A uniform slender rod of mass `m` and length `L` is released from rest, with its lower end touching a frictionaless horizontal floor. At the initial moment, the rod is inclined at an angle `theta=30^(@)` with the vertical. Thent he value of normal reaction from the from the floor just after release will be

Options

A. `4mg//7`
B. `5mg//9`
C. `2mg//5`
D. None of these

Explanation

Only two forces are acting on rod, normal reaction (vertically upwards) and weight (vertically downwards) since both forces are vertical centre of mass falls in vertical direction (downwards).
`OC=(1)/(2)sin30^(@)=(l)/(4)`
`alpha=(tau)/(I)`
`=((mg)((l)/(2)sin30^(@)))/((ml^(2))/(12)+(ml^(2))/(16)))`
`=(12)/(7)(g)/(l)`
`a_(C)=(OC)alpha`
`=((l)/(4))((12)/(7)(g)/(l))=(3)/(7)g`
Now, `mg-N=ma_(C)=(3mg)/(7)`
`thereforeN=(4mg)/(7)`