A rod of mass m and length L , pivoted at one of its ends, is hanging vertically . A bullet of the same mass moving at speed v strikes the rod horizontally a a distance x from its pivoted end and gets embedded in it. The combined system now rolates with rotates with angular speed `omega` about the pivot . The maximum angular speed `omega_(M)` is achieved for `x = x_(m)` . Then .

Options

A. `omega = (3vx)/(L^(2) + 3x^(2))`
B. `omega = (12vx)/(L^(2) + 12x^(2))`
C. `x_(M) = (1)/(sqrt(3))`
D. `omega_(M) = (v)/(2L) sqrt(3)`

Answer

A,C,D

Explanation

From angular momentum conservation
`mvx = ((ML^(2))/(3) + mx^(2))omega`
`omega = (3vx)/(L^(2) + 3x^(2))`
For maximum angular velocity
`(d omega)/(dx) = 0 `
`(d)/(dx) ((L^(2))/(x) + 3x) = 0 , (-L^(2))/(x^(2)) + 3 = 0 , x = (L)/(sqrt(3))`
` omega_("max") = (sqrt(3)v)/(2L)`