A particle begins to move with a tangential acceleration of constant magnitude 0.6m/s^2 in a circular path. If it slips when it's total acceleration becomes 1m/s^2, then the angle through which it has turned before slipping is?

Apr 25, 2018

$\frac{2}{3}$ radians

Explanation:

The angular acceleration $\alpha$ is related to the tangential acceleration ${a}_{t}$ by

$\alpha = {a}_{t} / R$

where $R$ is the radius of the circular orbit.

The angular velocity at a time $t$ is given by $\omega = \alpha \setminus t = {a}_{t} / R t$.

At this instant, the two components of the acceleration are

• the tangential acceleration ${a}_{t}$
• the centripetal acceleration ${\omega}^{2} R = {a}_{t}^{2} / R {t}^{2}$

The magnitude of the acceleration is

$\sqrt{{a}_{t}^{2} + {a}_{t}^{4} / {R}^{2} {t}^{4}}$

The time at which the particle slips is given by

${a}_{t}^{2} + {a}_{t}^{4} / {R}^{2} {t}^{4} = {a}_{0}^{2}$

where ${a}_{0} = 1 \setminus {\text{ms}}^{-} 2$

Thus

${t}^{4} = \frac{{a}_{0}^{2} - {a}_{t}^{2}}{a} _ {t}^{4} {R}^{2} \implies {t}^{2} = \sqrt{{a}_{0}^{2} / {a}_{t}^{2} - 1} \frac{R}{a} _ t$

and the angle through which the particle has turned at this time is

$\theta = \frac{1}{2} \alpha {t}^{2} = \frac{1}{2} {a}_{t} / R {t}^{2} = \frac{1}{2} \sqrt{{\left({a}_{0} / {a}_{t}\right)}^{2} - 1}$

Substituting the values of ${a}_{0}$ and ${a}_{t}$, we get

$\theta = \frac{1}{2} \sqrt{{\left(\frac{5}{3}\right)}^{2} - 1} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$