Tangential acceleration of a particle moving in a circle

Asked by admin @ 17/08/2021 in Physics viewed by 286 People

tangential acceleration of the particle moving on a circular path of radius 1 metre varies with time as shown the time after which total acceleration makes an angle of 45 degree with the velocity of the particle is (particle starts from rest)

Answered by admin @ 17/08/2021

Answer:

$(\frac{4\sqrt{3}}{3})^{1/3}$

Explanation:

The equation of the line shown will be

$a_t=\sqrt{3}t$

Radius of the circular path = 1 m

We know that acceleration is rate of change of velocity

Therefore,

$\frac{dv}{dt}=\sqrt{3}t$

or, $dv=\sqrt{3}tdt$

or, $\int_0^v dv=\sqrt{3}\int_0^ttdt$

or. $v=\frac{\sqrt{3}t^2}{2}$

Therefore,

The radial acceleration

$a_r=\frac{v^2}{r}$

$\implies a_r=\frac{(\sqrt{3}t^2/2)^2}{1}$

$\implies a_r=\frac{3t^4}{4}$

The tangential acceleration and the radial acceleration will always be perpendicular

Therefore at any instant if we take the direction of tangential acceleration as x-axis and radial acceleration as y-axis then total acceleration can be written as

$\vec a=a_t\hat i+a_r\hat j$

$\vec a=\sqrt{3}t\hat i+\frac{3t^4}{4}\hat j$

The velocity will always be in the direction of the tangential acceleration

The angle made by the total acceleration with the direction of velocity is given by

$\theta=\tan^{-1}\frac{a_n}{a_r}$

According to the question this angle is 45°

tan45° = 1

Therefore,

$1=\frac{a_n}{a_r}$

or, $a_r=a_t$

or, $\frac{3t^4}{4}=\sqrt{3}t$

$\implies t^3=\frac{4\sqrt{3}}{3}$

$\implies t=(\frac{4\sqrt{3}}{3})^{1/3}$

Hope this helps.

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