# Moment of inertia of a whisk

1. Jun 22, 2007

### remz

Hi,

So I have a DC motor powering a whisk (http://upload.wikimedia.org/wikipedia/commons/8/85/Schneebesen1.JPG). Imagine the whisk on the end of the motor shaft. I'd like to know what torque requirements I have for the motor to ensure that it can successfully spin the whisk.

Background

To accelerate the whisk to a speed w the motor must generate a torque T such that:

Tm = Jl x dw/dy

where:

Tm = torque generated by the motor
Jl = inertia of the whisk
dw/dy = angular acceleration of the whisk

The Problem

In order to solve this equation I need to determine the inertia of the whisk, Jl, such that:

Jl = m x R

where:

m = mass of whisk (80gms)
R = distance of whisk from point of rotation (0cms?)

So my question is basically how to calculate the moment of inertia for a whisk?

The confusion arises as it appears the centre of mass of the whisk lies on the axis of rotation, there is therefore no moment of inertia. As there is no moment of inertia, the motor generates no torque (or an infinite torque).

So, should I apply the above formula assuming that the centre of mass of the whisk is focused at the outer edge of the whisk...or is there a more logical approach.

Last edited: Jun 22, 2007
2. Jun 22, 2007

### Staff: Mentor

I'm not sure what a whisk is (picture?), but just because its center of mass is at the center of rotation, does not mean that it has zero moment of inertia.

To calculate the moment of inertia, you add up all the contributions of the pieces that are off-axis. Like for a rotating disk -- even though the center of mass is in the middle, you add up all the parts of the disk (including the effect of the distance from the rotational axis) to get the total I.

http://en.wikipedia.org/wiki/Moment_of_inertia

3. Jun 22, 2007

### FredGarvin

You are incorrect in your reasoning regarding the center of mass. Then center of mass is a theoretical concentration of all surrounding mass. All symmetric objects will have a center of mass along the main axis. That does not mean that moment of inertia is zero. If you look here, many objects are symmetric that have the center of mass along the main axis but do not have a zero moi:
http://en.wikipedia.org/wiki/List_of_moments_of_inertia

Usually with complex shapes, there are two methods; computer model calculated or physically measure the moi.

4. Jun 22, 2007

### remz

Thank you for taking the time to respond.

I used point mass to derive the moi. This was not in fact for a whisk but for an agitator to stir the contents of a chemical reactor similar to this http://www.dynamixinc.com/images/titan_2.gif [Broken]

Using J = mR^2,

I made the simplifications that the shaft did not exist and made assumptions for the masses of each of the blades. I then simplified the blades to point masses at a distance of half the length of the blade from the imaginary shaft. As my agitator has 6 blades, 2 rows of 3 blades, my moi became:

Jl = 6 x (mass of a single blade x half of length of blade^2)

...which gives me a good worst case moi, at least I hope.

Now, to continue the torque requirements of the motor I have to get a good indicator for the frictional torque imposed upon from the motor to stirrer connecting rod spinning through a shaft. But i'll leave that for another thread when I start running into difficulties.

Last edited by a moderator: May 2, 2017
5. Jun 23, 2007

### NoTime

No.
Notice the ^2 in the equation.
Try workin out J for a point mass of 2 at a length of 2 and a point mass of 1 at a length of 4.

Besides, the power required to spin up the rotor will be trivial compared to the resistance of whatever goo you are trying to mix.

6. Jun 23, 2007

### remz

NoTime: I calculated the moi for a single blade using:

J = m x r^2

and as moi are additive and as I have 6 blades of equal mass and equal perpendicular distance from the axis of rotation, I simply multiplied this value by 6:

Jl = 6 x (m x r^2)

Could you please elaborate on the error in my calculation? I carried out your problems and derived the following values:

J = mr^2 = 2 x 2^2 = 2 x 4 = 8

J = mr^2 = 1 x 4^2 = 1 x 16 = 16

Which tells me that the second point mass requires greater torque to accelerate it to a certain value than the first point mass.

Absolutely...of tens of magnitudes I would imagine. There are in fact two areas in my stirrer where dynamic resistance is significant - the first is as you say, in the drag of the rotor spinning through the fluid and the second is the friction of two solid bodies; the rotor shaft spinning through a tube which is used to hold the shaft in a vertical position and counter the effect of wobble (not shown on original picture).

Getting accurate values for these resistances will prove difficult I imagine. To simplify, I will ignore static and coloumbic friction as I imagine these will be difficult to define. Instead I will concentrate on only the resistances which are proportional to velocity and ensure that my motor is able to generate a starting torque of nominal magnitudes greater.

I'll let you know how I get on and which equations I shall be using. Your input is and will continue to be of great value.

Last edited: Jun 23, 2007
7. Jun 23, 2007

### NoTime

Ok, you get 8 an 16.
What that means is that the motor needs to supply 2 times the energy to move the weight of 1 at a length of 4 as it does to move the weight of 2 at a lenght of 2.

Try splitting your blade up into little chunks of M, calculate J for each chunk at its particular R, then sum all the Js.
Compare that to the J of a point M_total at .5R you started out with.

Any reasonable bearing will not generate enough resistance to bother including as a factor.

The goo is by far the biggest factor.
Probably the simplest solution is to put the whisk in the goo.
Measure how much torque it takes to turn it.
Multiply that by 2 and get a motor that can supply that.

8. Jun 29, 2007

### Staff: Mentor

In fact, torque may not be just a function the goo viscosity - it may be more related to total mass of goo. So, a test using your worst case goo - most massive/most viscous is in order.

Kitchen Aid mixers (tilt-head) come with a whisk, and they are rated by what their marketeers call 'flour-power'. A 250 watt direct drive motor can handle dough from up to 8 cups of flour, 325 watts handles 9 cups.

9. Jun 29, 2007

### FredGarvin

Indeed. If your bearings are adding that much resistance that you need to consider it in your motor calcs, you have larger issues at hand.

Best advice I have seen all day. Don't get mired down in the minute, hard/impossible details to describe and take care of the major elements.