# Rotating Rockets

1. Oct 14, 2010

### Tina20

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

Far out in space, a mass1=146500.0 kg rocket and a mass2= 209500.0 kg rocket are docked at opposite ends of a motionless 70.0 m long connecting tunnel. The tunnel is rigid and has a mass of 10500.0 kg.
The rockets start their engines simultaneously, each generating 59200.0 N of thrust in opposite directions. What is the structure's angular velocity after 33.0 s?

2. Relevant equations

center of mass = (m1x1 + m2x2 + m3x3)/m1 + m2 + m3
where m1 = mass of rocket 1, m2 = mass of tunnel, m3 = mass of rocket 3
x(number) = distance to the origin (in this case, rocket one used as origin (set at 0)

Moment of Inertia = m1r1^2 + m2r2^2 (m=mass, r= distance to center of mass)
Moment of inertia of thin, long rod about center = 1/12 ML^2 (M=mass, L=length of rod)

Torque = lengthxForce

angular acceleration = Torque/inertia

angular velocity = angular acceleration x time

3. The attempt at a solution

So, I found the center of mass, and I know it's correct because the previous question asked for it and was found to be correct. It was 41m.

Now, I think I may be calculating my moment of inertia wrong???
I = m1r1^2 + m3r3^2 + 1/12 ML^2
I = 447 938 063.8 kg*m^2

Am is supposed to include the moment of inertia of the tunnel (rod) in this scenario? Obviously I use the mass and their relative distances to the center of mass for the rockets to find their moment of inertia...but do I just add on the moment of inertia of the rod?

My answer was incorrect, so I am trying to figure out what is wrong about my process. I am assuming my moment of inertia is incorrect.

The torque is 70m x 59200N = 4144000 N*m

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

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 15, 2010

### rl.bhat

(1/12)*M*L^2 is the moment of inertia of the rod about its center. So to find its moment of inertia about the center of mass, use parallel axis theorem of moment of inertia.

3. Oct 15, 2010

### Tina20

Ok, I did that, but the angular velocity still comes out to be 0.305 rad/sec which is wrong. I still don;t know what else I could be wrong. Any ideas?

Tnet = lF = (70m)(59200) = 4144000 N*m

angular acceleration = Tnet/Itotal
angular acceleration = 9.24x10^-3 rad/s^2

w(omega) = angular accelerationxtime
9.24x10-3 rad/s^2 x 33s

Any help will be very much appreciated :)

Last edited: Oct 15, 2010
4. Oct 16, 2010

### rl.bhat

Net angular acceleration is gives by

$$\alpha = F(\frac{1}{m_1r_1} + \frac{1}{m_2r_2} + \frac{1}{m_3r_3})$$

5. Oct 16, 2010

### Tina20

Ok,

so F = lF which is torque, or is it the force of thrust?

I tried both Forces and substituted it into the equation you gave me. I am still not getting a correct answer. I assume r in the equation below is the distance to the center of mass correct? or is it the distance to the centre of the tunnel?

6. Oct 16, 2010

### rl.bhat

F is the force of the thrust. F= ma. So a = F/m. angular acceleration α is equal to F/(m*r). All the distances are measured from the center of the mass.

Show your calculations and expected result.

7. Oct 16, 2010

### Tina20

Ok, so F = 59200N

angular a = F (1/m1r1 + 1/m2r2 + 1/m3r3)
angular a = 59200N (1/m1r1 + 1/m2r2 + 1/m3r3) where m1=rocket 1, m2=mass of tunnel, m3=mass of rocket 2. r1 = rocket 1 to centre of mass, r2= centre of tunnel to centre of mass, r3 = rocket 2 to centre of mass.

I solved for angular a and multiplied it by the time and got the right answer! Thank you so much. I don't understand why I was getting it wrong last night...probably too late and I was making a silly mistake.

Thank you for your help!