# 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?

##### 1 Answer

#### Explanation:

The angular acceleration

where

The angular velocity at a time

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

The time at which the particle slips is given by

where

Thus

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

Substituting the values of