Angular velocity and Switching Systems

This is an Illustration and not a science link because it just illustrates your thought processes. I'm sure that everything in here you could figure out yourself. Angela Merkel could (but possibly not D. Trump).

We use the picture from the "Hitting Something" subchapter. A sword rotates around an apparent (or instantaneous) center of rotation. How fast does some point P somewhere on the blade move?


Movement described by a rotation around an apparent center of rotation

We assume of course a constant rotational speed or velocity. That means that the sword will finish a full circle in some time T_cycle. It thus makes 1/T_cycle = n full rotations per time unit. As a unit for the rotational speed you thus could give the number of revolutions per minute and call that "rpm". You just as well could do the number of revolutions per second and then you would call that the cycle frequency n with the unit "hertz" [Hz]. And yes, 1 Hz = 1/s.

So how fast does a point P travel that is found at some distance r_P from the apparent center of rotation? Since it travels on a perfect circle with radius r_P, it covers a distance equal to the circumference C_P of a circle with radius r_P for every revolution. C_P is equal to 2pr_P as even D.T. might know. Since we have n rounds per second, the total distance covered in 1 second and thus the speed is 2pn · r_P.
That gives us an idea: We can save a lot of writing if we don't use the cycle frequency n_cycle = 1/T_cycle but something called the circular frequency (or circle frequency) w = 2pn.
It is a simple as that. Using w instead of n or 1/T or, God forbid, rpm, makes writing equations much more efficient. So everybody who is anybody uses the circular frequency and nothing else for rotational velocities.

So how fast does a point P at some distance r_P from the apparent center of rotation travel? Here it is:

v_P	=	w · r_P
	=	w · (d_P + d_r)

for the point P shown

So far, so easy. But what happens if we do not describe the movement of the sword as a rotation around the apparent center of rotation (COM) but, in the way discussed at length in the backbone , as a combination of a pure translation for the center of mass plus a pure rotation around the center of mass. This is schematically shown below.


Movement described by a translation of, and a rotation around the COM

The point P then moves with the speed of the COM which is v_COM = w · d_r plus the rotational speed for the rotation around the COM which is v_{rot, COM} = w_COM · d_P . The circle frequency w_COM for the rotation around the center of mass is identical to the one we had before since, as the picture shows, the same rotation angle applies, i.e. w_COM = w,
This leaves us with

v_P	=	w_COM · d_P + w · d_r
	=	w · (d_P + d_r)

Same as above. It also works for the kinetic energies E_kin They must the same in both cases. With = moment of inertia for rotation around the apparent center of rotation, and I_COM = moment of inertia for rotation around the COM, we get for the kinetic energy using the rotation around the apparent center of rotations:

E_kin	= ½	I_A · w²
	= ½	(I_COM + md_R²)w²

We used the parallal axis theorem for this; m is the mass of the sword.
Calculating the kinetic energy for the COM systen we get

E_kin	= ½	(I _COM w² + mv²_COM)
	= ½	(I_COM + md_R²)w²

Just as it should be. I rest my case.

With frame

12.4.4 Impact

12.4.3 Hitting Something