A unifrom rod of mass `m` and length `L` is kept on a horizontal table with `(L)/(4)` length on the table. The end `B` is tied to a string as shown in the figure. The string attached to the end `B` is cut and the rod starts rotating about point `C`. Find the normal reaction from the table on the earth as soon as the string is cut.

Explanation

The free body diagram of the rod is as shown in figure.
Here the normal force will shift to the point `C` and we are assuming that the rod is rotating about `C` with angular acceleration `alpha`. Since `C` is a stationary point, applying Newton's second law in rotation from about `C`.
or `sumtau=l alpha`
or `mg.(L)/(4)={(mL^(2))/(12)+m((L)/(4))^(2)}*alpha`
or `mg.(L)/(4)=(7mL^(2))/(48)*alpha`
`alpha=(12g)/(7L)`
`impliesa_(CM)=alpha.r=(12g)/(7L)xx(L)/(4)=(3g)/(7)`
Applying Newton's Second Law in translational form,
`sumF=ma_(CM)impliesmg-N=(3mg)/(7)`
`impliesN=(4mg)/(7)`

Alternate Method :
You can solve this questions by applying Newton's Second Law in rotational torm about CM of rod.
`Sigma tau_(CM) = l_(CM).alpha`
`rArr N.(L)/(4) = (mL^(2))/(12).alpha rArr N = (mL alpha)/(3)` ...(i)
Sigma F = ma_(CM)`
`rArr mg - N = ma` ...(2)
Since point C is in contact with the table, so its acceleration will be zero.
`a-(alpha L)/(4) = 0 rArr a = (alpha L)/(4)` ...(3)
Solving above three equations, we get
`N = (4mg)/(7)`
Note : The value of `alpha` about centre of mass of rod `alpha = (12g)/(7L)` is same as that about point C.