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Rotation and Angular Momentum: Solving for thrust of a rocket 
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#1
Oct1710, 08:32 PM

P: 4

1. A solid bar of length L = 0.5m and with W = 0.1m weighs 2kg. It also has two constantthrust 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. 


#2
Oct1710, 08:56 PM

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)? 


#3
Oct1710, 09:09 PM

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 


#4
Oct1710, 09:19 PM

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? 


#5
Oct1710, 09:26 PM

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.



#6
Oct1710, 09:49 PM

HW Helper
P: 2,155

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


#7
Oct1710, 09:57 PM

P: 4

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



#8
Oct1710, 10:03 PM

HW Helper
P: 2,155

Yep, just checking.



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