Pivoted rod - impulse applied so no horizontal force on pivot

[Aaron7]

Homework Statement

There is a rod of length L pivoted on one end. It is originally at rest horizontally then released. When vertical, an impulse is applied to bring the rod to rest.

Throughout the question I have worked out I = 1/3 mL^2

w= √(3g/L) at the vertical position and an impulse of m√(gL/3) is the minimum required to achieve bringing the rod to rest.

I am attempting to find the impulse and distance from the pivot so that no horizontal force is exerted at the pivot.

Homework Equations

G = I dw/dt
v=rw
a=r dw/dt
Angular impulse J = r x (Impulse) = change in angular momentum
Angular momentum L = Iw

The Attempt at a Solution

I have tried: impulse required = mv = mrw
=> m√(gL/3) = mr√(3g/L)
=> r = L/3

I am confused on how to find the impulse and its distance so that the horizontal force on the pivot is 0. Any help is much appreciated.

Many thanks.

 

[tiny-tim]

Hi Aaron7!

Find the change in (linear) momentum of the centre of mass … that gives you the magnitude of the impulse

then find the change in angular momentum of the whole rod … that give you the torque of that impulse