# Striking the End of a Triangle in Space

1. Feb 11, 2012

### harrietstowe

1. The problem statement, all variables and given/known data

http://img221.imageshack.us/img221/4861/figure1h.png [Broken]

Mod note: Fixed the image link.

Consider the rigid object shown in this image. Four masses lie at the points shown on a rigid isosceles right triangle with hypotenuse length 4a. The mass at the right angle is 3m, and the other three masses are m. They are labeled A,B,C and D as shown. Assume that the object is floating freely in outer space. Mass C is struck with a quick blow, directed into the page. Let the impulse have magnitude ∫F dt = P . What are the velocities of all the masses immediately after the blow?

2. Relevant equations

The one given in the problem statement
v=ω x r
Torque = r x F
ΔL=r ΔP

3. The attempt at a solution

I think computing the CM of this system would be a good place to start. I think I need to use the angular impulse formula but I'm wondering which mass I should go after first. My intuition leads me to believe the velocity of mass A will be opposite that of mass C. Is this all reasonable?

Last edited by a moderator: May 5, 2017
2. Feb 12, 2012

### tiny-tim

hi harriet!
that's right …

impulse = mtotalvc.o.m
don't assume anything, it's too dangerous

just find the initial angular velocity, then add the effect of that (on each mass) to the vc.o.m

3. Feb 13, 2012

### harrietstowe

ok so let position of the CM be R and lets work in the x y z coordinate system
Then R=[m(0,0,0)+m(2a,0,0)+m(4a,0,0)+3m(2a,2a,0)]/6m
which simplifies to
R=((4/3)a,(1/3)a,0)
Then using the equation Impulse=(mtotal)(vcm) we can write:
P=(6m)(vcm)
vcm=p/6m
Now to find the initial angular velocity ω I can write:
vcm=ω r
for r though do I use the distance from the cm to mass C?
In that case I would have:
P/6m=ω (a/3)(√65) and so
ω=P/(2(√65)a m)

4. Feb 13, 2012

### tiny-tim

nooo

you've used m instead of 3m

anyway, it would be much easier in this case to say that the c.o.m. of the bottom three is obviously 3m in the middle, and then the c.o.m. of that and the top one is equally obviously 6m halfway up the middle

(and why do you keep saying "cm"? there's nothing about cm in the question)
no, you need to use moment of impulse = change in angular momentum