Inelastic Collision and Rotational Inertia

[Victorzaroni]

Homework Statement

A mass M slides down a smooth surface from height h and collides inelastically with the lower end of a rod that is free to rotate about a fixed axis at P as shown below. The mass of the rod is also M, the length is L, and the moment of inertia about P is ML²/3.

The angular velocity of the rod about the axis P just after the mass sticks to it will be:
(A) √(gh/2)
(B) √(2gh)/L
(C) (3/4L)√(2gh)
(D) (3L)/√(2gh)
(E) (9√(2gh))/L

Homework Equations

There's a bunch.

The Attempt at a Solution

I started with conservation of energy, so mgh=(1/2)mv² and thus v=√(2gh). I don't really know where to go next. I know inelastic means momentum is conserved and kinetic energy is not.

 

[tiny-tim]

Hi Victorzaroni!

I started with conservation of energy, so mgh=(1/2)mv² and thus v=√(2gh). I don't really know where to go next. I know inelastic means momentum is conserved and kinetic energy is not.

angular momentum is also conserved

 

[Victorzaroni]

oh okay. So The angular momentum of the block when it gets to the bottom equals to angular momentum of the rod when it hits it? How do I express that in an equation? Is the r in w=v/r the height of the circle?

 

[tiny-tim]

oh okay. So The angular momentum of the block when it gets to the bottom equals to angular momentum of the rod when it hits it?

of the rod-plus-block, yes

How do I express that in an equation? Is the r in w=v/r the height of the circle?

(you mean half the height?)

what does r have to do with it? this is a rod, you'll have to use moment of inertia

 

[Victorzaroni]

Okay. So: rmv=Iw=>rmv=(M+M)r²w?

 

[tiny-tim]

(have an omega: ω )

Okay. So: rmv=Iw=>rmv=(M+M)r²w?

uhh?

what's the moment of inertia of a rod?

 

[Victorzaroni]

It's (1/3)ML^2 (about the end) I meant to write that.

 

[tiny-tim]

then do so!

 

[Victorzaroni]

then do so!

Do what?

 

[tiny-tim]

combine …

Okay. So: rmv=Iw=>rmv=(M+M)r²w?

It's (1/3)ML^2 (about the end) I meant to write that.

 

[Victorzaroni]

okay. So: rmv=Iw => rm√(2gh)=(1/3)(m+m)L²
I'm just confused about what the r is. Is that L?

 

[tiny-tim]

Yup!

 

[Victorzaroni]

Unless I made a math error, I end up with (3√(2gh))/(2L), and that's not a choice. :(

 

[tiny-tim]

oh wait, it's not …

(1/3)(m+m)L²

… the "1/3" is only for the rod, not for the other mass

 

[Victorzaroni]

Oh okay yea I didn't think that that made much sense. So I am just equating: Lmsqrt(2gh)=(1/3)mL^2

 

[tiny-tim]

but you still need to add on the angular momentum for the other mass

 

[Victorzaroni]

But the other mass is sliding? So how does it have angular momentum?

 

[tiny-tim]

But the other mass is sliding? So how does it have angular momentum?

ah

everything has angular momentum (about any point that isn't on its line of motion)

conservation of angular momentum always means conservation of angular momentum about a particular point

you've chosen, as that point, the pivot at the top of the rod

you chose it because there's an unknown force at the pivot, and the only way you can avoid it is to make its moment zero

so now you need to find the angular momentum of the sliding mass about the pivot

i thought you already knew that, and had done it? …

So: rmv=Iw=>rmv=(M+M)r2w?

 

[Victorzaroni]

Well that's what I did and it was wrong. You said the right side of the equation was wrong because of the (m+m). I get now why we can equate things because of their angular momentum, but I am missing a piece and I don't what it is

 

[tiny-tim]

Well that's what I did and it was wrong. You said the right side of the equation was wrong because of the (m+m).

yes, because the 2nd m was wrong, it should have been m/3

(the 1st m was right)