*Notes to a video lecture on http://www.unizor.com*

__Relative Rotation__

The subject of this topic is a rotation of an object around some axis,

while this axis, in turn, rotates around another fixed axis (we will

consider only a case when both axes, fixed and rotating, are parallel to

each other).

Before going any further, let's refresh our knowledge about concepts of

*angular speed*,

*linear speed*and

*linear velocity*of a point-object rotating around an axis along a circular trajectory within a plane perpendicular to an axis of rotation.

If a point

**rotates on a circular trajectory with a center**

*A***, radius**

*O***is turning by certain angle**

*OA***during certain time**

*φ***. If at time**

*t***we mark the position of radius**

*t=0***as the beginning of rotation, angle of rotation from the beginning position of radius**

*OA***to its position at time moment**

*OA***can be considered as a function of time**

*t***. The unit of measurement of an angle that we will use is**

*φ(t)**radian*, so a full circle is a rotation by

**.**

*2π*We can talk about

*average angular speed*of rotation that we

define as average angle of rotation in a unit of time and can calculate

by dividing an increment of angle of rotation

**Δ**

*φ(t+***Δ**

*t)−φ(t)=***divided by increment of time**

*φ(t)***Δ**

*(t+***Δ**

*t)−t=***.**

*t*We can also talk about

*instantaneous angular speed*

**at any moment of time**

*ω(t)***as a limit of the above ratio as Δ**

*t***becomes infinitesimally small, which is a derivative of angle of rotation by time:**

*t*

*ω(t) = φ'(t)*For our purposes in most cases we will deal with rotations having a constant

*instantaneous angular speed*, which we will call just

*angular speed*, unless otherwise is explicitly indicated. In this case, if

**then**

*φ(0)=0***.**

*φ(t)=ω·t*Ret's put this rotation with constant

*angular speed*into a system of coordinates with an origin at point

*O*- center of rotation, XY-plane coinciding with a circular trajectory

and X-axis coinciding with the radius to a point where the rotation

begins.

Obviously, Z-coordinate of a rotating point-object is always zero.

X- and Y-coordinates can be easily obtain using trigonometric functions:

*x(t) = R·cos(φ(t)) = R·cos(ω·t)*

*y(t) = R·sin(φ(t)) = R·sin(ω·t)*If constant

*angular speed*of rotation along a circular trajectory is

**and radius of this circular trajectory is**

*ω***, the length of a circular trajectory covered by a rotating point during time from**

*R***to**

*t***Δ**

*t+***, as known from Geometry, equals to**

*t***[**

*R·***Δ**

*φ(t+***]**

*t)−φ(t)***Δ**

*= R·ω·***.**

*t*So, we can define the

*average linear speed*of rotation along a circular trajectory during time interval from

**to**

*t***Δ**

*t+***as**

*t***Δ**

*R·ω·***Δ**

*t/***.**

*t=R·ω*As we see, it does not depend on time and is constant. Therefore, we will call it just

*linear speed*of a point-object circulating with constant

*angular speed*.

*Linear velocity*is a vector with magnitude equal to

*linear speed*

**and directed always along a tangential line to a circle.**

*R·ω*To prove it, let's approach this concept using XY-coordinates. As we know,

*velocity*of a point-object is a vector with coordinates {

**,**

*x'(t)***}, where**

*y'(t)***and**

*x(t)***are coordinates of a point-object as functions of time and**

*y(t)***and**

*x'(t)***are the derivatives of coordinate functions.**

*y'(t)*Knowing the coordinate functions in case a point-object moves along a circular trajectory with constant

*angular speed*, as expressed above, let's find the

*linear velocity*during this rotation:

*x'(t) = −R·ω·sin(ωt)*

*y'(t) = R·ω·cos(ωt)*Scalar product of this

*velocity vector*and coordinate vector {

**,**

*x(t)***} is**

*y(t)*

*x(t)·x'(t)+y(t)·y'(t) =*

= −R·cos(ωt)·R·ω·sin(ωt) +

+ R·sin(ωt)·R··ωcos(ωt) = 0= −R·cos(ωt)·R·ω·sin(ωt) +

+ R·sin(ωt)·R··ωcos(ωt) = 0

This proves that vectors {

**,**

*x(t)***} (which is directed along a radius to a rotating point-object) and {**

*y(t)***,**

*x'(t)***} (velocity vector) are perpendicular and, therefore,**

*y'(t)**velocity vector*coincides with a tangential line to a circular trajectory.

As for magnitude of this

*velocity vector*, it's equal to

*v = √(x'(t))²+(y'(t))² =*

=√R²ω²sin²(ωt)+R²ω²cos²(ωt)=

= Rω=√R²ω²sin²(ωt)+R²ω²cos²(ωt)=

= Rω

That is, the magnitude of

*velocity vector*equals the

*linear speed*, which justifies the usage of term

*linear velocity*to emphasize its vector nature and physical nature of magnitude reflecting the pace of the rotation.

Now let's move to problems of relative rotation.

Assume that we have an original

*inertial frame of reference*with all actions occurring within XY-plane, so Z-coordinate is always zero.

A point

**is a center of rotation for point**

*O***, that rotates within XY-plane with constant**

*A**angular speed*

**around**

*ω*_{a}**along a fixed circle of radius**

*O***.**

*R*A point

**rotates in the XY-plane with constant angular velocity**

*B***around point**

*ω*_{b}**along a circle of radius**

*A***.**

*r*Mechanics of this combined motion can be represented by this picture

Here rod

**of the length**

*OA***rotates around center**

*R***with**

*O**angular speed*

**, while point**

*ω*_{a}**is a center of a disk of radius**

*A***spinning with**

*r**angular speed*

**with point**

*ω*_{b}**positioned on the edge of this disk.**

*B*Our task is to analyze the motion of point

**. In particular, we will analyze the**

*B**linear speed*of point

**in the original**

*B**inertial frame of reference*, where point

**is fixed.**

*O*We will use XY-coordinates and assume that at moment

**rod**

*t=0***is positioned along the X-axis and point**

*OA***is also on the X-axis to the right of point**

*B***, so initial angles of rotation of both, the rod and the disk, are zero.**

*A*At time moment

**vector**

*t**in the original inertial frame of reference will turn around fixed point*

**OA(t)****by angle**

*O***, and the position of a center of a disk**

*ω*_{a}·t**will be**

*A*

*x*_{a}(t) = R·cos(ω_{a}·t)

*y*_{a}(t) = R·sin(ω_{a}·t)These coordinates represent vector

*.*

**OA(t)**During this time disk will turn by angle

**, and vector**

*ω*_{b}·t*, that has a magnitude*

**AB(t)****, will have the following coordinates in the frame of reference associated with point**

*r***and axes parallel to those of original inertial frame where point**

*A***is fixed:**

*O*

*x*_{ab}(t) = r·cos(ω_{b}·t)

*y*_{ab}(t) = r·sin(ω_{b}·t)These coordinates represent vector

*.*

**AB(t)**The sum of two vectors, position vector of point

**in the original inertial frame, which is**

*A**, and position vector of point*

**OA(t)****in the reference frame associated with a center of rotation**

*B***of the disk, which is**

*A**, will be a position vector of point*

**AB(t)****in the original inertial frame, where point**

*B***is fixed.**

*O**=*

**OB(t)***+*

**OA(t)**

**AB(t)**Coordinates of this vector are:

*x*_{b}(t) = x_{a}(t) + x_{ab}(t)

*y*_{b}(t) = y_{a}(t) + y_{ab}(t)or, using the values of coordinates calculated above,

*x*_{b}(t) = R·cos(ω_{a}·t) + r·cos(ω_{b}·t)

*y*_{b}(t) = R·sin(ω_{a}·t) + r·sin(ω_{b}·t)Differentiating the above equations by time, we can find the

*velocity vector*of point

**in the original inertial frame of reference, where point**

*B***is fixed:**

*O*

*x'*

− r·ω_{b}(t) = −R·ω_{a}·sin(ω_{a}·t) −− r·ω

_{b}·sin(ω_{b}·t)

*y'*

+ r·ω_{b}(t) = R·ω_{a}·cos(ω_{a}·t) ++ r·ω

_{b}·cos(ω_{b}·t)Ret's now determine the magnitude of this

*velocity vector*, the

*linear speed*of point

**, as it goes along its trajectory.**

*B*

*v*

= √R²ω_{b}= √(x'_{b}(t))²+(y'_{b}(t))² == √R²ω

_{a}²+r²ω_{b}²+2Rrω_{a}ω_{b}cos((ω_{b}−ω_{a})t)
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