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

__Rotation__

Let's consider a

*rotation*of an object within XY-plane around Z-axis along a circular trajectory of radius

**with a center at the origin of coordinate.**

*R*Obviously, its Z-coordinate will always be zero, that is

**, but X- and Y-coordinates will somehow depend on time**

*z(t)=0***.**

*t*We would like this motion to be

*uniform*in some sense, but we have to define what we really mean calling this motion a

*uniform rotation*.

Recall that the term

*uniform motion*we used in a sense that an

object moves along a straight line with constant velocity, covering the

same distance in the same intervals of time.

When dealing with

*rotation*, the concept of distance along a straight line should be replaced with a concept of

*angle of rotation*. Inasmuch as we cover the same linear distance in

*uniform motion*during the same intervals of time, in case of

*uniform rotation*we cover the same

*angle of rotation*during the same intervals of time.

Assume that in the beginning of rotation our object is at point

**}**

*R, 0, 0***from the origin of coordinates. Let's introduce a new variable -**

*R**angle of rotation*as a function of time -

**- an angle from the positive direction of the X-axis counterclockwise to a**

*φ(t)**position vector*of our object at time

**.**

*t*Since during the rotation along a circular trajectory the distance between an object and the origin of coordinates is constant

**, we can express coordinate functions as functions of an**

*R**angle of rotation*, like in polar coordinates:

*x(t) = R·cos(φ(t))*

*y(t) = R·sin(φ(t))*Since we are analyzing the

*uniform rotation*, and this uniformity

means that we rotate by the same angle during the same intervals of

time, we define this motion as a rotation with constant

*angular speed*

**- constant first derivative of an**

*φ'(t)**angle of rotation*by time.

Assume,

*angular speed*

**(constant).**

*φ'(t)=ω*Then

*angle of rotation*

*φ(t) = ω·t + φ*_{0}

*φ*_{0}is an angle from the positive direction of the X-axis to a vector from

the origin of coordinates to a position of our object at the beginning

of motion at

**. Since we assumed that the object starts from a point**

*t=0***}**

*R, 0, 0***and the**

*φ*_{0}=0*angle of rotation*equals to

*φ(t) = ω·t*Now we can easily find the coordinate functions of the position of an object:

*x(t) = R·cos(ω·t)*

*y(t) = R·sin(ω·t)*

*z(t) = 0*Taking the first derivative from these functions, we find the components of the

*velocity vector*:

*x'(t) = −R·ω·sin(ω·t)*

*y'(t) = R·ω·cos(ω·t)*

*z'(t) = 0*Out of curiosity, let's calculate the scalar product of the

*position vector*

**}**

*x(t), y(t), z(t)**velocity vector*

**}.**

*x'(t), y'(t), z'(t)*It's equal to

[

**]**

*R·cos(ω·t)***[**

*·***]**

*−R·ω·sin(ω·t)***[**

*+*

++

**]**

*R·sin(ω·t)***[**

*·***]**

*R·ω·cos(ω·t)*

*=*

= 0= 0

As we know, if the scalar product of two vectors is zero, they are perpendicular to each other. Therefore, a

*velocity vector*in

*uniform rotation*is perpendicular to a

*position vector*.

Since the latter is a radius from a center of a trajectory circle to a

point where the object is located, the perpendicular to this radius -

the

*velocity vector*- is a tangential line to this circle at this point.

So, we have found an interesting property of a

*velocity vector*during a

*uniform rotation*- it's always tangential to a trajectory at a point of position of a rotating object.

Taking derivative from velocity functions, we find the components of the

*acceleration vector*:

*x'(t) = −R·ω²·cos(ω·t)*

*y'(t) = −R·ω²·sin(ω·t)*

*z'(t) = 0*Notice that this vector is equal to a

*position vector*multiplied by

**. It means that its direction is exactly opposite to a**

*−ω²**position vector*, that is directed towards the center of rotation.

So, we have found an interesting property of an

*acceleration vector*during a

*uniform rotation*- it's always directed towards a center of rotation.

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