Angular Acceleration of a Model Rocket

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Homework Help Overview

The problem involves calculating the angular acceleration of a model rocket that is spinning in a horizontal circle at the end of a rigid rod, given its mass, thrust, and the length of the rod.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of torque and moment of inertia in relation to angular acceleration. There are attempts to equate torque and moment of inertia formulas, with some questioning the correctness of their calculations and units.

Discussion Status

Some participants have provided alternative methods for calculating the moment of inertia, while others express uncertainty about their results. There is an acknowledgment of potential errors in calculations, and one participant claims to have resolved the issue with assistance.

Contextual Notes

Participants are exploring different interpretations of the problem setup and the equations involved, with some noting the importance of units in their calculations.

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Homework Statement


I'm getting this problem wrong and I'm not quite sure why.

The 200 g model rocket shown in the figure generates 4.0 N of thrust. It spins in a horizontal circle at the end of a 100 g rigid rod.
What is its angular acceleration (in rad/s2)?

knight_Figure_13_24.jpg



Homework Equations



T= r x F x sinθ

T=mr2α

where T is torque and α is angular acceleration

The Attempt at a Solution



(.600m)(4.0N)(sin45) = (.100kg + .200kg)(.600m)2 α

I set these two equations as equal. I ended up with an answer of 15.7 and it is wrong according to masteringphysics... One possible source of error that I'm thinking of is that the question asks for the answer in rad/s2... would this be in those units or would I need to convert it somehow? More than likely my mistake lies somewhere else.
 
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Moment of inertia of rod and rocket system is (m*r^2)/3 +
M*r^2
Now try.
 
Last edited:
I used that method, I = (.300kg)(.600)2/3 + (.300)(.600)2
and got the moment of inertia to be 0.144.
I then used the formula T = I x α therefore (1.697) = (.144)α
and got an answer of 11.8... which was incorrect. What am I doing wrong?
 
Use I = (.100kg)(.600)^2)/3 + (.200)(.600)^2
 
Problem solved. Thank you very much for the help rl.bhat
 

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