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

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Answer 1 · 2018-04-25T03:17:05

$\frac{2}{3}$ $2/3$ radians

The angular acceleration $α$ $alpha$ is related to the tangential acceleration $a_{t}$ $a_t$ by

$α = \frac{a_{t}}{R}$ $alpha = a_t/R$

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

The angular velocity at a time $t$ $t$ is given by $omega = alpha\ t = a_t/R t$ .

At this instant, the two components of the acceleration are

The magnitude of the acceleration is

$sqrt(a_t^2+a_t^4/R^2t^4)$

The time at which the particle slips is given by

$a_t^2+a_t^4/R^2t^4 = a_0^2$

where $a_0=1\ "ms"^-2$

Thus

$t^4 = (a_0^2-a_t^2)/a_t^4 R^2implies t^2 = sqrt(a_0^2/a_t^2-1) R/a_t$

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

$theta = 1/2 alpha t^2 = 1/2 a_t/R t^2 = 1/2sqrt((a_0/a_t)^2-1)$

Substituting the values of $a_0$ and $a_t$ , we get

$theta = 1/2sqrt((5/3)^2-1)= 1/2times 4/3 = 2/3$

A particle begins to move with a tangential acceleration of constant magnitude 0.6m/s^20.6ms20.6m/s^2 in a circular path. If it slips when it's total acceleration becomes 1m/s^21ms21m/s^2, then the angle through which it has turned before slipping is?