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QUESTION: Hi Steve,

How are you doing? I hope you are well. Thank you for all your helpful answers in the past.

A while ago you gave me some help with calculating how fast on object would rotate, and at that time you gave me the following formula

alpha = tau/I

You also gave me some help calculating the torque and the rotational inertia for the particular object I want to rotate - alpha = (F*d/2) / ((1/12)*m*d^2)

I thought I had done the calculations correctly, which were

alpha = 4.01e-16/ 1.7150246e-11 = alpha = 0.00002338158 radians per second,

but then found out that "Alpha should have units of radians per second^2" Are you able to tell me what I have done wrong? I simply divided the torque by the rotational inertia, but I did not know the answer should be angular velocity in radians per second^2. My torque is so weak compared to the rotational inertia that I just assumed 0.00002338158 radians per second was the right answer - should that be 0.00002338158^2 instead? This is the first time I have ever done this type of calculation..

Your advice would be much appreciated as always.

Eddie

ANSWER: Hello Eddie,

Your result is good. Unless you went wrong on what you plugged in for F, d, and m. F should have units of Newtons, d should have units of meters, and m should have units of kilograms.

Follow me as I explain how the units of alpha develop. I'll just put an n for the numerical part and deal with the units in detail.

Alpha = (F*d/2) / ((1/12)*m*d^2) = n Newtons.meters / kilograms.meter^2

The Newton is equivalent to kilograms.meters/sec^2 so I'll make that substitution and show you how things cancel.

Alpha = n kilograms.(meters/sec^2).meters / kilograms.meter^2

Cancel kilograms first

Alpha = n (meters^2/sec^-2) / meters^2

Cancel meters^2. And then, because radians isn't technically a unit, you can put it in there. Sorry, I don't know how to properly explain that bit.

Alpha = n radians/sec^2

So your result was alpha = 0.00002338158 radians/sec^2

I hope this helps,

Steve

[an error occurred while processing this directive]---------- FOLLOW-UP ----------

QUESTION: Hello Steve,

Thanks for you answer. I think I understand well enough. What I am trying to do now is determine the degrees per second the object would rotate.

When the formula gives me 0.00002338158 radians/sec^2 what can I do to find out how many degrees per second my object will rotate?

My first thought would be to just square the number and then convert it to degrees per second using one of the many online converters available, but then I was not sure if it would be that simple...

Is it possible to convert radians/sec^2 to degrees per second?

Thanks again for all your help,

Eddie

ANSWER: Hello Eddie,

Constant angular acceleration of the object will cause its rotation speed to increase linearly. The increase will be in a manner analogous to the way your car's speed increases when you ask the engine for acceleration.

Assuming the value of alpha is constant as the object rotates, the rotation speed will increase according to the formula

Wf^2 = Wi^2 + 2*alpha*theta

where Wi = 0, alpha = 0.00002338158 radians/sec^2, and theta is the expected rotation. I believe the expectation is that the object will rotate 90 degrees. If that's true, use theta = pi/2 radians. So plug in the data and work the equation. The square root of the value you get for Wf^2 is the final rotation speed as it reaches the stop.

The units of Wf will be radians/sec. To convert that to degrees/sec:

multiply the speed in radians/sec by (180 degrees/pi radians).

I hope this helps,

Steve

---------- FOLLOW-UP ----------

QUESTION: Hello Steve,

I have done the calculations and have come up with a rotation speed of 0.00237449581 degrees/sec. Is it fair to assume that a rotation speed this low means that in reality the object will not actually rotate at all?

Also, this may sound like an extremely basic question, but squaring a number less than 1, e.g. 0.00002338158, results in a smaller number. Is this what should happen in these calculations? Am I missing something important/basic here?

One final question, and I am sorry to keep peppering you with questions here. The value I got for Wf^2 was 1.71750332e-9, (should that be 1.71750332e-9^2) and the square root of 1.71750332e-9 is 0.00004144277. Have I done that correctly?

Thanks again, and have a merry Christmas!

Eddie

My calculations have been included below in case you want to see them...

Wf^2 = Wi^2 + 2*alpha*theta

Wf^2 = 0 + 2 * 0.00002338158 radians/sec^2 * pi/2 radians

Wf^2 = 0 + 2 * 5.4669828e-10 * 1.57079632679

Wf^2 = 0 + 1.09339657e-9 * 1.57079632679

Wf^2 = 0 + 1.71750332e-9

Wf^2 = 1.71750332e-9

The square root of 1.71750332e-9 is 0.00004144277

Converting to degrees/sec

0.00004144277 radians /sec * (180 degrees/pi radians = 57.2957795131)

0.00004144277 radians /sec * 57.2957795131 = 0.00237449581 degrees/sec

Hello Eddie,

Your question (about 0.00237449581 degrees/sec being no rotation) doesn't need a reply because I spotted an error in your math. Your question about numbers less than one getting smaller when you square them is yes. Consider this:

one half squared = one fourth

(1/2)^2 = 1^2 / 2^2 = 1/4

The value you got for Wf^2 is wrong. The fact that the units on alpha is radian/sec^2 does not mean that you should square 0.00002338158. Think of the value of g, the acceleration due to gravity:

g = 9.8 m/s^2

The fact that the units have seconds squared in the denominator means that each second, the speed of a falling body increase by 9.8 m/s. So assuming no air resistance

time...speed

0......0

1 s....9.8 m/s

2 s....19.6 m/s

3 s....29.4 m/s

4 s....39.2 m/s

etc.

So you should not have squared 0.00002338158. The fact that the sec term is squared allows you to have units of radians/sec (no "^2") after you take the square root of Wf^2. Note that I leave the units in place while doing math so I can see if things go awry.

Wf^2 = Wi^2 + 2*alpha*theta

Wf^2 = 0 + 2 * 0.00002338158 radians/sec^2 * pi/2 radians

Wf^2 = 0 + 2 * 2.338158e-5 radians/sec^2 * 1.57079632679 radians

Wf^2 = 0 + 7.345539996e-5 radians^2/sec^2

Wf^2 = 7.345539996e-5 radians^2/sec^2

Wf^2 = 0.734559996e-4 radians^2/sec^2

The square root of 1.71750332e-9 radians^2/sec^2 is 0.857064756e-2 radians^2/sec^2

So Wf = 0.00857064756 radians/sec

Converting to degrees/sec

Wf = 0.00857064756 radians/sec * (180 degrees/pi radians = 57.2957795131 degrees/radian)

Wf = 0.00857064756 radians /sec * 57.2957795131 degrees/radian = 0.491061933 degrees/sec

I hope this helps,

Steve

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I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.

I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.**Education/Credentials**

BS Physics, North Dakota State University

MS Electrical Engineering, North Dakota State University