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

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A particle is moving on a circular path of radius 1m. It starts from rest, its tangential acceleration is $\sqrt{3} t$ . The time after which the acceleration of the particle makes an angle of 30⁰ with centripetal acceleration is ${(n)}^{2 / 3} s$ . Find the value of n.

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answer is 2.

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Detailed Solution

$a \cos 30^{0} = a_{t} {and a sin 30}^{0} = a_{t} ∴ \tan 30^{0} = \frac{a_{t}}{a_{t}} = \frac{\sqrt{3} t}{a_{t}} a_{t} = 3 t o r \frac{v^{2}}{1} = 3 t ∴ v^{2} = 3 t$

But $a_{t} = \frac{d v}{d t} = \sqrt{3} t$ or $\int_{0}^{v} d v = \sqrt{3} \int_{0}^{t} t d t or v = \frac{\sqrt{3}}{2} t^{2}$
$∴ \frac{3}{4} t^{4} = 3 t o r t^{3} = 4$
Given
$\begin{array}{l} t = 2^{2 / 3} s \\ ∴ n = 2 \end{array}$