Two rockets rotating attached by rod

[Esoremada]

Homework Statement

http://puu.sh/5nKxl.png

Homework Equations

α = ω/t
τ = I*α

The Attempt at a Solution

CM = [232200*99 + 99/2*12500] / (107100 + 232200 + 12500)
= 67.102 m


τ = 43320*67.102 + 43320*(99 - 67.102)
= 4288680 N

α = ω/t
τ = I*α
τ = I*(ω/t)

ω = τ/I * t
= 4288680 / [107100*67.102² + 1250*(67.102 - 99/2)² + 232200*(99-67.102)²] * 28.3
= 0.1688

 

[haruspex]

Your answer would have been correct if the CoM had been fixed in space on an axle. What will happen instead?

 

[Esoremada]

Your answer would have been correct if the CoM had been fixed in space on an axle. What will happen instead?

The CoM moves through space, but how can that affect this question when the distance from the centre of mass to the ships remains the same regardless of orientation and speed relative to space.

 

[haruspex]

Yes, thinking about this again I believe your method is correct, but in the calculation you treated the tunnel as a point mass at its centre. That slightly underestimates the moment of inertia of the system. I get 0.166. Is that error enough to explain your answer's rejection?

 

[Esoremada]

Yes, thinking about this again I believe your method is correct, but in the calculation you treated the tunnel as a point mass at its centre. That slightly underestimates the moment of inertia of the system. I get 0.166. Is that error enough to explain your answer's rejection?

That was correct! So what is the correct way to calculate the moment of a inertia of a rod about a given pivot? I only know that it is 1/12*ML^2 and 1/3*ML^2 for pivots about the centre or end.

 

[haruspex]

That was correct! So what is the correct way to calculate the moment of a inertia of a rod about a given pivot? I only know that it is 1/12*ML^2 and 1/3*ML^2 for pivots about the centre or end.

Use the parallel axis theorem. The MoI about one end is simply a special case of that.

 

[Esoremada]

Oh cool I never even noticed that. I get how to derive 1/3*ML^2 now but I try doing it in this problem and get it wrong

1250*(67.102 - 99/2)^2 + 1/12*1250*99^2

That's the moment of the rod rotating about its centre plus the moment of the rod's centre of mass rotating about the systems centre of mass. What am I doing wrong?

 

[haruspex]

Oh cool I never even noticed that. I get how to derive 1/3*ML^2 now but I try doing it in this problem and get it wrong

1250*(67.102 - 99/2)^2 + 1/12*1250*99^2

That's the moment of the rod rotating about its centre plus the moment of the rod's centre of mass rotating about the systems centre of mass. What am I doing wrong?

Isn't it 12500?

 

[Esoremada]

...I need more sleep. Thanks again :P

 

[rcgldr]

What about treating the rod as two rods rotating about one end, where that commen one end is the center of mass?