Question

A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. An impulse p is applied to the end B. The time taken by the rod to turn through a right angle is
A. `(2pi(ml)/p`
B. `(2pi)(p/(ml)`
C. `(piml)/(12p)`
D. `(pip)/(ml)`

Answer 1

c) the angular impulse about center of mass is
`J = L_(f) - L_(i) = Iomega-0`
`therefore` J = `lomega`
or `pl/2 = (ml^(2))/12omega` or `p=(ml)/6 omega`
`therefore` `omega= (6p)/(ml)`
`therefore` `theta=omegat`
`therefore` `t=(theta)/(omega) = (pi)(2omega) = (pi)/(2 xx (6p)/(ml) = (piml)/(12p)`