Calculating Rotational Energy of a System with Three Masses and an Impulse

[LindaNguyen]

Homework Statement

Suppose three masses are arranged as shown, connected by a rodless mass.

(2 kg)-------(6 kg)--------(4 kg)

2m 3m

a) If this object is free to rotate in space, about what point will it spin?

Cm = 2(0) + 6(2) + 4(5)
_______________ = 2.7m
12

b) A small mass (m=0.1 kg) drops vertically onto the2 kg mass traveling 100 m/s and buries intself into the mass. What is the period of rotation of the system immediately after impact?

I'm not sure how I would begin setting this problem up.

c) What is the rotational kinectic energy of the system after the impact?

 

[cepheid]

Welcome to PF, LindaNguyen!

Homework Statement

Suppose three masses are arranged as shown, connected by a rodless mass.

(2 kg)-------(6 kg)--------(4 kg)

2m 3m

a) If this object is free to rotate in space, about what point will it spin?

Cm = 2(0) + 6(2) + 4(5)
_______________ = 2.7m
12

Looks good, except, are you sure that there wasn't some rounding error in your final answer? It's just a bit off from what I got. Oh, and you wrote "rodless mass", which I got a bit of a kick out of. (Not making fun of you, just thought it was a neat transposition).

b) A small mass (m=0.1 kg) drops vertically onto the2 kg mass traveling 100 m/s and buries intself into the mass. What is the period of rotation of the system immediately after impact?

I'm not sure how I would begin setting this problem up.

The key physics principle here is that angular momentum around the centre of rotation has to be conserved. Even the small mass, when it is just moving through space, has a certain amount of angular momentum. The amount depends on the point around which you measure the angular momentum, since L = r x p = r x mv. At the moment of impact, the vector r between the centre of rotation and the little mass is perpendicular to the vector v, so this cross product is simple to evaluate. Since angular momentum is conserved, you can equate this to the angular momentum of the whole system (rod + embedded mass) after the impact. You should be able to compute this easily enough (you'll need to find the moment of inertia of the system around the rotation point). EDIT: once you know the angular momentum, the angular speed should follow form that.

c) What is the rotational kinectic energy of the system after the impact?

This is just a matter of applying the right equation.

 

[LindaNguyen]

Thank you, Cepheid! I'm not very good at posting on forums so bare with me. I'll learn how to quote and etc. soon enough!

Part A) Sorry for the typo, hehe. I have a million things going on right now. As for the rounding error, I'm still getting 2.666666. I'm trying to find the center of mass here, correct? The 2m was supposed to be the distance between the first 2 masses and the 3m was supposed ot be the distance between the second 2 masses. I'm not sure why it came out like that, but...

[ 2(0) + 6(2) + 4(5) ] / 12 ----> 32/12 = 2.7m.

I'm working out the other parts right now, will get back to you!

 

[cepheid]

The problem was that I can't do arithmetic.

Carry on.

 

[LindaNguyen]

I'm still a tad bit lost on part B. Should I be finding the angular momentum about its center of mass?

 

[cepheid]

I'm still a tad bit lost on part B. Should I be finding the angular momentum about its center of mass?

Yes, you want to find the angular momentum of the little mass around the centre of mass of the rod system at the moment of impact. Then equate this to the angular momentum of the entire system around this point after the collision. This is because angular momentum is conserved, so it should be the same before and after the collision.

I gave you the equation for finding the angular momentum of a single point mass around a point in space. The position vector r is a vector giving the position of the point mass relative to the location around which you're computing the angular momentum (which is the centre of mass of the rod in this case).

 

[LindaNguyen]

Ah, I get it now (kind of)
L=IW (true for any object spinning around it's cm)
L=RxP (around an exterior point)
-----------------------------------------------------------
so, after I find the angular momentum about the small mass, I can set it to the whole system using L=IW, correct?

also.. if I'm finding the angular momentum of the small ***, would it look something like: (2.7i, 0j, 0k) x (0i, -10j, 0k) ---> -10 coming from mv (velocity is going down?)

 

[LindaNguyen]

Here's what I've come up with:
r x p = I'm getting 27.

if I do Ii Wi = If Wf
--->using I=mr^2 ; w=v/r ; r=CM=2.7m
I end up with 27 also.

now If Wf = 2(2.7)^2 + 6(0.7)^2 + 4(2.3)^2 * w

I know I solve for W after this but I'm not sure if I have the numbers in the ( ) correct.