# 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 $R$.

• Thanks for marking this as homework; I'm sure someone will be able to give you hints. (I would myself, but it's rather late here!) – Noldorin Nov 4 '10 at 1:08
• Maybe I haven't understood the problem, but have you looked at en.wikipedia.org/wiki/List_of_moments_of_inertia (particularly to the rod with the axis of rotation at the end of the rod)?. That's according to your point 2). For using that equation, you need the angular momentum, however, you know the torque and you know the time in which the torque is being applied. As far as I can see, that's all the information that you need to calculate the angular acceleration. – Robert Smith Nov 4 '10 at 1:54
• @Robert Smith I think you're mistaking the L in that equation for angular momentum, when it's actually length of the rod. – ZachMcDargh Nov 4 '10 at 2:31
• @ZachMcDargh: Argh, you're so right. – Robert Smith Nov 4 '10 at 4:15
• Very strange that they chose to call that L. I was confused both by the OP's post and that equation on Wikipedia for a second. – ZachMcDargh Nov 4 '10 at 4:49

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?

• Hmm not really sure..so there is actually a way to get an actual value for this thats not in terms of L or R? – Snowman Nov 4 '10 at 1:44
• I think there's a way to calculate L or R (they're the same thing) from the the information you can get from the video. So again, think about what you can get from the video. Hint: look for the amount of time it takes for something to happen. – David Z Nov 4 '10 at 4:33

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.