
A particle is moving along a circle of radius $\dfrac{{20}}{\pi }$ with constant tangential acceleration. The velocity of the particle is 80m/s at the end of the second revolution after motion has begun. Then tangential acceleration is
Answer
579.6k+ views
Hint: First we need to calculate the distance travelled by the particle in two revolutions. Then that distance and given velocity can be used in one of the equations of motion from which tangential acceleration can be calculated.
Formula used:
Distance covered in one revolution is equal to the circumference of the circle given as
$S = 2\pi r$
The equations of motion are given as
$
v = u + at \\
{v^2} - {u^2} = 2aS \\
S = ut + \dfrac{1}{2}a{t^2} \\
$
Here the symbols have their usual meaning.
Complete step by step answer:
Tangential acceleration is just the linear acceleration of the particle at every point of the trajectory which is given by a tangent drawn to the circle in which the particle is moving.
We are given the radius of this circle as
$r = \dfrac{{20}}{\pi }$
Now the particle makes two revolutions which means that it covers distance which is equal to twice the circumference of the circle of radius r. Therefore, the total distance travelled by the particle after two revolutions is given as
$S = 2 \times 2\pi r = 4\pi r = 4\pi \times \dfrac{{20}}{\pi } = 80m$
We are also given the final velocity of the particle to be
$v = 80m/s$
Now if we consider that the particle has started from rest then the initial velocity is $u = 0$. Now using these values in second equation of motion we get
$
{v^2} = 2aS \\
\Rightarrow a = \dfrac{{{v^2}}}{{2S}} = \dfrac{{{{\left( {80} \right)}^2}}}{{2 \times 80}} = 40m/{s^2} \\
$
This value of acceleration is nothing but the tangential acceleration of the particle. This is the required answer to the question.
Note:
When a particle moves in a circular orbit, the direction of the linear velocity is changing at every point of the trajectory of the particle. Due to this change in velocity the particle is said to be accelerating when it moves in a circular orbit.
Formula used:
Distance covered in one revolution is equal to the circumference of the circle given as
$S = 2\pi r$
The equations of motion are given as
$
v = u + at \\
{v^2} - {u^2} = 2aS \\
S = ut + \dfrac{1}{2}a{t^2} \\
$
Here the symbols have their usual meaning.
Complete step by step answer:
Tangential acceleration is just the linear acceleration of the particle at every point of the trajectory which is given by a tangent drawn to the circle in which the particle is moving.
We are given the radius of this circle as
$r = \dfrac{{20}}{\pi }$
Now the particle makes two revolutions which means that it covers distance which is equal to twice the circumference of the circle of radius r. Therefore, the total distance travelled by the particle after two revolutions is given as
$S = 2 \times 2\pi r = 4\pi r = 4\pi \times \dfrac{{20}}{\pi } = 80m$
We are also given the final velocity of the particle to be
$v = 80m/s$
Now if we consider that the particle has started from rest then the initial velocity is $u = 0$. Now using these values in second equation of motion we get
$
{v^2} = 2aS \\
\Rightarrow a = \dfrac{{{v^2}}}{{2S}} = \dfrac{{{{\left( {80} \right)}^2}}}{{2 \times 80}} = 40m/{s^2} \\
$
This value of acceleration is nothing but the tangential acceleration of the particle. This is the required answer to the question.
Note:
When a particle moves in a circular orbit, the direction of the linear velocity is changing at every point of the trajectory of the particle. Due to this change in velocity the particle is said to be accelerating when it moves in a circular orbit.
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