Question

In the figure shown,a small block of mass m moves in fixed semicircular smooth track of radius R in vertical plane .It is released from the top .The resultant force on the block at the lowest point of track is.
The correct ans is 2mg (explain)

Answer 1

Let us consider a particle of mass M sliding in a hemisphere of radius R. Let the normal reaction be N and the velocity at the given point be v. Resolving Mg along N and balancing with the centripetal force we have:
N−mgcos(θ)=mv2Rat P, θ=0Thus, N − mg = mv2RThus, the net force experienced by the particle is given by the centrifugal forceFNet= mv2Rit is experiencing. Again from the conservation of energy, the velocity at P is given by:  mgR=12mv2=>v=2gR‾‾‾‾√FNet= mv2R=m2gRR=2mg