Question

starting from rest,a particle moves along a circular path with constant angular acceleration of π/4 rad/se2. the time at which, the angle between velocity v and acceleration a of the particle becomes 45°is
a.2/
$\sqrt{\pi}$
b.4/
$\sqrt{\pi}$


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Answer 1

Answer:

$(a) \frac{2}{\sqrt{\pi}}$

Explanation:

1) We have,

$\alpha =\frac{\pi}{4} rad/s^2$

Angle between velocity and acceleration of particle is $\theta$

This is same as angle between net acceleration and tangential acceleration .

Then,

$tan(\theta)=\frac{a_r}{a_t}$

2) We know,

$v=u+at = 0+r\alpha t$

$tan(45^{\circ})=\frac{v^2/r}{r\alpha } \\ \\=> r\alpha =\frac{(r\alpha t)^2}{r}\\ \\=>t=\frac{1}{\sqrt{\alpha }}=\frac{2}{\sqrt{\pi}}$

Hence, At this time  the angle between velocity v and acceleration a of the particle becomes 45°.