# A question about kinematics of a rigid rotor

1. Apr 21, 2016

### dorratz

1. The problem statement, all variables and given/known data

Hey guys,

http://athena.ecs.csus.edu/~grandajj/me143/1_Introduction_Tires_WM2D/1_4_2_Exercise_2.pdf

2 questions:
1. How can I undersdant that I should sum up the angular velocity(w) of both bodies into one w, so that the the disk itself has the total w? why the disk doesn't have just w2?
2. Why the solution for contains reference of Vc, but the addition of relative velocity is from E to D, and not from C to D?

Thanks,
Dor

2. Apr 22, 2016

### CWatters

It's been a long time since I did this but...

It's because you have one rotating disc mounted on another rotating disc. Suppose you stand at the north pole and turn around slowly with respect to planet earth. Then your angular velocity relative to the sun is equal to..

w1 + w2 + w3
where
w1 is your angular velocity relative to the earth.
w2 is the angular velocity of the earth relative to the sun (approximately 360 degrees per day)
w3 is the angular velocity of the earth as it orbits the sun (approximately 360 degrees per year)

The velocity of D with respect to B = (Velocity of C with respect to B) + (Velocity of D with respect to C)

Points E and C are on the same axis. So the velocity of E relative to B is the same as the velocity of C relative to B.

3. Apr 23, 2016

### dorratz

Hi,
As for the second question, I don't undersdand, because although the two points are on same axis, they aren't on the same cooardinate, so the distance vector between them and point B is different, so the vector product of angular velocity and distance vector will be different.

4. Apr 23, 2016

### CWatters

What matters is that they are the same distance from the axis of rotation (0.15m).

If you connect a motor to one end of a cylindrical shaft and rotate it at say 10rpm then all points on the surface of the shaft rotate at the same velocity regardless of how far along the shaft they are. The radius of the shaft affects the velocity of a point not position along it's length.

There is probably better way to explain this mathematically but I'm afraid very rusty.

5. Apr 25, 2016

### dorratz

Thanks a lot

If there is anyone who knows how to explain it mathematically I would thank him, cause by the formal math it should be different.

6. Apr 25, 2016

### BvU

$$| DE \times CE | = | DE |\ |CE | \\ | DC \times CE | = | DC |\ |CE | \sin(\angle DCE) = | DC |\ |CE |\ { |DE|\over |DC| } = | DE |\ |CE | \quad \rm ?$$and of course both are in the z-direction.