Summary

In a uniform circular motion, the particle executing circular motion has a constant speed, and the circle is at a fixed radius, but the speed of the particle is changing (not constant) then, there will be an additional acceleration that is tangential acceleration which acts in the direction tangential to the circle. Tangential acceleration is defined as the rate of change of magnitude of tangential velocity of the particle in a circular motion, and its direction is in the direction of the tangent to the circular path.

The tangential acceleration formula is given as, \({a_t} = r\alpha .\) The tangential acceleration vector and centripetal acceleration vector are mutually perpendicular. The vector sum of tangential and centripetal accelerations gives the net acceleration of the particle. i.e \(\left( {\vec a = \overrightarrow {{a_r}} + \overrightarrow {{a_t}} } \right).\)

FAQs on Tangential Acceleration Formula

Students can check the below frequently asked questions on tangential acceleration formula:

Q1. What do you mean by tangential acceleration?
Ans: The rate of change speed of the particle in the circular path is known as tangential acceleration. It is equal to the product of angular acceleration \({\left( \alpha \right)}\) and the radius \((r)\) of the circular path. i.e \(\left( {{a_t} = r\alpha } \right).\)

Q2. What is the formula of centripetal acceleration and tangential acceleration?
Ans: (i) We can find tangential acceleration with the help of the tangential acceleration formula, which is given as:
\({a_t} = \frac{{dv}}{{dt}}\) or \({a_t} = r\alpha \)
(ii) And, we can find centripetal acceleration with the help of the centripetal acceleration formula, which is given as:
\({a_r} = \frac{{{v^2}}}{r} = r{\omega ^2}.\)

Q3. In which direction the tangential acceleration works?
Ans: A tangential acceleration works in the direction of a tangent at the point of circular motion. Its direction is always in the perpendicular direction to the centripetal acceleration of a rotating object.

Q4. What force causes tangential acceleration?
Ans: The tangential force component will create tangential acceleration, which will cause the object to accelerate along the tangent. Then, the object will undergo non-uniform circular motion as both the direction and magnitude of the velocity of the object change.

Q5. Give an example of both centripetal and tangential acceleration.
Ans: Suppose you are holding a thread to the end of which is tied to a stone. Now when you start whirling it around, you will notice that two forces have to be applied simultaneously. One which pulls the thread inwards and the other which throws it sideways or tangentially. Both these forces will generate their respective accelerations. The one-pointed inwards will generate centripetal or radial acceleration, and the one pointing sideways will generate tangential acceleration.

Q6. What is centripetal acceleration?
Ans: Centripetal acceleration can be defined as the component of acceleration in the radial direction (towards the centre).

Q7. What is the difference between centripetal and tangential acceleration?
Ans: Tangential acceleration is in the direction of the tangent to the circle, whereas centripetal acceleration is in the radial direction of the circle pointing inwards to the centre.

We hope you find this article on the Tangential Acceleration Formula helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them.