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 it's total acceleration becomes `1m/s^2`, then the angle through which it has turned before slipping is?

Answer 1

`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\ 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^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`