Answer

Verified

388.2k+ views

**Hint**Tangential acceleration can be defined as the rate of change of velocity along the tangent when the body is moving in a circular path. In a way, the tangential acceleration is nothing but linear acceleration that is viewed from a reference point that is tangential to the circle. Since the tangential acceleration is linear acceleration from a reference frame, the equations of motion hold true to a body executing tangential acceleration.

Formulas used: We will be using the formulas from the equations of motion. We have velocity and radius of the circular path, thus we can use the equation, ${v^2} - {u^2} = 2as$ where $v$ is the final velocity of the body, $u$ is the initial velocity of the body, $a$ is the tangential acceleration of the body, and $s$ is the displacement covered by the body.

**Complete step by step answer:**

We know that when a body executes circular motion along a circular path of a fixed radius, then the motion causes a tangential acceleration on the body. This tangential acceleration shows linear characteristics from a reference frame.

Thus, all the equations of motion corresponding to the linear motion will hold true for tangential acceleration. Now from the problem we can infer that the radius of the circular path is $r = \dfrac{{20}}{\pi }$ and the velocity of the particle at the end of the second revolution is $v = 80m/s$ .

Now we know that the circumference of the circle will be the distance travelled by the particle for each revolution. In the second revolution the distance will be, $s = 2\left( {2\pi r} \right)$

Substituting the known values, we get,

$s = 2\left( {2 \times \pi \times \dfrac{{20}}{\pi }} \right)$

$ \Rightarrow s = 80m$

Now that we have the distance travelled, to find the tangential acceleration let us use the equation of motion, ${v^2} - {u^2} = 2as$ .

Let us consider the particle starts from rest, so the initial velocity $u = 0$ .

${\left( {80} \right)^2} - {\left( 0 \right)^2} = 2 \times a \times 80$

$a = \dfrac{{{{\left( {80} \right)}^2}}}{{2 \times 80}}$

Solving for $a$ we get,

$ \Rightarrow a = 40m/{s^2}$

**Thus, the tangential acceleration of the particle during the second revolution is $a = 40m/{s^2}$ .**

**Note:**The tangential acceleration is closely related to angular acceleration and is given by,

${a_T} = \alpha r$ where $\alpha $ is the angular acceleration experienced by the body.

Recently Updated Pages

Which of the following is correct regarding the Indian class 10 social science CBSE

Who was the first sultan of delhi to issue regular class 10 social science CBSE

The Nagarjuna Sagar project was constructed on the class 10 social science CBSE

Which one of the following countries is the largest class 10 social science CBSE

What is Biosphere class 10 social science CBSE

Read the following statement and choose the best possible class 10 social science CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Why is the Earth called a unique planet class 6 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

One cusec is equal to how many liters class 8 maths CBSE