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Asked by didpayeshivam | 12 Apr, 2024, 11:19: PM

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

Answered by Thiyagarajan K | 15 Apr, 2024, 07:32: AM
NEET neet - Physics
Asked by didpayeshivam | 12 Apr, 2024, 11:19: PM
NEET neet - Physics
Asked by purabidey2001 | 07 Feb, 2024, 07:54: AM
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Asked by roshnibhati921 | 03 Nov, 2023, 01:15: PM
NEET neet - Physics