Question

A particle starts from rest and moves on a circular path with constant tangential acceleration of $0.6{\text{m/s}}^2$. If the particle slips when its total acceleration becomes $1{\text{m/s}}^2$, then the angle moved by it before it starts slipping is

• A. $\frac{2}{9}\text{rad}$
• B. $\frac{2}{3}\text{rad}$
• C. $\frac{2}{5}\text{rad}$
• D. $\frac{2}{7}\text{rad}$

Answer 1

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

We know that total acceleration,
$a=\sqrt{(a_t{)}^2+(a_c{)}^2}$, here, $a_t$ is tangential acceleration and $a_c$ is centripetal acceleration.
$\Rightarrow 1=\sqrt{(0.6{)}^2+(a_c{)}^2}$
$\Rightarrow a_c=0.8{\text{m/s}}^2$
Now, on applying equation of motion with constant angular acceleration,
${\omega }_f^2-{\omega }_i^2=2\alpha \theta$
$\Rightarrow {\omega }^2-0=2\times \frac{a_t}{R}\times \theta$
$[R$ is radius of the circular path and $a_t=\alpha R]$
$\Rightarrow {\omega }^{2}R=2a_t\theta$
$\Rightarrow a_c=2a_t\theta$
$\Rightarrow \theta =\frac{a_c}{2a_t}=\frac{0.8}{2\times 0.6}=\frac{2}{3}\text{rad}$