A rod `AB` of mass `M` and length `L` lies on smooth horizontal table impulse `J` is applied to end `A` as shown in the figure immediately after imparting the impulse :

Options

A. The radius of curvature of trajectory of `A` as seen from the ground is `(8)/(9)L`.
B. The radius of curvature of trajectory of `B` as seen from the ground is `(2)/(9)L`.
C. The instantaneous axis of rotation is at a distance of `L//6` from the mid point of the rod.
D. the mid point of the rod will move along a straight line.

Explanation

`R=(v^(2))/(a_(c))`
`v=(J)/(M)`
`omega=((JL)/(2))/((MI^(2))/(12))=(6J)/(ML)`
`rArr v_(A)=v+(omegaL)/(2)=(J)/(M)+(3J)/(M)=(4J)/(M)`
`a_(c_(A))=(omega^(2)L)/(2)=(18J^(2))/(M^(2)L)`
`rArrR_(A)=(v_(A)^(2))/(a_(c_(A)))=16(J^(2))/(M^(2)).(M^(2)L)/(18J^(2))=(8)/(9)L`
`v_(B)=v-(omegaL)/(2)=(2J)/(M)`
`rArrR_(B)=(4J^(2))/(M^(2)).(M^(2)L)/(18J^(2))=(2)/(9)L`