In the figure shown, a small ball of mass 'm' can move witgout sliding in a fixed semicircular track of radius R in vertical planet. It is released from the top. The resultant force on the ball at the lowest of the track is

Options

A. `(10 mg)/(7)`
B. `(17 mg)/(7)`
C. `(3 mg)/(7)`
D. zero

Answer

A

Explanation

From conservation of energy, the kinetic energy of ball at lowest portion is (`v_(c ) =` speed of centre of ball)
`(1)/(2)mv_(c )^(2) + (1)/(2) xx (2)/(5)mv_(c )^(2) = mg R`
or `(7)/(10)mv_(c )^(2) = mgR`
Since net tangential force on sphere at lowest point is zero, net force on sphere at lowest position is
`= (mv_(c )^(2))/(R ) = (10)/(7) mg`