A particle is moving in a circular path of radius 1 m. It starts from rest and its tangential acceleration is 3t. The time after which the acceleration of the particle makes an angle of 300 with radial acceleration is (n)2/3 seconds. Find the value of n.

a cos300=ar and a sin300=at ∴tan300=atar=3tarar=3t or v2r=3t ∴v2=3tBut at=dvdt=3×tor ∫0vdv=3∫0ttdt or v=32t2 t3=4Given t=22/3s n=2

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A particle is moving in a circular path of radius 1 m. It starts from rest and its tangential acceleration is $\sqrt{3} t .$ The time after which the acceleration of the particle makes an angle of $30^{0}$ with radial acceleration is ${(n)}^{2 / 3}$ seconds. Find the value of $n .$

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answer is 2.

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Detailed Solution

$a \cos 30^{0} = a_{r}$ and    $a \sin 30^{0} = a_{t}$
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$∴ \tan 30^{0} = \frac{a_{t}}{a_{r}} = \frac{\sqrt{3} t}{a_{r}}$
$a_{r} = 3 t$ or    $\frac{v^{2}}{r} = 3 t$
$∴ v^{2} = 3 t$
But $a_{t} = \frac{d v}{d t} = \sqrt{3} \times t$
or    $\int_{0}^{v} d v = \sqrt{3} \int_{0}^{t} t d t$ or    $v = \frac{\sqrt{3}}{2} t^{2}$
$t^{3} = 4$
Given $t = 2^{2 / 3} s$
$n = 2$