Conservation of angular momentum of a thin rod

[xfreedom]

Homework Statement

A thin rod of mass M and length L rests on a frictionless table and is struck L/4 from its CM by a clay ball of mass m moving at speed v. The ball sticks to the rod. Determine the translational and rotational motion

Homework Equations

Irod=1/12 mr^2
Irod=1/3 mr^2
L=mrv=Iw

The Attempt at a Solution

I fount out the translational velocity, used the conservation of linear momentum.
But I'm kind of stop in the second part.
I found out:
m(1/4 l)v=Iw
but what is I?
I guess perhaps we needs to find out the new center of mass, is it (1/12 Ml^2)+(mr^2) where r is calculated with the new center of mass? Is it(m 1/4 l)/(M+m)? And again which Irod should I use?

thank you for help.

 

[tiny-tim]

hi xfreedom! welcome to pf!

… but what is I?
I guess perhaps we needs to find out the new center of mass, is it (1/12 Ml^2)+(mr^2) where r is calculated with the new center of mass? Is it(m 1/4 l)/(M+m)? And again which Irod should I use?

for conservation of angular momentum, you can measure angular momentum about any point

(it is only for τ = Iα that you are restricted to the centre of mass or centre of rotation)

so you may as well use the centre of mass of the rod itself

 

[xfreedom]

hi xfreedom! welcome to pf!


for conservation of angular momentum, you can measure angular momentum about any point

(it is only for τ = Iα that you are restricted to the centre of mass or centre of rotation)

so you may as well use the centre of mass of the rod itself

Thanks for your reply.
So just to be sure, is the correct formula should be

m(1/4 l)v=Iw

where I equal to 1/12 ML^2 + [(m 1/4 l)/(M+m)-1/4 l]^2 m?

thank you.

 

[tiny-tim]

hi xfreedom!

… where I equal to 1/12 ML^2 + [(m 1/4 l)/(M+m)-1/4 l]^2 m?

(isn't that second term negative? )

no, the second term is just the moment of inertia of the clay at a distance L/4

(and now I'm off to bed :zzz:)

 

[xfreedom]

hi xfreedom!


(isn't that second term negative? )

no, the second term is just the moment of inertia of the clay at a distance L/4

(and now I'm off to bed :zzz:)

...and I believe the clay and the rod will rotate with the new center of mass? So the I for clay should be the distance between the clay and the new CM?

Thank you.

And have a good dream lol

 

[tiny-tim]

hi xfreedom!

i'll expand on my original post (it's too late to edit it), in case you'e not clear about what formula to use:

for conservation of angular momentum, we can measure angular momentum L about any point P

provided we use the full formula L_P = I_c.o.mω + mr_c.o.m x v_c.o.m

(it is only for L_P = I_Pω (and τ_P = I_Pα) that we are restricted to P being the centre of mass or centre of rotation)

so you may as well use the centre of mass of the rod itself ​

...and I believe the clay and the rod will rotate with the new center of mass?

it depends what you mean

the centre of rotation is well below the centre of mass …

that is (by definition) the point about which the clay and the rod rotate

but if you mean, do you use the new centre of mass to calculate the angular momentum of the clay-and-rod, the answer is yes

(but why do that when it's easier to calculate the two individual angular momentums, and then add them?)

So the I for clay should be the distance between the clay and the new CM?

(I ? )

as i said, you can use any point P

if you use the centre of mass, you get m(u-v)R = Iω

if you use a point distance d above the centre of mass, you get m(u-v)(R-d) = Iω - MVd

subtracting those, and dividing by d, gives m(u-v) = MV …

which is automatically true, since it's the equation for conservation of ordinary momentum! ​

finally, instead of the centre of mass, you might in this case find it more convenient to use the point of impact as P, since that gives you ω independently of the masses