In the figure shown,a small block of mass m moves in fixed semicircular smooth track of radius R in vertical - Physics - Laws Of Motion

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)

Somnath Mazumder · Accepted Answer

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"https://s3mn%20mnimgs%20com/img/shared/ck-files/ck_5661dbfd26198%20png"
style="width: 427px; height: 216px;">
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=2gRFNet= mv2R=m2gRR=2mg

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)

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)