Correct Answer - (i) `(3)/(9)` mg N , (ii) `(5 sqrt3)/(8) m//s^(2)`
In the figure `angleCPO = 60^(@) = angle OPC , therefore OP = R`
The extension of spring = R = `(3R)/(4) = (R)/(4)`
`therefore " " F = kx = (mg)/(R) xx (R)/(4) = (mg)/(4)`
(i) Resolving the forces along the radius of semicircle , we have
`N + F "cos" 60^(@) = "mg cos" 60^(@) `
or `" " N + (mg)/(4) "cos"60^(@) = "mg cos" 60^(@)`
`therefore " " N = (3)/(8) mg N`
(ii) Tangential force
`F_(t) = " mg cos" 60^(@) + F "sin" 60^(@)`
=`"mg cos" 60^(@) + (mg)/(4) "sin" 60^(@)`
= `( 5 sqrt3 "mg")/(8)`
Thus tangential acceleration `a_(t) = (F_(t))/(m) = (5 sqrt3 g)/(8) m//s^(2)`