Rotation from Transient Load 2

[MegaStructures]

What would be the best way to calculate how far this object would rotate if a force acted on it for 3-seconds only?

“The most successful people in life are the ones who ask questions. They’re always learning. They’re always growing. They’re always pushing.” Robert Kiyosaki

 

[MegaStructures]

WARose: Your first method yields a rotation that is greater than the static solution. Seems unlikely to me, but I suppose that means 3-seconds is a long enough acting force to achieve max displacement of the system

“The most successful people in life are the ones who ask questions. They’re always learning. They’re always growing. They’re always pushing.” Robert Kiyosaki

 

[BAretired]

Static solution: e = F*h/W; e = 1500*60'/100,000 = 0.9'

So after everything comes to rest, it moves less than a foot.

How is the force applied? If it is applied gradually from 0 to 1500#, and if the rod is a rigid body, its center of gravity should not move more than 0.9', but if it is applied fully from the start, it will give the rod some acceleration, which will send it beyond 0.9'.

If the rod bends as a result of the moment, the center of gravity will not be at mid-length, but to account for that, you would need to know the mechanical properties of the rod which have not been given.

The sketch in the OP did not suggest a long, slender rod. It appeared to be a stocky block which would normally be considered a rigid body.

BA

 

[WARose]

WARose: Your first method yields a rotation that is greater than the static solution. Seems unlikely to me, but I suppose that means 3-seconds is a long enough acting force to achieve max displacement of the system

It's really not a static problem. (Technically, no problem is.) But that second solution I posted was (IIRC) independent of time. So you've got more than one tool.

 

[BAretired]

WARose,

The second problem you posted has a 50# force pushing a 20# bar with the force perpendicular to the bar. In that case, the bar would continue to rotate around the pin and, in fact, would accelerate. The angular velocity at theta = 60^o was found. How is that independent of time?

In the current case, the weight of the rod is huge compared to the applied force. It moves a very small amount. If it moves beyond the point of static equilibrium, it will swing like a pendulum until it stops. I'm not sure exactly where it is at 3 seconds, but my guess is close to 0.9' right of the pin.

BA

 

[WARose]

The second problem you posted has a 50# force pushing a 20# bar with the force perpendicular to the bar. In that case, the bar would continue to rotate around the pin and, in fact, would accelerate. The angular velocity at theta = 60o was found. How is that independent of time?

It's independent of time because you can set the energy of the applied force equal to the raised height. [EDIT: Actually it's not quite that simple....I had to use a trial and error solution when I did the part in the strikethrough.] With enough calculations you can also determine (angular) acceleration, velocity and so forth at various points. And yes, in that particular example it will go a full 360 (I don't get that here).....but we are talking using this as a method to get where he wants to go.

Personally, I prefer the first solution I posted [on 7 Aug 21 18:01]. That accounts for all of it.

 

[BAretired]

Well WARose, I looked at it briefly but, in spite of your neat handwriting, it came out as Greek to me. It has been a long time since I looked at dynamic formulas; I get confused enough when things stay in one place.

BA

 

[WARose]

Well WARose, I looked at it briefly but, in spite of your neat handwriting, it came out as Greek to me. It has been a long time since I looked at dynamic formulas; I get confused enough when things stay in one place.

Are you calling my handwriting sloppy?

(Kind of bad ain't it?)

But in any case....I tried it (i.e. MegaStructures problem) myself by the methods posted on [7 Aug 21 18:01] & [7 Sep 21 22:49], and came out with a rise height of 0.4" (@ the c.g. of the hanging mass).

I had to go with a guess on the inertia of the hanging body.

And as I noted in a edit in my last post, doing it by the solution in the post on [7 Sep 21 22:49].....you need some trial & error to solve that one.....because solving for θ was tough (for me at least).

In any case, hopefully we have given the OP enough tools to do the job. (Or at least get something conservative.)

 

[BAretired]

Are you calling my handwriting sloppy? rofl

Wouldn't even dream of it! You should see mine.

So you found the same result as Tomfh. Thanks for taking the trouble to check it out.


BA

 

[Tomfh]

So it will swing past 0.9' (the static solution as noted by BAretired), and must come to a halt by 2'. The exact answer, ugh...

These questions remind me of why I much preferred statics over dynamics at university!

 

[BAretired]

Tomfh,

Do we know that it will come to a halt at 2'? That's where it is at 3 seconds, but perhaps it will travel further at 3.5 seconds; or perhaps it has already gone further and is on its way back toward the equilibrium position at 3 seconds.

BA

 

[WARose]

So it will swing past 0.9' (the static solution as noted by BAretired), and must come to a halt by 2'.

I came out with less. (As noted: 0.4 inches @c.g.) It shouldn't swing any higher than that. (By conservation of energy.)

 

[BAretired]

I came out with less. (As noted: 0.4 inches @c.g.) It shouldn't swing any higher than that. (By conservation of energy.)

I'm sorry, but I have trouble believing that it hit the magical maximum at precisely 3 seconds. MegaStructures, the OP, would have to be a psychic to know that. Either that or he had it all figured out before asking the question.


BA

 

[WARose]

I'm sorry, but I have trouble believing that it hit the magical maximum at precisely 3 seconds.

According to my calculations: it didn't. It was shortly thereafter. But it didn't rise much beyond that.

 

[Tomfh]

Do we know that it will come to a halt at 2'?

It will halt at (or before) 2'

Another (crude) approach:

The force could accelerate an object of this size to 0.44 m/s in 3 seconds, giving it a kinetic energy of 4390 J.

4390 J could raise the object by 9.9mm, which equates to a lateral displacement of 0.6m, which is 2'.

Again, this is conservative. The real answer is likely going to be less.

I'm sorry, but I have trouble believing that it hit the magical maximum at precisely 3 seconds.

I don't think anyone's saying that? Just that it can't rise any higher than 0.4" (and hence cannot move further than 2'), due to the energy being exhausted by then.

 

[BAretired]

What would your answer be if the question said 3 hours instead of 3 seconds?

BA

 

[Tomfh]

My crude method above doesn’t work if the force is applied indefinitely.

 

[WARose]

My crude method above doesn’t work if the force is applied indefinitely.

I have to wonder how appropriate my angular impulse-momentum solution is for 3 seconds. BUT since it gave only slightly more than the conservation of energy solution (see post on [7 Sep 21 22:49]).....I'm not sweatin' it.

 

[BAretired]

I interpreted the original question to be essentially "Where would the object, in this case a rod, be after holding the force for 3 seconds?". If I understand the answer correctly, it would move a maximum of 24" horizontally and rise about 0.4" vertically but not necessarily at 3 seconds.

At maximum rise, the net force acting on the rod would be 1500# - 100,000*2/60 = -1830#. It must have been decelerating before maximum rise, then started accelerating in the opposite direction at that instant. After 3 seconds, the applied force is removed, so the rod would swing to a vertical position in due course, but we don't know where it is at precisely 3 seconds. Interesting problem! Makes me wish I knew a little more about dynamics.

Years ago, I assisted a surveyor who was surveying a tunnel at Keno Hill Mines in the Yukon. To get his bearings, he dropped two plum-bobs down a 900' deep shaft, then aligned his instrument with the two strings. That method seemed to lack precision to me, because the plum-bobs seemed to take forever to settle down if they were out of plum. And it didn't help if someone higher up the shaft accidentally kicked one of the strings.

BA

 

[Tomfh]

At maximum rise, the net force acting on the rod would be 1500# - 100,000*2/60 = -1830#. It must have been decelerating before maximum rise, then started accelerating in the opposite direction at that instant. After 3 seconds, the applied force is removed, so the rod would swing to a vertical position in due course, but we don't know where it is at precisely 3 seconds. Interesting problem! Makes me wish I knew a little more about dynamics.

I did an iterative model in excel. At 3 seconds when the force stops the rod is at 1.4', it continues to a maximum of 1.6' (at 3.6 seconds) before swinging back.

If the force continues to be applied the rod swings to 1.8', at 4.3 seconds, before swinging back.


That's all frictionless, no damping, etc.

 

[BAretired]

Very nice! Thanks Tomfh.

BA