Rigid Bodies Equations of Motion Rotation (Learn to solve any question)
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- Опубліковано 3 сер 2024
- Learn about dynamic rigid bodies and equations of motion concerning rotation about a fixed axis with animated examples. Learn to figure out the moment about a point, writing normal and tangential acceleration equations, and using kinetic diagrams to figure out unknowns.
🔹Rigid Bodies - Equations of motion: Translation: • Rigid Bodies and Equat...
🔹Rigid bodies and rotation about a fixed axis: • Rigid Bodies: Rotation...
Intro (00:00)
Kinetic Diagram (01:39)
Equations of Mass Moment of Inertia (02:30)
The uniform 24-kg plate is released from rest at the position shown (02:44)
The two blocks A and B have a mass of 5 kg and 10 kg (06:47)
The 30-kg disk is originally spinning at ω = 125 rad/s (08:27)
Find more at www.questionsolutions.com
Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2014.
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Thank you for the comment, really glad they help you out. Best of luck with your studies!
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Really glad to hear that. I wish you the best with your studies!
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1. Read the chapter and take notes
2. Try some fundamental problems
3. Watch your video and try all problems before seeing solution
4. Do ALL fundamental problems from textbook WITH EASE
5. Get A's on every test......
Any desire to do hydromechanics?
That's an awesome way to use these videos as a supplemental tool to do well in school. This is pretty much exactly what I envisioned students would do when they watch these videos. Well done!
No plans to do hydromechanics, but I will add it to my list of things to do :)
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You're very welcome! Really glad to hear they helped.
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Thank you very much! I am glad these helped :)
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I am glad to hear it's helpful :) Keep up the good work!
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You're very welcome! Really glad to hear it helped!
Another awesome video!! Will be taking thermo next semester, and I was very glad to see you have videos for it as well :)
Great! I hope the thermodynamics videos will be helpful to you as well :)
I like how you skip all the 'simple' calculation steps that we can do ourselves and only focus on the concept you're trying to teach us (eg. 5:46, you don't waste time going over this calculation, which ultimately decreases the time needed to learn a concept)
I try my best to keep these videos as concise as possible so it won't take too much time for students. If something is required before hand, I usually create a video to showcase that separately so if anyone needs a refresh, they can always take a look at those videos. I appreciate you noticing the little things, it's nice! 👍
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I've been watching literally all of your Dynamics videos over the last few weeks in preparation for finals and you're literally saving my life. I'll edit this comment in 2 weeks to see if all this studying worked
I am really glad to hear these videos have been helpful! I wish you the best with your finals and, yes. please, let me know how it goes 👍
How did it go?
@@hanigharaibeh It went great, ended up with a B in the class. I literally grinded these youtube videos and it saved me
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Thanks
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u are amazing, God bless u. Can u make videos about THERMODYNAMICS for mechanical engineering?
It's on my list to do :) What book do you use for thermodynamics?
@@QuestionSolutions
THERMODYNAMICS An Engineering Approach (ninth edition)
@@EngSereneJibreen Okay many thanks!
For question number 2, because we are using the point at the center of the pulley, wouldnt the moment only equal I*(angular acceleration), or is incorporating ma a different method
If I understand you correct, you are saying that there is only one moment and that's about the center which equals Iα? However, we have 2 weights attached to the pully. Each will create a moment about the center. So you have to make sure to keep the perpendicular distances to point O from each weight in mind, so in this case, that's 0.15 m.
for problems on finding the angular acceleration, how can you determine the appropriate approach in solving for either translational or centroidal (like in this video)?
Usually, you can approach these questions in multiple ways. When you do enough questions, you sort of get a feel for the way you should go at it, and if you're wrong, or you get stuck, you can always backtrack and do a different method.
This might be a simple question but for the last exercise, I first attempted to solve it by myself with the n-t coordinate system, but then realized the x-y system should be used. I was just confused because the procedure for analysis for the book also tells us to establish the n-t coordinate system in this chapter. Are there any simple ways to identify which coordinate system should be used?
How did you establish the n-t coordinate system for the last question? Usually, you have a pretty good idea of what type coordinate system to use based on what the question is asking and the information provided. I think it comes naturally after you do a few questions.
8:03 The formula for the kinetic diagram has the accelerations for both cylinders as positive. Why is one of them not negative if they are accelerating in opposite directions?
Btw Thank you for all your videos, I couldn’t pass this class without you!
Here, we are looking at the moment created by the 2 accelerations. Both create clockwise moments, and we picked clockwise to be positive, so they are both positive values.
Hi sir, at 9:20, is the force AB drawn acting on the member or the disk?
It's on the pin that is connected to the disk, so the member. This is the force on the member because of the disk.
@@QuestionSolutions I see, thank you sir!
@@kanhchanaly6445 You're very welcome!
Dear sir, at 9:19 how could you assume that force on pin A was acting along the line joining A and B?
It is a 2 force member. This comes from statics, but if you have time, please take a look at this video: ua-cam.com/video/IxkwZYXECkk/v-deo.html (time 1:36)
hi Sir I have a question why is the force pointing to A I know that if I would make a free body diagram of the wheel I would have in position B two reactions in x und y-direction but these reactions created from the pin are constraining reactions so I could draw them in any direction
If you draw a free body diagram and drew out the components of the forces at the pin, you'd have an x-component and a y-component. The force that goes towards A is the resultant of those 2 components. I understand what you are saying, that you can actually assume any direction for your reactions, however, with a bit of intuition, you can make a very educated guess that the resultant force has to be towards point A. I think that comes from doing a lot of questions, and mostly from your statics class, where this should have been covered. In order for this wheel to be attached and held in place, rod AB has to support it, so that's how you can assume that force AB has to be from B to A. I hope that makes sense :)
Hey, thanks for the clear explanations! I'm not sure how quick I'll get a reply here, but for the problem at 6:47, are the accelerations at both block A and B still the same if the pulley at O is made of two different plates, both with different masses and radiuses? For example, in the problem I have, the pulley is set up the way as I described and block A is tied to the plate with the bigger radius and mass, and block B is tied to the one with the smaller radius and size.
So it depends on how many cable there are. Please see this video first: ua-cam.com/video/IudPPGIV5QM/v-deo.html
If you still have questions, let me know, thanks!
@@QuestionSolutions Thanks for the quick reply! The video you linked isn't exactly my type of problem though, the pulleys aren't spread out, the pulleys are at the same location, O. There's still only one cable. Would this mean they are still different accelerations?
I tried solving the problem myself and got that block A, which is tied to the pulley with the greater mass and radius has a higher acceleration than Block B, which is tied to the pulley with the smaller mass and radius. I should mention too that Block A has a smaller mass than block B, so block B is accelerating down and block A is accelerating upwards.
Again, thank you for the quick reply! Your videos are immensely helpful!
@@TheSpokenBanana It's really hard to give an answer without seeing a problem and the wording of a question. I especially don't want to guess and say yes or no because I don't want you to get it wrong. It might be better to see your TA or professor during office hours and ask your question. Is the problem you're asking something like this? ua-cam.com/video/mP8XTfFmQeo/v-deo.html
@@QuestionSolutions Still not exactly, but I understand! I'll ask around - thanks for your help, anyways! Like I said, your videos have helped me out countless times in dynamics!
Hi sir, why didn't Ax component and Ay component be drawn in the free body diagram at point A in the third question at 9.20 ? It is because Fab is the resultant force of component Ax and Ay aldy? Same for point B also?
That is correct, Fab is the resultant force, so the math is easier to keep it that way instead of breaking it into components. It's actually a 2 force member. 👍
hi! I had a question regarding the coordinate axes. For the kinetics of particles (translation), when choosing the positive coordinate system, when we solved problems we would need to set the acceleration in the positive direction no matter what. Even if it was known to be going in the negative direction, if we set the equation up as so, the answer would be wrong. Indeed, the math would only work if we put the acc in the positive. However, i was wondering if it is the same case for rigid bodies (translation, rotation, gpm). Do we necessarily need to put the acc in the positive direction that we set it as, or can we put the acc in the negative direction (if it is known to be going in the negative direction) and still get the correct answer? For example, if we did not know where it was going, we could put it in either direction and it would give a positive if guessed correctly and negative if guessed wrong. Thank you for your help!!
I am unsure of your question. When you say for kinetics of a particle, we always pick a positive coordinate system, and you get the wrong answer otherwise, this isn't true. For example, if we take projectile motion of a particle, and we assumed up to be positive, and gravity was affecting the particle, the gravity there is negative. Is that what you mean or maybe I am not following your question too well 😅 For rigid bodies, you can pick whatever side you want to be positive. So if you have a rigid object with forces facing to the left, you can still pick right to be positive. So if all the forces are facing right, the acceleration would also be to the right, and you will still get the correct answer. The mistake students make is choosing a positive side and then switching that side halfway through the question. If you stick to the side you picked to be positive, for the whole duration of the question, even through multiple substitutions, you will end up with the same answer, because at the end, all we are doing is assuming a positive direction. I hope that makes sense. If your question is asking for something different, please give me an example or a follow up question so I can better answer you :)
@@QuestionSolutions sorry for being unclear haha thank you so much!
@@QuestionSolutions so just to confirm, for kinetics of rigid bodies, we take the positive coordinate axes to be in the direction of acceleration?
@@nikkisharp1559 I would say yes, because it makes the math easier, don't have to play around with those pesky negative signs 😅 (but you don't have to if you don't want to, you will still get the same answer as long as you follow through with your positives and negatives).
@@QuestionSolutions perfect thank you so much!
If there is a frictional moment at the pivot of the pulley, will it be correct to subtract it from the right hand side of the equation? [6:55]
Yes, if you account for friction, then it will create an opposite moment.
@@QuestionSolutions will it be correct if you subtract it from the left hand side of the equation instead of the right?
@@arjungovender3248 Since its a moment, it'll be right side. 👍
@@QuestionSolutions If its a frictional moment of the pivot O, and the pivot is where the moments are taken will it still be on the right side? I am just querying because in a question that I am was doing when I insert the frictional moment on the left side I arrive at the answer but not when I insert it on the left.
@@QuestionSolutions Could you please explain what the right side of the equation means and then what the left side of the equation means. Apologies for the inconvenience
for 3:36 how did you find the 45 degree angle was that just tan inverse.
So normal and tangential components are always perpendicular to each other. In other words, it creates a 90 degree angle between them. Here, we just have it bisected, so you end up with 45 degrees on each side.
Can we solve the first example using virtual power: Fgi*Vg + Fg*Vg + Mig*w + Me*w = 0 ? There is no initial omega nor velocity so could it be solved that way?
There are many ways to solve these problems, I only showcase one method. You can get to the same answer in so many ways :)
I dont understand the equation that you put in 8:06. What does mean that "5a * 0,15 + 10a * 0,15 + I * (a/15)"? where does that formula come from?
Thanks!
So you're talking about the other side of the equation. For that, you're looking at the kinetic diagram. We have 2 mass times accelerations, one for block A and one for block B, both with create a moment. So it's just mass times acceleration of each block times the perpendicular distance to the pulley. We also have the moment about point O of the pulley, which is just the mass moment of inertia times and angular acceleration (this is shown in pink on the diagram). So all of this is a moment equation. The left side based on the free body diagram, and the right side is using the kinetic equation.
Question Solutions when you talk about moment? What moment are you talking about? Angular momentum? Im a little bit lost
@@tarekhirmas218 Simply the moment about the center of mass. Momentum is completely different so don't confuse the two :) A moment is force times perpendicular distance, it should have been covered in the statics class before dynamics. Sometimes, a moment is also called torque. Please see: simple.wikipedia.org/wiki/Moment_(physics)#:~:text=In%20physics%2C%20moment%20of%20force,rotation%2C%20plays%20an%20important%20role.
Tanx
You're welcome!
at 4:30 our rotation occurs at pin A right not at the center of gravity of the plate ? why don't we use parallel axis theorem for our right hand side of the equation
Where it rotates doesn't matter to us. Parallel axis theorem is used when we find the mass moment of inertia about a different point other than the center of mass. So if we found the mass moment of inertia about point A for the plate, then we would need to use the parallel axis theorem. Here, we are finding it about the center of mass.
at 4:53 I would love if there was a line of text relating the (M_k_)_A_ part to the (24)(0.3535alpha)(0.3535)+1alpha
I feel like there's a line there that could be showing it, from the KE, but I'm struggling to find where the numbers came from.
or.. I could keep watching X3 heck. Maybe flip it so the explanation comes first. Having the symbological equation under it would help. I'm slowly piecing together where the second r comes from.
@@ROTSTarge I am not sure I understand what you are asking 😅 I go over where all the numbers come from right after I show the equation. Which part are you confused about? The second r is simply the perpendicular distance from point A, so the place where rotation occurs to the center of mass. If I explained everything first, without writing anything down, most students will not understand. Most of the time, it's easier to follow once you see the numbers on screen, rather than looking at a white screen while an explanation is going on. Let me know which part is confusing, and I'll do my best to help.
@@QuestionSolutions Yeah this was written before I realized you went over where the numbers came from, sorry X3
@@ROTSTarge No worries, as long as you got it 👍
@@QuestionSolutions I think the thing that would have made the biggest difference for my understanding is if you had included the units. That's how I check my work and it's something that most professors drill into us, so having those would make the numbers make a lot more sense when we try to figure it out without waiting for your explanation.
In the first example, why is the angular acceleration about the centre and the hinge is same?
The hinge is stationary, it has no angular acceleration. We are only finding the x and y reactions at the pin. Maybe I am misunderstanding your question, please provide a timestamp so I can take a look, thanks!
5:43 you can't add I(G) for the summation of moments about A. Don't you need to use parallel-axis theorem and convert to I(A) before adding?
Never mind, I see what you've done
Okay great!
hi good evening. i still confuse how to determined the position for summation of moment whether it is negative or positive😅 is there we can see from the rotation to the point 0?
You can pick any point you want to sum the moment about. It's totally up to you, but it's best to pick a spot where you can easily figure out distances and also where you can eliminate as many unknowns as possible.
@@QuestionSolutions wah thankyou 🙏🏼 so if the point we assume at the left side, then K.diagram Force downward , so the Force is Clockwise right?
I am not too sure about your question. But maybe this video will help? ua-cam.com/video/QNNnPZ68STI/v-deo.html
All moments work the same.@@irfnkml..
4:30 isn't that equation valid only if the plate is rotating around point G?
nvm I get it. You added the md^2 to the moment of intertia in another step. usually I do that all in one step
Do it whichever way you like, many ways to get to the same answer :)
Hello sir, should I use kinetic diagram for every problem?
It really depends on the question. For example, if you're using relative motion or instantaneous center of zero velocity, then there is no need for a kinetic diagram. But if you need to solve the problem using multiple equations, you need a kinetic diagram.
@@QuestionSolutions Well, I prefer using relative velocity equation. By the way, I'm seriously struggling in this chapter although got relatively a good score dynamics midterm but currently rigid bodies' chapters are extremely exhausting. Statics was rewarding but these rotating machines are really hard to keep up with. And almost all student agree that moment of inertia and rigid bodies are unbeatable! 😂
Just to confirm for my sanity I guess. The rotation of the center of mass is about the pivot, and the normal acceleration points towards the pivot, right?
You are missing fundamentals 😅 But yes, normal acceleration always points towards the center of curve, while tangential acceleration is tangent to the path of travel. Please, if you have time, go back and watch some of the previous videos, you will find these much easier to understand if you do that.
@@QuestionSolutions No I just wanted to make sure since the first diagram had the normal arrow pointing slightly to the side of the pivot X3 So I just wanted to make sure that there was no major error on my part
@@ROTSTarge 😅 okay
I love you
❤
Thx from 2022
You're very welcome!
How do we know that the acceleration is zero as shown in the third question?
Are you talking about 10:06? If so, the wheel doesn't go left, right, up or down, so it can't have an acceleration.
🐐
😄
I'm just wondering why Ax is to the left
I don't know where you're referring to, please use timestamps.
a normal why zero ? in rotating objects
I don't understand your question.
Can you point me to a resource on the web that derives the fact that: The equation of motion taken about a random point is equal to the equation of motion taken about the center of mass? Preferably using vectors. I wish I could pay you tuition rather than my incompetent instructor!
I'd love to help, but what do you mean by "The equation of motion taken about a random point is equal to the equation of motion taken about the center of mass?" Which equation of motion?
@Question Solutions Aren't there 2 different sets of equations for summing moments about the center of gravity vs. Summing moments about an arbitrary point on a rigid body that is not the center of mass?
@@r2k314 Hmm, I am not sure what you mean here, is there an example of the 2 cases you can provide for me to take a look? In most cases, if not all, you can write a summation of moments about any point you want.
@Question Solutions exactly! Is there a proof that shows how you get the same result regardless of the point you pick. Thank you for your time.
@@r2k314 Did you check your textbook, on the section on moments?
difficult topics
I think the more questions you do, the easier it becomes.
@@QuestionSolutions yes teacher you are right, I took the exam today , Thanks to you, I was able to answer the questions