NEET Class neet Answered
when the block is at the top point D of loop , we have
First term in RHS of above expression is weight " mg " and N is normal force .
LHS of above expression is centripetal force required for circular motion .
v is speed of block at D and r is radius of loop .
Minimum speed v is determined by equating normal force N = 0 .
hence we get ,
Hence total mechanical energy of block at D
( Kinetic energy + potential energy) ≥ [ (1/2) m vmin2 ] + [ m g (2r) ]
total mechanical energy of block at D ≥ [ (1/2) m g r] + [ m g (2r) ]
total mechanical energy of block at D ≥ [ m g (5/2) r ]
At the point A, where the block is released , by conservation of energy,
total mechanical energy ≥ [ m g (5/2) r ]
At the point A, total mechanical energy is only due to potential energy , we get
m g h ≥ [ m g (5/2) r ] or
h ≥ (5/2) r