Applied Angular Velocity & Acceleration

[TJS]

Hey guys, I'm working on a product that rotates cups around a shaft, I am trying to calculate the torque required to do so. I have given it my best shot below, does this seem right to you?
The Main Shaft Motor will need to rotate holder one from the filling cups position to the pressing lids position within a 2 second interval (see attached for an illustrtaion). The angle of rotation between these two points is 29deg. As the motor will be required to rotate the shaft from a static equilibrium, an acceleration is required, the acceleration will be at it's max at the center point of this rotation just before it begins to decelerate. Therefore, selecting the motor to cope with the worst-case scenario of ∝ = maximum I will use the angular velocity at the point of half rotation.
The angle at the half rotation point is :
θ = 14.26 deg
The time required to get there will be half the time of a full rotation, therefore:
∆t=1-0
To find ω, θ needs to be in radians, therefore converting degrees to radians.
θ rad=14.26*0.01745329= 0.2489rad
Finding the angular velocity ω at the center point.
ω =( 0.2489)/(1 )
Using ω to calculate ∝ = maximum:
∝ = ( ∆ω )/(∆t )= ( 0.2489-0)/(1-0 )= 0.2489 m⁄s^2
Stepper motors can be selected based on Torque and Power requirements. Using angular acceleration and velocity the power and torque requirements can be calculated using the formulas below;
τ = I × ∝ and P = τ × ω
From the solid-works CAD model, the value for moment of inertia about the axis of the motor shaft is taken to be 0.197 Kgm^2 and therefore;
Multiplying the moment of inertia by a safety factor of five to ensure reliability, the moment of inertia I will take is:
0.9850 Kgm^2
τ = I × ∝ = 0.9850×0.2489=0.2452Nm

P = τ × ω= 0.2452 ×0.2489= 0.0610 W
Based on these calculations I have selected the NEMA Stepper Motor model number 42BYGHW60. This stepper motor can produce a torque between 0.16-0.56Nm.

 

[JRMichler]

Finding the angular velocity ω at the center point.
ω =( 0.2489)/(1 )

NO. To understand why, draw a couple of simple sketches. First, acceleration vs time with constant acceleration. Second, velocity vs time. Hand sketches are good enough, you don't need CAD for this.

Otherwise, it looks good. I prefer to put in a safety factor at the very end by putting it on the torque requirement. Don't forget to include friction.

 

[TJS]

NO. To understand why draw a couple of simple sketches. First, acceleration vs time with constant acceleration. Second, velocity vs time. Hand sketches are good enough, you don't need CAD for this.

Hi, do you think you could elaborate on what is incorrect here? Do you mean that ω =( 0.2489)/(1 ) is not the velocity at the center point but just the average velocity?
Does this effect my calculations or is it just a writing error?

 

[JRMichler]

The two sketches I recommended will communicate better than 1000 words. Sketch them, post them, and then ask.

I have done hundreds of these type calculations, and I still make simple little sketches of acceleration, velocity, and position vs time.

 

[TJS]

The two sketches I recommended will communicate better than 1000 words. Sketch them, post them, and then ask.
I have sketched these two graphs very quickly on MATLAB, i think I see what you mean, my worst case acceleration is actually as t approaches zero and not at t=1. If my graphs are correct and this is true, now I don't know what acceleration to take as I can't take it when t=0 as this would give me a value of infinity.
What do you think?
Acceleration Vs Time

Velocity Vs Time

 

[JRMichler]

You show a plot with constant velocity. A constant velocity move requires infinite acceleration for an infinitely short time (an impulse) at the start, and an equal and opposite acceleration at the end of the move.

A typical motion profile for an application such as yours would be a constant acceleration move. Accelerate halfway at constant acceleration, then decelerate the second half at the same constant acceleration. The sketch below shows what I mean. The horizontal axis is time, the vertical axis is acceleration. The next step is to make a similar sketch of the velocity, then start putting real numbers (and units) into both the acceleration and velocity sketches. You may want to add a third sketch for position vs time.

The purpose of these sketches is to help understand exactly what you are calculating. Until you have that understanding, you are trying to solve a problem by blindly plugging numbers into randomly selected equations. After you have that understanding, you will not only know what equations to use, but you will be able to solve more complex cases where canned equations are not available.

 

[TJS]

You show a plot with constant velocity. A constant velocity move requires infinite acceleration for an infinitely short time (an impulse) at the start, and an equal and opposite acceleration at the end of the move.

A typical motion profile for an application such as yours would be a constant acceleration move. Accelerate halfway at constant acceleration, then decelerate the second half at the same constant acceleration. The sketch below shows what I mean. The horizontal axis is time, the vertical axis is acceleration. The next step is to make a similar sketch of the velocity, then start putting real numbers (and units) into both the acceleration and velocity sketches. You may want to add a third sketch for position vs time.

View attachment 213795

The purpose of these sketches is to help understand exactly what you are calculating. Until you have that understanding, you are trying to solve a problem by blindly plugging numbers into randomly selected equations. After you have that understanding, you will not only know what equations to use, but you will be able to solve more complex cases where canned equations are not available.

This has been extremely helpful, thank you, it makes a lot of sense now. I have plotted the sketches and they have really helped my understanding.