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 $1m/s^2$, Find the angle through which it would have turned before it starts to slip.

• A. $\frac{2}{3}$ radian
• B. $\frac{2}{3}$ degree
• C. $\frac{4}{3}$ radian
• D. $\frac{4}{3}$ degree

Answer 1

The correct option is A $\frac{2}{3}$ radian

Answer is A.
Here, $a_{net}=\sqrt{{a_t}^2+{a_c}^2}\Rightarrow {\omega }^2={{\omega }_0}^2+2a\theta$
Therefore, ${\omega }_0=0,so{\omega }^2=2a\theta$.
${\omega }^{2}R=2\left(aR\theta \right)$
$a_c={\omega }^{2}R=2a_t\theta$
$1=\sqrt{0.36+{(1.2\times \theta )}^2}\Rightarrow 1-0.36={\left(1.2\theta \right)}^2$
$\Rightarrow \frac{0.8}{1.2}=\theta \Rightarrow \theta =\frac{2}{3}rad$
Hence, the angle through which it would have turned before it starts to slip is 2/3 rad.