Piyush Kumar Behera
Last Activity: 8 Years ago
mass = 0.1 kg
drawing the fbd at that instant,we get the following equation,
T-mgcos30=mv2/r
because this only provides the centripetal force.
and here v is the instantaneous speed of the particle =1m/s and r=1
So T=mv2/r+mgcos30=0.1N+0.866N=0.966N (substituting the values)
to complete the loop it must have a total mechanical energy of 2.5mgr (reference for potential enrgy is the lowest point in the circle,the minimum velocity is ) total mechanical enrgy at that point is m/2+1.34m=0.84m (taking r=1 and m the mass)
so it cannot complete the circle.