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