Striking the End of a Triangle in Space

[harrietstowe]

Homework Statement

http://img221.imageshack.us/img221/4861/figure1h.png

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?

Homework Equations

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

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?

 

[tiny-tim]

hi harriet!

I think computing the CM of this system would be a good place to start.

that's right …

impulse = m_totalv_c.o.m

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.

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 v_c.o.m

 

[harrietstowe]

ok so let position of the CM be R and let's 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)

 

[tiny-tim]

ok so let position of the CM be R and let's 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)

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)

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)

no, you need to use moment of impulse = change in angular momentum

 

[paranormal]

I would first clarify the problem by asking for more information. In particular, I would ask for the units of P and if the masses are all in the same plane or if the object is allowed to rotate in three dimensions. This information can greatly affect the solution and the approach to solving the problem.

Assuming that P is in Newton-seconds and the masses are all in the same plane, I would start by finding the center of mass of the system. This can be done by taking the average of the x and y coordinates of the masses, weighted by their masses. From there, I would use the laws of conservation of momentum and angular momentum to solve for the velocities of the masses after the blow.

To find the angular velocity of the system, I would use the equation ΔL = rΔP, where ΔL is the change in angular momentum and ΔP is the impulse. This would give me the angular velocity of the system.

Next, I would use the equation v = ω x r to find the velocities of the masses A, B, and D, where v is the velocity, ω is the angular velocity, and r is the distance from the center of mass to each mass. This would give me the velocities of these masses immediately after the blow.

For mass C, I would use the equation Torque = r x F to find the torque on the mass. Since the mass is struck with a quick blow, we can assume that the torque is constant and the mass will rotate at a constant angular acceleration. From there, I can use the equations of rotational motion to solve for the final angular velocity and the final velocity of mass C.

In conclusion, the key to solving this problem is to first find the center of mass and then use the laws of conservation of momentum and angular momentum to solve for the velocities of the masses. It is also important to clarify any uncertainties or missing information in the problem statement to ensure an accurate solution.