What Is the Final Angular Velocity of a Rod After Being Struck by a Bullet?

[Abhishekdas]

Please help me with this Question based on angular momentum...

Homework Statement

A uniform rod of length L rests on a frictionless horizontal surface.The rod is pivoted about a fixed frictionless axis at one end.Initially at rest the rod is hit by a bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v.It hits the rod at the centre and becomes embedded in it.
Mass of rod = 6m
mass of bullet = m

Find the final angular velocity of the rod...?

Homework Equations

Impulse*r=Change in angular momentum.
angular momentum abt a point =I*w

The Attempt at a Solution

I went about by considering the axis as my reference n used d above eqn..
mass of body after impact = 7m..

So, MI of the body about axis = 7mL²/3

my eqn is
mvL/2=(7mL^2*w)/3

I am getting w=3v/14L

ans in buk is 2v/9L...

Thank you...

 

[Beaker87]

So, MI of the body about axis = 7mL²/3

The moment of inertia of this body should be the moment of inertia of the uniform rod about the axis plus the moment of inertia of the bullet.

Think of the bullet as a point particle of mass m embedded in the center of the rod.

 

[Mandeep Deka]

Whenever we consider the Moment of Inertia of a two or more joint bodies about a point, we consider the net moment of inertia as the sum of the MIs of each of the bodies about that point and not the sum of the masses!

So, you tell me what should be the Moment of Inertia of the body about the point of suspension, after the bullet gets embedded in it? (I mean your mistake is in calculating the MI of the final body)

 

[vissh]

hi abhi (^.^)

I went about by considering the axis as my reference n used d above eqn..
mass of body after impact = 7m..
So, MI of the body about axis = 7mL²/3

>The place where you went wrong is in Calculating Moment of inertia.
>Think again, didn't the formula you used to calculate M.O.I is used for "Uniform" rod .But the system we get after collision isn't uniform {check urself}.
>If u agree with me, then i give you a suggestion :- Think of the rod and particle as separate 'bodies' for calculating M.O.I. ; find out their respective contribution at the chosen axis and then, add these 2 M.O.I.s to get the M.O.I. of "rod + particle" system :D
>Try it >:)

 

[Abhishekdas]

Firstly...Thanks a lot guys for replying...

Now I am going to reply to all of you at once...Coz u all basically mean dat dere is a problem wid my MI formula...

I had a doubt about dat so dat was my error i gues...But stilll

Correct my thinking...im thinking dat d bullet is embedded into it ryt...so doesn't it form a part of the rod? Lyk in questions where dey say the body "sticks" to it i us d formula of MI taking d body's MI separately...coz its lyk its sticking out its head kinda thing...but wen its embedded den isn't it a diff case? Or maybe d rod loses its uniformity and so 7mL^2/3 is invalid...Im still confused ...This embedding park is makin me think lyk d uniformity is distorted n stuff...I dunno...please reply...

Thanks again...

 

[vissh]

>Hmm this is what i thinks {I cud be wrong But u said to reply} :-
-->Hmm in questions , the particle gets sticked or embedded means the same thing[as much i use myself in many questions].
>So u can use ur "sticking out its head" thinking and then "i us d formula of MI taking d body's MI separately" :D
--->Hmm i will say we can to an approximation take the rod "even after striking " To Be uniform with same mass as before.{Keep the particle out for rod's calculation} .
>So,take the rod separately , apply formula for getting its M.O.I at the desired axis ;
take the particle and find its controbution and just add both And you will get the desired/required M.O.I.

 

[Abhishekdas]

Thanks...lets see what d odrs have to say...Even i think dat d buk wants us to go n think lyk dat...Answer matches by dat ...lol...neway...