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:

• A. $\frac{1}{3}rad$
• B. $\frac{2}{3}rad$
• C. $\frac{4}{3}rad$
• D. $2rad$

Answer 1

The correct option is B $\frac{2}{3}rad$

We know

$a_{net}=\sqrt{a_{tangential}^2+a_{centripetal}^2}$

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

given ${\omega }_o=0$ and $a_{tangential}=0.6m/s^2$

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

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

$a_{centripetal}=2a_{tangential}\theta [a_{tangential}=\alpha R]$

$1=\sqrt{0.36+(1.\theta {)}^2}$

$\to (1.2\theta {)}^2=1-0.36=0.64$

$\Rightarrow \theta =\frac{0.8}{1.2}=\frac{2}{3}$ radian