Question

A particle begins to move with a tangential acceleration of constant magnitude $0.6m/s^2$ in a circular path. If it slips when its total acceleration becomes $1m/s^2$, then the angle through which it would have turned before it starts to slip is $\frac{a}{b}$radian. Find$\left(ab\right)$.

Answer 1

Find centripetal acceleration

Formula used: $a_c={\omega }^{2}r,a_t=\alpha r$

Given: $\alpha =1\frac{m}{s^2},a_t=0.6m/s^2$

${\omega }^2-{\omega }_o^2=2\alpha \theta$

${\omega }^2=2\alpha \theta \dots \dots$

(particle starts from rest)

$a_c={\omega }^{2}r$

$=2\alpha \theta r=2a_t\theta$

Find angular displacement

Formula used: $a=\sqrt{a_c^2+a_t^2}$

$1=\sqrt{{(2\times 0.6\times \theta )}^2+{\left(0.6\right)}^2}$

$1-0.36=(1.2\theta {)}^2$

$\theta =\frac{0.8}{1.2}=\frac{2}{3}rad$

$a=2$ and $b=3$

so $ab=6$

Final answer: $6$