Question

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

Answer 1

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.