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# Physics/Calculating the speed of a rotation

Question
QUESTION: Hello,

Thank you for all your helpful answers in the past. I am interested calculating motion. I am not sure if my question falls under Newtonian Mechanics, having never studied the subject myself, but I was wondering if you might be able give me some advice.
I have a hypothetical object that I want to rotate, and I want to calculate how fast it will move.
The object is a metal cube that is suspended on a rod running through a hole in the middle of the cube. The turning point that the cube rotates on is always fixed. Underneath the cube is an electromagnet. The electromagnet attracts the cube to rotate it 90 degrees.
I know the weight of the cube, and I know the strength of the electromagnet, in Newtons. What I want to do now is to calculate the time that this 90 degree rotation would take. If I know the weight of the rotating object, and the degree of force which causes this rotation, is there a way I can calculate how fast the rotation would be? Is it possible to calculate this rotation speed with only the two pieces of data listed above, or do I need something more?
Any advice would be greatly appreciated!

Thanks again, and best regards,
Eddie

I'll assume that the cube will experience a constant angular acceleration, alpha. 90 degrees of rotation is an angular displacement theta = pi/2 radians. The angular velocity will accelerate during the pi/2 radians rotation. We can get an expression for the final angular velocity. Wf, sing the angular kinematic formula
Wf^2 = Wi^2 + 2*alpha*theta
where Wi = 0 and theta = pi/2. So we find that
Wf^2 = 2*alpha*pi/2

So we need an expression for alpha. Alpha is related to the torque, tau, that we get from the force, F, by
tau = I*alpha
where I is the rotational inertia. So we need expressions for tau and for I.

The electromagnet will develop the torque, tau, that causes the angular acceleration, alpha. To get a valid expression for tau may require knowing more about how the electromagnet acts on the cube. But perhaps it would be useful to find a starting point using an assumption that the force, F, is applied at a distance d/2 from the pivot. (This might be true when the nearest face is perpendicular to the magnetic field. But I have doubts that it would be valid after 30 degrees of rotation.)
tau = Force * lever arm = F*d/2

The rotational inertia, I, is related to the geometry of the object. I assume the hole through the cube is from the center of one face to the center of the opposite face. According to the site
http://en.wikipedia.org/wiki/List_of_moments_of_inertia
the formula for the rotational inertia of such a cube is
I = (1/12)*m*d^2
where m is the mass and d is the length of an edge. And now we need an expression for the torque, tau.

We now can write an expression for the alpha in the formula of the 1st paragraph. From the 2nd paragraph,
alpha = tau/I
And from the 3rd and 4th paragraphs
alpha = (F*d/2) / ((1/12)*m*d^2)
Plugging that into the 1st formula
Vf^2 = 2 * [(F*d/2) / ((1/12)*m*d^2)] * pi/2
See if the information you have, mass and force etc., will work in that formula.

I hope this helps,
Steve

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

Electromagnetic rotati
QUESTION: Hello Steve,

Thanks very much for your very thorough answer, appreciate it. I have been trying comprehend what you told me, and after thinking about it for a week I think I might be starting to get a handle on it. I am just wondering if I could ask you more about the point you mentioned regarding the force being applied at a distance d/2 from the pivot when calculating the torque.
I have included some drawings of how the electromagnet is envisaged to work on rotating the cube, shown in the bottom three images. (I am not sure if the images show it properly, but the rotation of the cube is stopped at 90 degrees by the plastic housing of the electromagnets, which physically prevents the cubes
from rotating further)
Assuming I have reasoned this motion correctly, I am wondering where/how I should measure in order to get the figure for the  "Force * lever arm = F*d/2" you told me about in calculating the tau. Would the distance
from the center of the ferrous metal section to the center of the electromagnet be correct, or would the edge of the metal section to the closest point of the electromagnet be better? (Or something else I have not thought of, which seems more likely!)

Also, and this is really revealing the extent of my ignorance here, I am not sure what units of measurement to use in these calculations. For example, would the value for Force be written in Newtons or gram force? How about the Distance and the Mass?  I have read that the Angular Velocity is written in radians per second, but having never done anything like these types of calculations before, I am not sure what unit to use for the other numerical expressions.

Thanks again for all your help. In ancient cultures people used to think of  science as magic. I am now starting to understand they felt!

Best regards,
Eddie

Vf^2 = 2 * [(F*d/2) / ((1/12)*m*d^2)] * pi/2

If F is in Newtons, d is in meters, and m is in kilograms, then Vf will be in radians/second. A radian is about 57 degrees. A rotation of 90 degrees is a rotation of (pi/2) radians.

About the problems I mentioned. I assumed that the cube was a uniform cube. The formula for rotational inertia I used
I = (1/12)*m*d^2
was for a solid, uniform (or homogeneous) cube. The rotational inertia of your sort-of cube would not be determined using that formula. Some kind of test would be the way I would go about finding its rotational inertia.

The other thing that I suspect is a problem is the orientation of the electromagnet versus the ferrous metal portion of your cube. You said you know the strength of the magnet in Newtons. The force can't really be constant throughout the rotation of the cube. The magnetic field would not be uniform in the location of the ferrous metal portion of the cube as it rotates. So I think the force would vary with position.

I feel like I don't know enough about the situation to allow you to put much faith in the answers I give. I suggest that building a model and testing it is the best approach.

Regards,
Steve

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

QUESTION: Thanks Steve,

I understand. You are right, I should build a model, and I actually plan to do so in the near future. Before I build the model, it would be really useful to get a sense of whether the angular velocity might fall into a certain range, because if it falls outside this range, I may have to alter the design - something it would be good to know before building a test model.

At this point, I am really just trying to determine whether the angular velocity would fall into an acceptable range for my project, or rather, determine that it does not go below a certain level - above/faster is fine. For that reason, I am wondering if I could still use the  I = (1/12)*m*d^2 formula for rotational inertia you gave me for the homogenous cube.
Given that the mass of my cube-like object is actually less than I originally led you to believe (sorry), is it possible to say whether the moment of inertia will be lower or higher than rotational inertia for the homogenous cube shape you told me how to calculate?

Regarding the torque, if I can calculate the force exerted on the ferrous metal section by the electromagnet when the distance between the two objects is furthest and therefore weakest, could I use this value to calculate a minimum value for the torque?

Actually, when I calculate the torque using the F*d/2 formula you gave me, something confusing happens. I was playing around with the numbers and did some calculations like this –
A. 0.001 gram force x 0.02 meters = 0.00002
B. 0.001 gram force x 0.002meters = 0.000002
The distance in calculation A is larger than for calculation B, but the torque is greater in A. Have I done this correctly? I assumed that a smaller distance would result in more torque.

Thank for very much again for all your help. I would say it has been invaluable, but it I am finding out that it seems absolutely everything can be quantified! ;-)

I hope you had an enjoyable vacation.
Best regards,
Eddie

Hello Eddie,

That seems like a good plan. I'll try to make some reasonable approximations and see if I can help you predict the angular velocity of your device after 90 degrees of rotation. Your cube-like object, I'll call it the rotor, will not have the rotational inertia of a pure cube, but I'll try to make some adjustments to get an expression that may not be far off.

There's a detail that I don't know about your rotor. Does the shaft rotate with the rotor? If so, the rotational inertia of that should be considered too. If not, the fact that the rotor has a hole through it affects the rotational inertia. However, the material close to the axis of rotation has a smaller contribution than material farther from the axis. The density of the plastic, the shaft through the center, and the ferrous metal section would be good to know, but I'll estimate that the metal parts are twice as dense as the plastic. Of course if you have details on that, let me know. Now I have to admit something: When I looked up the formula for the rotational inertia of a cube, what I found was the formula for a rectangle. And I made an error adjusting that for a cube. The formula should have been
I = (1/6)*m*d^2

Now I'll try to make adjustments on that formula. Because one corner of your cube is rounded off, it's as if about 1/4th of your shape is a cylinder. The formula for the rotational inertia of a cylinder is
I = (1/2)*m*r^2
where in this case the radius r = d/2. So the rotational inertia is given by
I = (1/2)*m*(d/2)^2 = (1/8)*m*d^2
But then the ferrous metal section is where the corner of the cube would have been, so that tends to offset the fact that the shape is cylindrical in that portion of the rotor. So it may seem that I'm trying to make this too easy, but I think you can use
I = (1/6)*m*d^2

So that changes the formula for Wf to that I gave you a few followups ago.
Wf^2 = 2 * [(F*d/2) / ((1/6)*m*d^2)] * pi/2

If you plug in the data F in Newtons, d in meters, m in kilograms, then Wf is in radians/s. You may like a prediction of the time to rotate 90 degrees, it would be
theta = (Wf + Wi)*t/2
where theta = pi/2 and Wi = zero for this situation. The time will be in seconds.

There are some things I don't think I know enough about to be sure how close to reality this will be. The validity of my approximations above is uncertain of course. But I'm also concerned about my lack of knowledge about the friction this will have to overcome. It could be that it doesn't move at all if the friction is too high. If the friction is half as great as the force exerted on the ferrous metal section, it would cut the acceleration in half and decrease the final velocity by a factor of 0.707. And I don't really expect that the torque throughout the rotation will be constant.

1. If you can calculate a minimum value for the torque using the force for the greatest distance between the electromagnet and the ferrous metal section. There's a complication in the way torque is calculated. Think of a wrench. It's best to apply your force perpendicular to the handle. Sometimes when working on an engine, there are obstacles that prevent me from applying my muscles perpendicular to the handle. In that case
tau = Force * lever arm * sin(angle between Force and lever arm)
And that will happen as the rotor rotates. The force may get larger because the ferrous metal section gets closer to the electromagnet. So the force may increase as sin(angle between Force and lever arm) decreases, minimizing the variation in the torque. But of course sin0=0, so the torque will approach zero as the rotor approaches 90 degrees of rotation. Hopefully momentum will carry it to the stop.

2. You looked at the change in torque as the lever arm changes length. You did that correctly. Your assumption about what would happen to the torque as the length changes was wrong. Again think about a wrench. You get most torque when you grasp the handle at the end. If you want to loosen a bolt that is really tight, you can increase your torque by using a longer handle.

I hope this helps,
Steve
Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thank you! Clearly explained and thorough beyond reasonable expectation. You are the Physics Professor I never had.

Physics

Volunteer

#### Steve Johnson

##### Expertise

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.

##### Experience

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