Finding angular acceleration from torque We have to analyze this video
Givens:


*

*An applied net torque due to the wind on the windmill is equal to $1500 \mathrm{\ N \cdot m}$.

*Each (of the 3) propeller props weighs approximately $45 \mathrm{\ kg}$ and has a moment of inertia equal to about 65% of that of a rod of the same mass being spun about its end.

*This torque is applied for approximately 45 seconds before the explosion, prior to which the windmill was at rest.
Question: What was the angular acceleration caused by the torque?
So here's my attempt at it:
$$T=I\alpha$$
$$T=ML^2/3 \times \alpha \times .65$$
(due to the whole 65% thing. Actually not sure if I should put $3\times M$ for each propeller)
And so this is where I get stuck. I'm not given $L$, so I'm not sure how to work around this. I could also use $T=1/2MR^2\times a$, but then I don't know $R$.
 A: The first thing I would point out to you is that $\tau = \frac{1}{2} MR^2 \alpha$ is really just $\tau = I\alpha$, with a particular choice of $I$. Is that choice appropriate for this problem? (Ask yourself the same thing any other time you consider using $\tau = \frac{1}{2} MR^2\alpha$.)
Next, note that the moments of inertia of different parts of the windmill do add up to produce the total, just like with mass. You can't just use the moment of inertia of one propeller prop, you have to calculate the total moment of inertia.
Finally, consider this: what information can you get from the video, that could supplement the 3 "givens"? There's no length scale in the video, so you can't measure the length of a prop directly, but there is a time scale. What can you do with that?
A: When a torque is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia.
This relation can be thought of as Newton's Second Law for rotation. The moment of inertia is the rotational mass and the torque is rotational force.
Angular motion obeys Newton's First Law. If no outside forces act on an object, an object in motion remains in motion and an object at rest remains at rest.
Using Newton's second law to relate Ft to the tangential acceleration at = rα , where α is the angular acceleration:
Ft = mat = mrα

and the fact that the torque about the center of rotation due to Ft is: 
$\tau$ = Ftr , we get:
$\displaystyle\tau$ = mr 2α.
For a rotating rigid body made up of a collection of masses m1,m2.... the total torque about the axis of rotation is: 
$\displaystyle\tau_{{\:\rm total}}^{}$  =   $\displaystyle\sum$$\displaystyle\tau_{i}^{}$
=   ($\displaystyle\sum$miri2)α.
The second line above uses the fact that the angular acceleration of all points in a rigid body is the same, so that it can be taken outside the summation. 
The moment of inertia, I , of a rigid body gives a measure of the amount of resistance a body has to changing its state of rotational motion. Mathematically,
I = $\displaystyle\sum$miri2.    
Note: The units of moment of inertia are 
kg $\cdot$ m 2.
This allows us to rewrite Equation 8.9 as:
$\displaystyle\tau_{{\:\rm total}}^{}$ = Iα 
which is the rotational analogue of Newton's second law.
FOR MORE INFORMATION AND IMPORTANT NOTE CLICK ON LINK :
http://www.studygtu.com/2016/02/relationship-between-torque-angular-velocity-angular-acceleration-momentum.html
