Two particles A and B move on a circle. Initially particles A and B are diagonally opposite to each other and start with velocities $π$ $r a d / s e c$ , $2 π$ $r a d / s e c$ respectively in the same direction. Particle A has angular acceleration $\frac{π}{2}$ $r a d / s e c^{2}$ and $B$ moves with uniform angular velocity. Find the time after which both the particle $A$ and $B$ will collide. (in seconds)

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answer is 2.

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Detailed Solution

$θ_{A / B} = θ_{A} - θ_{B} = 0 - π = - π$

$ω_{A / B} = ω_{A} - ω_{B} = π - 2 π = - π$

$α_{A / B} = α_{A} - α_{B} = π / 2 - 0 = π / 2$

$θ_{A / B} = ω_{A / B} t + 1 / 2 α_{A / B} t^{2}$

$- π = - π t + \frac{1}{2} \times \frac{π}{2} t^{2} \Rightarrow - 1 = - t + \frac{t^{2}}{4}$

$t^{2} - 4 t + 4 = 0, (t - 2)^{2} = 0 \Rightarrow t = 2 \sec$

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