# Angular Acceleration of a Model Rocket

1. Dec 6, 2008

### x2008kwa

1. The problem statement, all variables and given/known data
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)?

2. Relevant equations

T= r x F x sinθ

T=mr2α

where T is torque and α is angular acceleration

3. 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.

2. Dec 6, 2008

### rl.bhat

Moment of inertia of rod and rocket system is (m*r^2)/3 +
M*r^2
Now try.

Last edited: Dec 6, 2008
3. Dec 6, 2008

### x2008kwa

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?

4. Dec 6, 2008

### rl.bhat

Use I = (.100kg)(.600)^2)/3 + (.200)(.600)^2

5. Dec 6, 2008

### x2008kwa

Problem solved. Thank you very much for the help rl.bhat