For the arrangement in the Figure, the particle `M_(1)` attached to one end of string which moves on a horizantal table in a circle of radius `l/2` (where `l` is the length of the string) with constant angular speed `omega`. The other end of the string attached to to mass `M_(2)` which rest on a vertical rod. When the rod collapse, the acceleration of mass `M_(2)` at that instant

Options

A. g
B. `(omega^(2) l )/(2)`
C. `( 2 M_(2) g - M_(1)l omega^(2))/(2 (M_(1) + M_(2)))`
D. `(M_(2)g + M_(1)l omega^(2))/(M_(1) + M_(2))`

Answer

C

Explanation

From Newton's second law

`T - M_(1) omega^(2) (l)/(2) = M_(1) a` and `M_(2)g - T = M_(2) a`
After solving above equations , we get
`a = (2 M_(2)g - M_(1) l omega^(2))/(2 (M_(1) + M_(2)))`