Rotating Rockets Homework: Angular Vel. After 33s

[Tina20]

Homework Statement

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?

Homework 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

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

Please help :)

 

[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.

 

[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
= 0.3052 rad/sAny help will be very much appreciated :)

 

[rl.bhat]

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
= 0.3052 rad/s


Any help will be very much appreciated :)

Net angular acceleration is gives by

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

 

[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?

 

[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.

 

[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!