On finding Torque & RPM

I’ve just got into making motors and trying to figure out useful equations…. Presently, three things I need clarity on, please.

Thing 1:

I’ll be using the following equation to arrive at fair approximations of torque and rpm, given various motor rotors of different radii, masses, and with different stator coils (electromagnets) providing the repelling force….

T = I x alpha = F x r x sin(theta):

By example…
A rotor Mass of 100 kg,
‘r’ of 2 meters,
A force of 75 Newtons,
point of force at a 90 degree angle.

First filling in T = F x r x sin(theta), as such:
T = 75 N x 2 m x sin(90) = 150 N*m
T = 150 N*m

Next, add the other part of the equation, T = I x alpha; I = 1/2Mr^2, hence:
T = 150 N*m = 1/2M x r^2 x alpha
Resolved progressively…
T = 150 N*m = 50 kg x 2^2 x alpha
T = 150 N*m = 50 kg x 4 x alpha
T = 150 N*m = 200 x alpha

Next, for alpha:
alpha = 150 divided by 200 = .75 rad/s^2

Next, convert .75 rad/s^2 to RPM using this equation:
([rad/s]/6.28318530718) x 60 = RPM
*** Notice, rad/s is not squared in this equation. That is correct? ***
(Note: 6.28318530718 was Pi x 2, I just filled it in….)
Solved progressively…
([.75 rad/s]/6.28318530718) x 60 = RPM
.11936620731891365 x 60 = RPM
.11936620731891365 x 60 = 7.161972439134819 RPM
.75 rad/s^2 = 7.161972439134819 RPM

Correct? Fixable?

Little side note: Seems to produce realistic results when using metric, but it becomes a tangled mess when I attempt imperial. Just something on the side I’m struggling with at present — converting newtons to lb-ft, oz-in; meters to inches, so on. If any direction on a solid means for converting metric to imperial (and visa versa) as concerns using equations like this…. I seem to falter miserably when putting in the conversions I come up with. Hadn’t planned on posting this issue here and now, informally, but there it is, anyway….


Thing 2: Please refer to attachment "thing 2".

With the angle known, and the distance between A & B known, can the length of line 3 be found; line 3 being perpendicular to line 2 (as needed).

I think I read somewhere that if the hypotenuse is multiplied by sin(theta), it will give the length of the ‘opposite’.

Thing 3: Some clarity on ‘r’, please.
> The ‘r’ in the torque equation is that magical perpendicular moment arm. Pretty sure I got that, but please set me straight if not.
> In finding moment of inertia, using I=1/2Mr^2, ‘r’ is strictly the radius of the mass in question; never in any way having to do with where the point of action (of the repelling force) is? Not 100% clear on that. The obvious answer seems ‘yes’, but uncertainty creeps in when, in ‘thing 1’, converting rad/s to RPM. I’m thinking that if it’s the circumference of the rotor that we are using to arrive at rad/s and/or RPM, the ‘r’ in this case might be the distance from the axis to the point of force, which is slightly outside the tangible rotor, where the actual repelling force created by the interaction of the rotor magnet and the stator coil occurs; and hence a slightly greater radius, circumference, distance…., see what I’m getting at? Leave it be if not. My capacity for these things is only so forgiving here at my level….

My apologies if the questions could have been clearer.

Sincerely grateful for your guidance here.

Attached Images
File Type: bmp Thing2.bmp (60.4 KB)


Leave a Reply

Name *
Email *