# Two rockets rotating attached by rod

1. Nov 19, 2013

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

http://puu.sh/5nKxl.png [Broken]

2. Relevant equations

α = ω/t
τ = I*α

3. 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

Last edited by a moderator: May 6, 2017
2. Nov 20, 2013

### haruspex

3. Nov 20, 2013

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.

4. Nov 20, 2013

### 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?

5. Nov 20, 2013

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.

6. Nov 20, 2013

### haruspex

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

7. Nov 21, 2013

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?

8. Nov 21, 2013

### haruspex

Isn't it 12500?

9. Nov 21, 2013