Rotation and Angular Momentum: Solving for thrust of a rocket

[engineer08]

1. A solid bar of length L = 0.5m and with W = 0.1m weighs 2kg. It also has two constant-thrust rockets attached on either end. Each rocket is small enough to be considered a point mass of 0.25kg. If the bar is initially at rest and in two seconds after the rockets are fired it achieves a rotation rate of 1000rpm, determine the thrust of each rocket.

Homework Equations

torque = I * angular acceleration
I = (1/12)*(L^2+W^2)*M



3. I found I, angular momentum, to be 0.0542. I am not sure what to do from there.

 

[diazona]

I is moment of inertia, not angular momentum.

What equation do you know that relates torque and angular momentum (which is usually written L)?

 

[engineer08]

Correct my apology, I is moment of inertia. An equation I know that relates torque with angular momentum 'L' is:

torque = dL/dt = d(Iw)/dt

The acceleration changed from 0 to 1000rpm in 2 seconds. and L=Iw, which is the angular velocity times the moment of inertia. I'm confused as to what to do

 

[diazona]

It's not acceleration that changed from 0 to 1000rpm in 2 seconds

Think about this: what else can you calculate from that 1000rpm figure?

 

[engineer08]

So, angular velocity changes from 0 to 1000rpm in 2 seconds, and we know I. I can therefore solve for angular momentum, right? Couldn't you also integrate the sum of the moments exerted by each rocket from time t to 0? I still am not sure how to translate all of this to the thrust of each rocket.

 

[diazona]

So, angular velocity changes from 0 to 1000rpm in 2 seconds, and we know I. I can therefore solve for angular momentum, right?

Right, try that.

By the way, you do know what kind of physical quantity thrust is, right?

 

[engineer08]

thrust must be in Newtowns (N), presumably? I was pretty certain of that.

 

[diazona]

Yep, just checking.