Question

A particle begins to move with a tangential acceleration of constant magnitude 0.6 m/s2 in a circular path. If it slips when its total acceleration becomes 1 m/s2 , then the angle through which it would have turned before it starts to slip is

Answer 1

When net acceleration=1

Tangential acceleration=0.6 and centripetal =0.8

Now w^2R=0.8 

So w^2=0.8/r

 

Tagential acceleration=0.6=rα 

So α=0.6/r

Now as acceleration is constant, we can use

w^2=2α theta 

You can get theta from here which equals 2/3rad