Doubt on conservation of angular momentum

[ubergewehr273]

Homework Statement

A uniform thin rod of length $2l$ and mass $m$ lies on a horizontal table. A horizontal impulse $J$ is given to the rod at one end. There is no friction. The total kinetic energy of the rod after impulse will be ?

Homework Equations

$Jl=I\omega$
$J=mv_{cm}$
$K=\frac {1} {2} mv^2_{cm} + \frac {1} {2} I_{cm}\omega^2$

The Attempt at a Solution

Using the 2nd equation it is evident that $v_{cm}$ is known. However while calculating $\omega$ after impulse is applied, my doubt is about which point are we supposed to conserve angular momentum? I get the answer while conserving about COM of rod but technically speaking doesn't impulse $J$ provide an external torque about the COM ? So, then about which point should we conserve angular momentum?

 

[Orodruin]

Please show your work.

 

[ubergewehr273]

Please show your work.

From 1st equation $v_{cm}=\frac j m$ and from 2nd $\omega=\frac {3J} {ml}$
Using the 3rd $(I_{cm}=\frac {ml^2} {12})$
$K=\frac{mv^2} {2} + \frac{m(2l)^2\omega^2} {24}$
Therefore $$K_{final}=\frac{2J^2} {m}$$

 

[PeroK]

my doubt is about which point are we supposed to conserve angular momentum? I get the answer while conserving about COM of rod but technically speaking doesn't impulse $J$ provide an external torque about the COM ? So, then about which point should we conserve angular momentum?

Angular momentum about any point is conserved. Taking the COM may be the best choice to solve the problem.

 

[haruspex]

doesn't impulse J provide an external torque about the COM ?

Whether it is external depends on your definition of "the system".
You wrote Jl=Iω. If we consider the J as external to the system then that equation acknowledges that the angular momentum is not conserved. It is changed by the external impulse J.
Or we can take the source of J as being part of the system so that J is internal. But now we have to take Jl as the initial angular momentum of the system. Either way we get the same equation.

 

[haruspex]

Angular momentum about any point is conserved

About any fixed point in space, yes, or about the COM as a variable point.
And to illustrate, we could take moments about the point in space where impact occurred, so angular momentum is zero throughout: Iω+mvl=0.

 

[ubergewehr273]

Ok, what if I wish to conserve angular momentum about impact point. Then what equation must I write because throughout it will be 0

About any fixed point in space, yes, or about the COM as a variable point.
And to illustrate, we could take moments about the point in space where impact occurred, so angular momentum is zero throughout: Iω+mvl=0.

I didn't quite understand the last point.

 

[haruspex]

Ok, what if I wish to conserve angular momentum about impact point. Then what equation must I write because throughout it will be 0

I didn't quite understand the last point.

After impact, the rod is rotating at rate ω, contributing Iω to its angular momentum, and with linear momentum mv. The line of motion is distance l from the reference point, so that linear momentum also contributes mvl to the angular momentum, bringing the total to Iω+mvl.

 

[PeroK]

Ok, what if I wish to conserve angular momentum about impact point. Then what equation must I write because throughout it will be 0

I think you are confused about "conservation of angular momentum about a point" and "taking moments about a point". In fact, you got me confused too! This problem is about analysing a sudden impulse, which is not directly concerned with conservation of angular momentum, as the impulse provides a change to the angular momentum (about any point).

Conservation of angular momentum applies to the motion after the impulse has been given.