A uniform rod of mass m and length l is at rest on smooth horizontal surface. An impulse P is applied to end B as shown.

The point about which rod rotates just after impulse being applied is located at

Options

A. `(l)/(3)` from A
B. `(l)/(6)` from A
C. `(l)/(2)` from A
D. `(l)/(4)` from A

Answer

A

Explanation

Using consevation of linear momentum
`P = mV`
`V = (P)/(m)`
let us suppose that the Rod will rotate about point C. which is at a distance y from COM
`v = omega y = 0 rArr (P)/(m) - omega y = 0` .....(1)
from angular momentum conser vation.
`P xx (L)/(2) = (ML^(2))/(12) omega rArr omega = (6P)/(mL)`
Putting (2) in (1)
`(P)/(m) - (6P)/(mL) y = 0 rArr (6Py)/(mL) = (P)/(m) rArr y = (L)/(6)`
So `AC = (L)/(2) - (L)/(6) rArr AC = (L)/(3)`