1
$\begingroup$

Physics has never been my strong point... I have a task to size a motor. The question is: Based on information below, how do I calculate required motor torque to accelerate a mass with the information below? Please explain the formulas and where they originate from. I understand how to calculate the force required. I don't understand how I compute the required torque to achieve this force.

Variables:

Motion Direction:  Horizontal - Perpendicular to gravity
Mechanics:  Ballscrew & Linear Slide mechanism.

(W) Weight of Load to move:           100KG
(D) Distance load should move:        1 m
(A) Speed load should move:          .04 m/s
(C) Coef of Friction of Mechanism:    .003
(p) all Screw Pitch:                 2.75 mm
(d) Ball Screw Dia:                   20mm
(T) Min motor torque required:         ?

*Please arbitrarily choose variables for any missing information and declare them in your post.

Thank you for your time.

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
1
$\begingroup$

assuming constant acceleration, a, we need to cover 1 meter with an average speed of $$04m/s =\frac{0 +V_{final}}{2} \\ V_{final}= 0.8m/s $$ Force, F, needed for this acceleration is:

$V_{final}= \sqrt{2a*x}: 0.8 = \sqrt{2a*1}\\ a=0.8^2/2= 0.32 m/s^2 $

$$ F = ma =100*0.32m/s^2 =32kgm/s^2 \\ and\ 32* 1.003 = 32.09kgm/s^2$$ This is force demand increased by friction coefficient.

and the torque is $$\tau = 32.09kgm/s^2/(mechanical\ advantage\ of\ screw)=\frac{32.19}{(\pi*0.02 )/0.00275)}\\= 1.4kgm*m = 1.4* 9.8 =13.76Nm$$

We have assumed $sin( small\ angle\ \alpha)=\alpha$ on the screw pitch.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Hi Thanks for reply. When I try and convert to SAE units, I do not get correct results and the reason I belive is in the acceleration calculation. The value becomes drastic. This is not correct. .... Example::: wt(lbs) = 45...... Distance (in) = 39.37...... Desired Velocity (in/min) = 1183 ........ Friction = .003...... Pitch (in) = .109....... Dia (in) = .789...... ----- VFinal = 1183*2 = 2366 in/min....... a = VFinal^2/2 = 2798978 in/min...... F = 126331872 in/lbs...... Mech Adv = 22.844..... T = 12801xxxxxxx in-lbs..... What am I missing? Thanks very much, $\endgroup$
    – ExcelMania
    Commented Jan 17, 2019 at 1:52
  • $\begingroup$ @ExcelMania, I will check it later, you check for the difference between lbs weight and lbs force, if I remember correctly you need to divide by 32.2. $\endgroup$
    – kamran
    Commented Jan 17, 2019 at 4:52
  • $\begingroup$ Hi kamran. Thanks for your help... 1 lbf = 1lb weight. So I am confused. $\endgroup$
    – ExcelMania
    Commented Jan 18, 2019 at 1:17
  • $\begingroup$ Go metric, man. Go metric. It's all much simpler and it's international as is this site. $\endgroup$
    – Transistor
    Commented Oct 13, 2019 at 10:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.