# Rotation and Angular Momentum: Solving for thrust of a rocket

by engineer08
Tags: angular, momentum, rocket, rotation, solving, thrust
 P: 4 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. 2. Relevant 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.
 HW Helper P: 2,155 I is moment of inertia, not angular momentum. What equation do you know that relates torque and angular momentum (which is usually written L)?
 P: 4 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
 HW Helper P: 2,155 Rotation and Angular Momentum: Solving for thrust of a rocket 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?
 P: 4 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.
HW Helper
P: 2,155
 Quote by engineer08 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?
 P: 4 thrust must be in Newtowns (N), presumably? I was pretty certain of that.
 HW Helper P: 2,155 Yep, just checking.

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