## Monday, April 23, 2018

### Unizor - Physics - Mechanics - Kinematics - Rotation

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 R with a center at the origin of coordinate.

Obviously, its Z-coordinate will always be zero, that is z(t)=0, but X- and Y-coordinates will somehow depend on time 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} - on the X-axis at distance R from the origin of coordinates. Let's introduce a new variable - angle of rotation as a function of time - φ(t) - an angle from the positive direction of the X-axis counterclockwise to a 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 R, we can express coordinate functions as functions of an 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 φ'(t) - constant first derivative of an angle of rotation by time.

Assume, angular speed φ'(t)=ω (constant).

Then angle of rotation φ(t) = ω·t + φ0, where φ0
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 t=0. Since we assumed that the object starts from a point {R, 0, 0} on the X-axis, φ0=0 and the 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)} and the 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

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 −ω². It means that its direction is exactly opposite to a 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.