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This question relates to a previous question (How to calculate the required torque in a static equilibrium system) where I have enquired about the start-up torque required to rotate a system. However, the parameters have changed slightly since then, and as a result I am not entirely sure if the system is still considered to be in "static equilibrium" - somehow I believe it still is, and I am speculating that the only factor that has changed is gravity, considering the additional weight added.

However, I do not know what other forces are now at play in the new system.

Question: What is the required start-up torque to rotate the system with the parameters provided below ?

This is what I have:

2x bars perpendicular to each other, intersecting each other at their centres. Something like this:

                  10kg (F3)
                    |
                    |
                    |
                    |
10kg (F1) ----------+---------- 10kg (F2)
                    |
                    |
                    |
                    |
                  10kg (F4)


Bar weight      : 0.5kg (500g)
Axle weight     : 0.5kg (500g)
Max RPM         : 3 RPM
F1, F2, F3, F4  : 10kg each
Time to Max RPM : 10sec

*** Ignore Friction and Drag.

The motor torque specification is given in kg.cm.

A motor (whose start-up torque requirements I need) is attached to an axle. The motor is required to turn the axle at a constant speed of about 3RPM. What is the required start-up torque to reach 3RPM after 10 sec ?

(I am hoping to hear from Kamran again)

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  • $\begingroup$ please indicate the direction of the forces, F1, F2...F4. if they are in the plane of rotation or pointing up/down.? $\endgroup$ – kamran Feb 10 at 18:58
  • $\begingroup$ @Kamran: Good to hear from you again. The orientation of the system is vertical. Up/down ? $\endgroup$ – Tino Fourie Feb 10 at 19:08
  • $\begingroup$ So the 10kg forces are pointing down? and the axel is horizontal. or if it's small we ignore it? what is the length of the bars? just assume something? As it is the forces just add 40kg load to the axel, irrelevant to the moment of inertia. $\endgroup$ – kamran Feb 10 at 19:41
  • $\begingroup$ Sorry forgot to add the length (1 meter for each) of the bars. Sorry for the oversight. Weight of the axle (horizontal orientation) is minor at 500g (0.5kg). $\endgroup$ – Tino Fourie Feb 10 at 19:49
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This is a quick answer till further detail by the OP.

The 10 kg forces directed downward at the ends of the bars cancel out as for the rotation of the system, they just impart 40kg force at the axel. Let's say the length of the bars is 100cm. and we ignore the axel for now because it is too close to the rotation center.

Then the moment of inertia of each bear if it's a uniform thickness rod is:

$$I= \frac{1}{12}mL^2 \ for \ two \ I =\frac{1}{12}2mL^2$$

$$\tau = I\alpha= \frac{1}{12}1kg10000cm *3.33*2\pi $$

I let you do the rest. Please clarify If I misunderstood your question.

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  • $\begingroup$ Kamran, Thank You and apologies for very late reply. Things out of my control and relates to Telecoms company. I'm a little confused: Is there perhaps a typo at 1kg10,000cm? (In my defence, I did mention the bar lengths to be 1meter each which is 2x500mm, and yes they are all uniform) $\endgroup$ – Tino Fourie Feb 22 at 10:07
  • $\begingroup$ @Kanran: 100(sqr) = 10,000. Not thinking straight! $\endgroup$ – Tino Fourie Feb 22 at 12:43

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