# Angular acceleration of one rotating object fixed to another rotating object

1. Nov 28, 2008

### p21bass

1. The problem statement, all variables and given/known data
Given that a wheel with w(1) = -44i, is mounted on a turntable with w(2) = 35k, determine overall angular velocity and acceleration. Give both magnitude and direction.

2. Relevant equations
a = w(1) X w(2) ?

3. The attempt at a solution
I got the correct angular velocity is 56.2 rad/s, at 38.5 degrees from the x-axis towards the z-axis. But my text (and in-class instruction) only showed how to get the angular acceleration by way of either angular velocity or the angle itself. Since the radii of the wheel and turntable are not given, and since there is no time given, I do not know how to go about solving this. Per the answer in the back of the book is 1540j rad/(s^2), the only way I can figure they did this is they simply crossed the two angular velocities. Is that correct? Or is there something else that I've completely missed?

2. Nov 29, 2008

### alphysicist

Hi p21bass,

The angular acceleration they are asking about is related to the change in the combined angular velocity vector for some time period, so you can calucate it directly.

You found that the angular velocity makes is in the xz plane. As time goes on, that vector will rotate about the z axis, and so will change.

So what you can do here is to find what the combined angular velocity vector is at a time dt after the "start time", subtract the angular velocities and divide by dt:

$$\alpha = \frac{ \omega (\mbox{at time t = dt}) - \omega(\mbox{at time = 0})}{dt}$$

The first step is to find out what w(1) becomes at time dt; after that you can mostly follow what you have already done before.

3. Nov 30, 2008

### p21bass

That doesn't help. The only info given are the angular velocities of the wheel and turntable. There is no time period info given, or able to be derived. There is no radii info given, or able to be derived. There is no linear velocity or acceleration given. What I don't understand is how to determine angular acceleration when none of that information is available.

4. Nov 30, 2008

### alphysicist

You can evaluate the expression in my last post with the information given. For the instantaneous angular acceleration the time period is dt, which will later be cancelled out of the expression.

The important thing is that the angular velocity vector of the wheel is rotating around the z-axis; it's rotating in the x-y plane at a rate of 35 rad/s because the turntable is carrying it around.

After a time dt, the wheel's angular velocity, which was completely in the x direction, now has an x and a very small y component (again, because the turntable is carrying it around). Once you determine those components at the time dt, you can use that in the derivative formula in my last post and evalute to get the answer.

So the wheel's angular velocity was (- 44 i ) at t=0; what is it at t=dt?

5. Nov 30, 2008

### p21bass

No, you can't. The only information given is the wheel's angular velocity is -44i rad/s, and the turntable's angular velocity is 35k rad/s. There IS NO "dt" - as mentioned in both of my posts. If t, or "dt", is given, or any other info was given, I wouldn't need to be here.

Again, the ONLY information given is the angular velocities of the wheel and turntable at a particular instant. Since there's only one value for each, your equation (which is technically the average angular acceleration over dt) is inappropriate for the problem.

Last edited: Nov 30, 2008
6. Nov 30, 2008

### alphysicist

My equation was just the definition of the derivative (I left the limit of dt going to zero implied) and so gives the instantaneous acceleration. You are not "given" a value of dt to use the formula.

Instead of using the definition from my post, if you want another route, just write down what the angular velocity of the wheel will be at any time t. At t=0 it's given (44 in the -x direction); the rotation around the z axis tells you how long it takes to go completely around. So what is the x and y components of the wheel's angular velocity at an arbitrary time t?

(In other words, you could find the period of its motion if you wanted--it's rotating around at 35 rad/second, so you could easily find how long it takes to go 2 pi radians. After a quarter period it's pointing in the -y direction, after a half period it's pointing in the +x direction, after 3/4 period in the +y direction, and after a full period it's back pointing in the -x direction. So what are it's components at any arbitrary time t?)

Then use that to find a time dependent formula for the combined angular velocity at any time t, and take the time derivative to get your answer.

(Again, I stress that the two angular velocities are all you need; the turntable's angular velocity tells you how the other's angular velocity moves in the x-y plane.)

So here is the starting point: find an expression for the wheel's angular velocity at any arbitrary time t. At t=0 it is -44 i; what is it at any time t?

Last edited: Nov 30, 2008
7. Nov 30, 2008

### djeitnstine

This is indeed correct. When crossing I get the same answer as in the back of your book.

8. Nov 30, 2008

### alphysicist

Hi djeitnstine,

p21bass recognized in the original post that the cross product gives the correct answer, and they were asking if this was always true, or if it was just an accident.

It is not always true; it only happens to give the correct result here because the angular velocity vector of the wheel is constant in the rotating frame of reference.

The procedure I described in my last post (find the angular velocity as a function of time and then take the derivative) shows why that result is correct, and also gives the correct result even if the vector is not constant in the rotating frame.

9. Dec 1, 2008

### p21bass

That's all I was asking. The rest may be accurate, but was unnecessary for this problem, as stated. I'm sure I'll get a better explanation tonight from my instructor, as he goes over our homework.

Also, Hibbeler's Engineering Mechanics: Dynamics, 11 Edition, p 540-541 (different text, as my Physics class uses Giancoli's Physics for Scientists & Engineers, 4th edition) gives a good example of this sort of problem.

10. Dec 1, 2008

### alphysicist

Let me be clear: it looks like your answer is incorrect to me.

And I don't see how the procedure is unnecessary for this problem; it is the problem! Do you understand why the cross product needs to be in a certain order for this problem? or why the cross product gives the time derivative in the first place?

Why are you trusting the book's answer without finding out why its answer is right or wrong? Textbooks are by no means infallible--I've found plenty of errors in textbooks.

I don't understand why you are so against just trying to do the problem. But here are two older threads with the same kind of question you have:

11. Dec 2, 2008

### p21bass

The answer is correct. The method explained in the Hibbeler text gives the same answer as the answer in my class' text. And, it allows you to solve the problem in one step. I gave the method you defined a go, and came up with the same answer. This method is likely the beginning of the derivation for the equation shown in Hibbeler's text. However, it is a little more complicated, as it requires three steps - definition of the position of the wheel, differentiation, and then solving.

The angular velocity of the wheel (for this instance) can be shown to be w = -44i(cos35tk + sin35tk) - where 35t ends up being the angular position of the wheel with reference to the turntable. And the derivative: w' = -44i(-35sin35tk + 35cos35tk). Since t = 0 for the instant asked, sin(0) = 0, and cos(0) = 1. Therefore, w' = -44i(35k) = -1540jrad/(s^2).

This is the same answer as listed in the Giancoli text, and same answer found in using the method of the time derivative shown in the Hibbeler text.

I was not against trying to do the problem. My only question was if the cross-product of the two vectors was all that was necessary to solve the problem. As shown by the Hibbeler text, no - there's a little more to it than just that. But, when you define two coordinate systems (XYZ - fixed, and xyz - rotating), the other factors become zero, so it ends up that you are left with the cross-product.

I do, however, appreciate your help.

12. Dec 2, 2008

### alphysicist

That was my point: in your original post you said the answer was

angular acceleration = w(1) X w(2) = 1540j rad/(s^2)

and that is not correct. (That cross product does not give the angular acceleration, and 1540j rad/(s^2) is not the answer to the original problem.) I agree that w(1) X w(2) does equal 1540j rad/(s^2), but that is not the angular accleration.

angular acceleration = w(2) X w(1) = -1540j rad/(s^2)

because the angular acceleration is in the negative y directions (at t=0 the angular velocity vector is in the -x direction and rotating counter-clockwise). The order of a cross product is very important (as I'm sure you know if you are reading through a dynamics text) and the original problem specifically asked for the direction.

13. Dec 2, 2008

### p21bass

You're right. In typing my original post, I accidentally omitted the "-" in front of the magnitude.

In looking back at what I did by following the method I found in the Dynamics text - yes, it was w(2) x w(1). I'm pretty sure I understand that method - and I realized what I was doing - but somehow had a brain-fart in relaying that information.

Thanks for catching that for me.

14. Dec 2, 2008

### alphysicist

Oh, I see. I was hoping we had found a mistake in the textbook--that's always fun to do!