Angular Acceleration of a Model Rocket

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The discussion revolves around calculating the angular acceleration of a model rocket generating 4.0 N of thrust while spinning in a horizontal circle. The user initially misapplied the torque and moment of inertia equations, leading to incorrect answers. After receiving guidance, they corrected their moment of inertia calculation and used the appropriate formula, T = I x α, to find the angular acceleration. The final resolution involved using the correct moment of inertia formula, which clarified the calculation process. This highlights the importance of accurately applying physics principles in problem-solving.
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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.
 
Physics news on Phys.org
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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