Dude. No lie, I had literally no idea what was going on in class and I had an exam the next day. Like, I didn't know how to even start the HW, didn't understand lectures, etc. I didn't even know how much I didn't know. So the day before the exam, I watched your videos and took really good notes. After the exam, the grade comes out and I got a 98% after the curve, the second highest score for that exam. You are a literal lifesaver.
Wow, that's actually pretty crazy. I am really happy (not at the fact that you didn't do your HW), but at the fact that you were able to take notes from these videos and get a 98%. That's amazing! Keep up the great work and I wish you the best with your studies.
Your animations make the problems more intuitive where the book and Mastering cannot/will not. Thank you. I wish I had found these sooner, might have less desk imprints in my face.
Glad to hear they are helping! If possible, please share these videos with friends/classmates who may also find them helpful. That will help this channel and help them as well, win win 👍
Thank you thank you so much for this video. You earned yourself a subscriber 1 minute in. I will be watching many more of these for my upcoming exam. Thanks you thank you!!
I’m going to a private university and UA-cam is teaching me better… you’re amazing. Wouldn’t it be nice if you did chapters in sections? I’d love to watch your lectures then go to class for a full comprehension
I'm happy to hear UA-cam has been helping you. Could you elaborate on what you mean by "chapters in sections?" I like feedback so if it's something I can keep in mind for the future, I'll incorporate it.
3:24 I-components: 0=8sin45-2wbc, answer you have wab = 2.282rad/s cw, J-components: -2wab = -8cos45, answer you have is wbc = 2.828 rad/s ccw. I think subscript is switched. I-component is wbc=2.828 rad/s cw, and J-component is wab = 2.828rad/s ccw.
No, what's shown is correct. So imagine this to be a fully moving system. Imagine the block at C actually sliding down. When it's sliding, is it possible for rod AB to rotate counter-clockwise? No, right? Now look at rod BC, and imagine it rotating about point B while block C is sliding down. Can it rotate clockwise? No. So it must be counter-clockwise. You can actually see the angular velocity arrows drawn in black shown on the diagram. Once we find the angular velocity, the i, j, k components are irrelevant to us since we get a scalar answer.
At 2:25, why is the angular velocity defined to be in the -k direction? If it's moving clockwise, shouldn't it be moving in the -j direction because it is moving towards -y axis?
Good question, one where a lot of people can get confused. The vector for the angular velocity must be found using the right hand thumb rule. So you have to curl your fingers in the direction of rotation, and point your thumb out. The direction of the thumb shows the angular velocity vector. This image shows it well: scripts.mit.edu/~srayyan/PERwiki/index.php?title=File:AngularKinematics02.png
@@ryanbragg4334 You're very welcome! You can sort of think of angular velocity as how fast something rotates about an axis, but just keep in mind that the vector for it has to be found using the right hand rule. And most problems, we see it in 2D, but it's really a 3D problem, so you sort of have to imagine the pin where something rotates around is actually a 3D pin, and it'll become easier to visualize in your mind. Anyways, best of luck with your studies!
The vector for the angular velocity must be found using the right hand thumb rule. So you have to curl your fingers in the direction of rotation, and point your thumb out. The direction of the thumb shows the angular velocity vector. This image shows it well: scripts.mit.edu/~srayyan/PERwiki/index.php?title=File:AngularKinematics02.png
Thank you so much for this video mate, great explanation and great video quality too. Btw at my university our prof. talked about something called "Stübler Theorem" or Stübler's Theorem" when she was explaining relative acceleration on rigit bodies. I couldn't find anything about it online, have you ever heard of such a thorem?
Thanks! Hmm, I don't think I heard of Stübler Theorem. If you do ask your professor again on what it is, please share with me also, I am curious to know what it is. I searched for it but couldn't get any hits.
For the future, please kindly use timestamps so I know where you are referring to. I assume you're asking about the k component of the angular velocity? If so, the vector actually points into the screen. You have to use the right hand rule and see where the thumb points. So here (2:24), the angular velocity is clockwise, so if you curl your fingers so that they are clockwise, the thumb will point into the screen, which is the negative k direction. Here is a diagram illustrating the vector directions. scripts.mit.edu/~srayyan/PERwiki/index.php?title=File%3AAngularKinematics02.png
Vc in the question is given to us as 8m/s. All we did there was just break that into components. Since the 8m/s is at an angle of 45°, we can use cosine and sine to break the velocity into x and y components.
So you're trying to solve for va, which is why it's isolated on the left side. You can do it the other way, make sure your position coordinates and values are properly set.
Very good video! May I please ask how do u know where to start please? For example, U start with Velocity of B in the first example, but in the last one, u start with velocity of e. Thanks in advance!
Thank you. To answer your question, I think it comes from intuition. Just a hunch that going this way will work our better and give an answer faster. Sometimes, it's wrong but usually, it's right. As you do more questions, your intuition becomes better and you can just visualize the flow of the question and answer. :)
Please see 2:46. We are breaking the velocity into x and y components using the 45 degree angle given to us. So the x-component can be found using sine and y-component using cosine. Y-component is negative because we chose up to be positive, but the component is downwards. I hope that helps!
So Ve is found by multiplying the angular velocity of gear A by the radius, where as Vc is found by multiplying the angular velocity of the rod BC by the length. So each have their own corresponding angular velocities.
So the vector would point into the screen. Curl your right hand fingers in the direction of the rotation and look to see where your thumb points. Since it points into the screen, that's the negative k-direction, or negative z-axis.
You can usually make an educated guess as to the the direction of velocity, you just have to imagine the system moving in your mind. If not, that's alright, in the end, if you end up with a negative value, that means the assumption you made for the direction of velocity is incorrect, in other words, it's opposite to your assumption.
So clockwise means that if you use the right hand rule, where you curl your fingers to match the direction of rotation, you will see that your thumb points away from you. If it's away from you, it's in the negative z direction so it's a negative k component. Keep in mind that while the rotation is clockwise/counter-clockwise, the vector for angular velocity is found using the right hand rule. See: hyperphysics.phy-astr.gsu.edu/hbase/rotv.html
Hii there love your videos, I just want to clear up some of my confusion, at the third example (6:19) how do we know that we should calculate the VE and VC, like what would be the reasoning ? I understand what you are doing I just struggle to understand how to know in an exam that, that is what i should be doing. I hope this makes sense 🙈
I am not too sure how to answer that question. :( It should sort of "appear" in your head what steps to take next. If it doesn't happen, that's okay, but it just means you need to solve more problems and visualize the method to solve that problem in your head. Everyone does it at a different pace, but the more questions you solve, the easier it is to know how to tackle that problem.
Hi there :) in the example at 3:24 , from the equations of the i and j components, don't Wab and Wbc have the same sign and therefore direction of rotation?
So the position vector goes from B to A. In other words, to the left and up. We chose up and to the right to be positive. Since the x-component goes to the left, it's negative.
So the position vector goes from B to A. In other words, to the left and up. We chose up and to the right to be positive. Since the x-component goes to the left, it's negative.
So it's rotating clockwise, right? When we look for the vector, we have to use the right hand rule. So you curl your right hand fingers in the clockwise direction and see where your thumb points. It points towards you, in other words, out of the screen, so that's the z-axis, (k component).
Greatly explained as usual! I've got a question though, a simple one rather. At 6:51 , for the relative velocity equation. What if we wrote it for C with respect to E, so Vc/e instad. So [ Vc=Ve+Vc/e] I've done that and my answer came out to be negative (-105), what is this supposed to mean? Becaus clearly it cant go the other direction. The gear does infact rotate and has to rotate in a clockwise manner I believe haha for the system to work in the first place, atleast in this question per se.
Most likely a positive/negative error. For your r_C/E, was it -0.05? Because you switched it around, so it'll now be negative. If you get 105, but the sign is wrong, it's almost always just a positive/negative error, maybe when you did the cross multiplication, you forgot a positive or negative, etc. It's hard to say 😅
@@QuestionSolutions Ah yes, it was a simple sign error haha. To my luck, my errors are mostly always related to the signs. It was the R_c/e as u said. Thank u once again!! Have a graat day
hi i have a question, i will really appreciated if you can responde quickly. the thing i wawnt to ask why in the 2:32 we call it -k. its spoused to be -i
So to figure out the direction of the vector, you have to use the right hand rule. While the angular velocity is clockwise, this is a 3D vector. If we curl our right hand fingers the same direction as our angular velocity, we see that our thumb points into the screen. So that's the negative z axis, or -k axis.
Thanks for the videos they are very helpful. One mistake you have is you have labelled the angular velocities in rad/s but your values have been calculated in degree mode. This means some of your values have come out incorrect w.r.t the units you have used
You are incorrect. The answers shown are correct. So I can see where the confusion comes from, but the degrees were used to break velocity into components. What you end up as units is still m/s. So the components of 8m/s can be 8sin45 degrees, but the answer that comes out is 5.66 m/s. So there is no correlation between the degrees and rads for the final answer. The units shown are correct :)
@@QuestionSolutions hm interesting. I never knew you could use rads and metres interchangeably. I was always taught to convert the units, when I presented this problem to others they also said the units were mixed up. Maybe this is a difference in method of teaching
Can someone please explain to me the following. In my class in a similar exercise to 1:50 when we defined Vb=Va+w*r(b/a) we also include Vb= V (relative) +Va +w*r(b/a) , and as for r(b/a) we also changed and make it relative. When are the cases that need to add relative velocity and make r(a/b) also relative ??
It's really hard to answer that question without actually seeing what example you took up in class. At 1:50, at A, there is no velocity since it's just pinned there, so it would be zero. And since VB is a rotation about a fixed axis, (here the fixed axis is pin A), we can figure it out by multiplying angular velocity by a vector from A to B. To use the relative equation, you need 2 velocities to compare. So at 2:43, we are comparing the velocity at B to velocity at C. So in simple terms, if we need a velocity and its a rotation about a fixed axis, we can find it by multiplying the angular velocity by "r" and if we are comparing one velocity to another, then we need to use the relative velocity equation. Vb = ωr is not a relative velocity equation, its just an equation to find the velocity at a point if it's a rotation about a fixed axis. Also, you mentioned r(b/a). So again, if we are just using V = ωr, then its the distance from the axis to the location where we are finding the velocity. If it's the relative velocity equation, then its a vector from the location of the second velocity to the first velocity. I think it'll help a lot if you do the question first, and then if you get stuck, stop and view what step I did next. Sometimes, doing it yourself is super helpful. 👍
@@QuestionSolutions So If I am getting it right it there was an external velocity acting at point b (for example a motor pushing b) then we would use the relative velocity equation?
@@melekmnif1933 Yes, you definitely can use it like that, but its not limited to just those cases. So let's look at just example 1 in the video. In that example, the block at C moves down because of gravity. When it moves down, it has a velocity, which is Vc. Now that block is connected to rod CB. So what that means is, velocity at point C is the same velocity as the block. The goal is to use that to figure out the velocity of the other parts that's connected. You can see that to rod CB, there is another rod connected, which is AB. You can also see that because block C slides down, rods CB and AB both rotate. So whenever you have a velocity with connected pieces, we can use relative velocity to figure out the missing angular velocities and linear velocities. Since in that example we know the velocity at C, we can use that to figure out the other parts. So instead of thinking about an external velocity, think of it like this: If we have connected parts, and we know the velocity of one of those parts (or even angular velocity of one of those parts), we can use that velocity to figure out the velocity of another part. That's why it's called relative velocity, the velocity with respect to another point. Does that make sense?
@@QuestionSolutions I got your point but I think I asked a wrong question. I hope I am not asking for too much but here is a link to the exercise : To answer the velocity part of the question we first used Vb=Va+w*r(b/a)= -300j and then Vb= V(relative) +Vc +w*r(b/c) to find w So my question is why did they add V(relative) to the formula? and how to find it in general? www.google.com/search?q=The+gear+has+the+angular+motion+shown.+The+peg+at+A+is+fixed+to+the+gear+and+the+vertical+distance+from+D+to+A+is+1.2ft.+A.+Determine+the+angular+velocity+and+angular+acceleration+of+the+slotted+link+BC+at+this+instant.+B.+Don%E2%80%99t+forget+to+draw+your+FULL+kinetic+diagram+of+all+non-zero+force+and+velocity+vectors&rlz=1C5CHFA_enUS810US810&sxsrf=ALeKk03arZ5Cjk2Ex4ovEQ5o21y8CulZbA:1605637945859&source=lnms&tbm=isch&sa=X&ved=2ahUKEwj6h5q7m4rtAhWNg-AKHZV4AYIQ_AUoAXoECAcQAw&biw=1920&bih=1089#imgrc=1PebBbBekr9sEM Thank you for the effort !!
I have a separate video on general plane motion. Were you able to take a look at that one? You will see that general plane motion questions can't be easily solved like these ones, you need a gazillion equations to get an answer.
At the end of solution of question 1, why are the rotation directions opposite ? both the values of omega are taken to be positive and must be anticlockwise according to the convention
They are positive, meaning the directions we assumed were correct. Notice how we assumed a clockwise rotation for AB and a counter-clockwise rotation for BC. Getting positive values indicates that those assumptions were correct. But regardless of sign conventions, with these problems, it's easy to figure out directions just by imagining how these contraptions would move. Here, block C moves down, and there is no way for that to happen if link BC was going clockwise with AB also going clockwise. The same is true for the opposite, if link AB was going counter-clockwise, then block C can't move down, it has to move up.
Could you provide me with the timestamps you're referring to? I am having a hard time understanding your question. I'll try my best to help. Also, the steps shown are correct, there are no mistakes in the solution.
In the first exercise (with the slider block C at 8m/s); when filling in the general equation of relative velocity it is confusing to have 8sin45.i . Shouldn't the i component be 8COS45 i and the j - 8 sin45° ? For this exercise it doesn't matter because cos and sin of 45° is the same, but I think the sin and cos should we switched, unless I'm wrong ofc.
So here, I can't give you a definite answer because I don't know whether you took the top angle or the bottom angle and both will be 45 degrees. If you took the top angle, then opposite to the angle give you the y-component, and adjacent gives you the x-component. If you took the bottom angle, opposite gives you the x-component and adjacent gives you the y-component. If you, for whatever reason have an idea that says x-components are always cosine and y-components are sine, you need to get rid of that before you fall into a lot of pitfalls and lose marks on your exams. Here is a video < 60 seconds that will really help you understand how it works: ua-cam.com/users/shortsvynnKlJD_Jo?feature=share
So you have to use the right hand rule to figure out the direction of the vector. Here, we have a clockwise rotation, so curl your right hand fingers clockwise and see where the thumb points. Notice that it will point into your screen, which is the negative z-axis.
For a given instant ICR has zero velocity and for the problem at 3:29 ICR is the point of contact between the surface i.e. poin C If this statement is true why have you used V,C as 5i during calculation of V,B at 4:17
The gear rack has a velocity of 5 m/s, this is a given in the problem. So all we are doing, in the simplest sense, is actually just comparing the velocity of one point to the velocity of another. We use -5i because that is the velocity of the rack. If it's zero, then the gear above will go on top, carrying the whole system with it. I think for these problems, you really have to use your imagination to see how this device would move in real life. What you're thinking of is a system that's making contact with a surface that's not moving, but here, that is not the case. I hope that helps!
Great video as always, but why did we not find VB in the second problem the way we found VE in the 3rd problem which all we did was multiply the angular velocity by the radius? Thank you
We can't do that because there is a gear rack on the bottom that's affecting the speed of the wheel. It's not an independent object, it's connected to something else. In the case of VE, you can see that it rotates about a fixed axis, in this case, point B, and point B doesn't move. In the VB question, the center moves along the x-axis. 👍
when we say relative motion for the second point on the rod how can we assume if it is only on x axis or both for example and when to do this assumption
Sorry, but I don't understand your question. Where are you referring to? Is there a specific part on the video? Let me know and I will try to help you out. Thanks!
Thank you. I think if students understand how to apply the new equations they learn, regardless of where the questions come from, they should be solvable. But I will keep in mind the Merriam and Kraige books as well. Thanks again for your feedback, I appreciate it.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
Dude. No lie, I had literally no idea what was going on in class and I had an exam the next day. Like, I didn't know how to even start the HW, didn't understand lectures, etc. I didn't even know how much I didn't know. So the day before the exam, I watched your videos and took really good notes. After the exam, the grade comes out and I got a 98% after the curve, the second highest score for that exam. You are a literal lifesaver.
Wow, that's actually pretty crazy. I am really happy (not at the fact that you didn't do your HW), but at the fact that you were able to take notes from these videos and get a 98%. That's amazing! Keep up the great work and I wish you the best with your studies.
the dynamics playlist single handedly thought me in one all nighter before my exam than the lectures of this semester have in total.
I am really glad to hear that! Keep up the great work and I hope your exam went well.
You are better than all doctors who teach this subject in my uni (combined).
Thank you very much. Best of luck with your studies! :)
@@QuestionSolutions Could you still do the slider problem using k if it wasn’t a 45 degree angle?
@@mrmagoo-i2l Can you give me a timestamp? I'll take a look and let you know 👍
@@mrmagoo-i2l What do you mean by k? Please elaborate.
probably because doctors don't know too much about engineering
The question looks so complex in the beginning but when you really focus I see how easy you have made this. Thanks dude, saving lives out here.
You're very welcome! Glad you could follow along, hopefully it helps you out :)
The animations and the overall production quality is great. Keep making these videos and I'm sure you'll succeed!
Thank you very much for your kind comment. I really appreciate it! I also wish you the best of success with everything you do.
I just had my engineering mechanics exam the other day. Thank you, I would not have been able to pass without your help.
You're very welcome. I hope you did amazingly on your exam :)
This video was truly revolutionary for me, I understand this section so much better now. Thank you
I'm so glad! Keep up the great work and best wishes with your studies.
Glad I found this channel. You may be the reason I get my degree this semester.
I wish you the utmost success with your degree and future endeavors!
@@QuestionSolutions I got my degree. Thank you 100!
@@donnymcjonny6531 AWESOMEEEEE! So happy to hear that!! 🎉🎉🎉🎉🎉🎉
@@donnymcjonny6531 Congratulations!
@@tuanphong6188 Thank you!
Your animations make the problems more intuitive where the book and Mastering cannot/will not. Thank you. I wish I had found these sooner, might have less desk imprints in my face.
You're very welcome! I am glad to hear the animations help :)
Seeing the problems animated helped so much when it came to solving.
Glad to hear that! :)
Best explanations of dynamics I've seen so far - even better than Jeff Hanson.
Thank you very much, I really appreciate it.
Better than my lecturer🙏
Thank you very much! Best wishes with your studies.
You are deserving of much gratitude. Thank you.
Thank you so much :)
we need more video. this are really getting me through dynamic
Glad to hear they are helping! If possible, please share these videos with friends/classmates who may also find them helpful. That will help this channel and help them as well, win win 👍
I AM GRATEFUL TO LEARN FROM THIS CHANNEL. IT HELPED ME MASSIVELY DURING EXAMS.
I am really glad to hear that :) I hope you did amazingly on your exams.
Great job. Really nice clean animations that help with visualizing the problems.
Thank you very much! Best wishes with your studies.
One of the best youtube channels out there
Thank you very much :)
Thank you sir for making these videos! I really appreciate your effort.
You're very welcome! Best of luck with your studies.
Which software you use for this animation???
I use after effects to animate.
You are the best teacher! I really appreciate your videos
Thank you very much, I really appreciate it. :)
This was extremely helpful, thank you
You're very welcome! Best wishes with your studies.
such a legend you'll be remembered forever mate.
Thank you very much!
Thank you thank you so much for this video. You earned yourself a subscriber 1 minute in. I will be watching many more of these for my upcoming exam. Thanks you thank you!!
You are very welcome and thank you for the sub. Keep up the great work and I hope all the videos help you out with your upcoming exam!
I’m going to a private university and UA-cam is teaching me better… you’re amazing. Wouldn’t it be nice if you did chapters in sections? I’d love to watch your lectures then go to class for a full comprehension
I'm happy to hear UA-cam has been helping you. Could you elaborate on what you mean by "chapters in sections?" I like feedback so if it's something I can keep in mind for the future, I'll incorporate it.
Thanks dude I have midterm in 2 days and I'm going to ace it all thanks to you :D
Glad to hear! I wish you the best with your midterms. :D
How did the midterms go
3:24 I-components: 0=8sin45-2wbc, answer you have wab = 2.282rad/s cw, J-components: -2wab = -8cos45, answer you have is wbc = 2.828 rad/s ccw. I think subscript is switched. I-component is wbc=2.828 rad/s cw, and J-component is wab = 2.828rad/s ccw.
No, what's shown is correct. So imagine this to be a fully moving system. Imagine the block at C actually sliding down. When it's sliding, is it possible for rod AB to rotate counter-clockwise? No, right? Now look at rod BC, and imagine it rotating about point B while block C is sliding down. Can it rotate clockwise? No. So it must be counter-clockwise. You can actually see the angular velocity arrows drawn in black shown on the diagram. Once we find the angular velocity, the i, j, k components are irrelevant to us since we get a scalar answer.
I just wanna let u know my professor with "A Ph.D." can't do half as good of a job as you. Thank You Truly!
I'm glad to hear these videos help :) You're very welcome and I wish you the best with your studies. Keep up the good work!
You are wonderful, my brother. so much effort to provide us with the best way to understand
Thank you very much
You are very welcome! I wish you the best with your studies.
At 2:25, why is the angular velocity defined to be in the -k direction? If it's moving clockwise, shouldn't it be moving in the -j direction because it is moving towards -y axis?
Good question, one where a lot of people can get confused. The vector for the angular velocity must be found using the right hand thumb rule. So you have to curl your fingers in the direction of rotation, and point your thumb out. The direction of the thumb shows the angular velocity vector. This image shows it well: scripts.mit.edu/~srayyan/PERwiki/index.php?title=File:AngularKinematics02.png
@@QuestionSolutions Thank you sir. I realised my understanding of these problems was completely off. Thanks for taking the time to help!
@@ryanbragg4334 You're very welcome! You can sort of think of angular velocity as how fast something rotates about an axis, but just keep in mind that the vector for it has to be found using the right hand rule. And most problems, we see it in 2D, but it's really a 3D problem, so you sort of have to imagine the pin where something rotates around is actually a 3D pin, and it'll become easier to visualize in your mind. Anyways, best of luck with your studies!
@@QuestionSolutions That's exactly why I was confused haha. I was thinking in 2D. Thanks again!
@@ryanbragg4334 No worries!
I got 4 in dynamics course at university all thanks to you
Nicely done! Keep up the awesome work and I wish you the best in your other courses.
The Videos is very helpful and the explanation is very good you are better than all doctors I've seen in this subject 💙💙
Wow, thank you! I am glad it was helpful to you. Keep up the great work and best wishes with your studies. ❤
Well Explained
Thank you very much!
what a saviour! man you are amazing!
Thank you very much!
Your Videos with helpfull Concepts are GOAT(Greatest of all time)🙏
Thank you very much, glad to hear it was helpful!
can someone please tell me why at min 2:30 is in the negative k direction ? i dont get it that part
The vector for the angular velocity must be found using the right hand thumb rule. So you have to curl your fingers in the direction of rotation, and point your thumb out. The direction of the thumb shows the angular velocity vector. This image shows it well: scripts.mit.edu/~srayyan/PERwiki/index.php?title=File:AngularKinematics02.png
wish i had these videos when i was taking dynamics in university
I hope they are helpful to new students :)
Thank you for your videos, they're immensely helpful.
You are very welcome!
Thanks for the great video bro
You're welcome and thanks for the comment!
Love the animations!
Thank you very much!
Thank you these are awesome!!!!
You're very welcome!
thank you so much you're videos are really helpful for my mechanics exam
Glad to hear. I wish you the best on your exam!
Concise, easy to follow channels like yours make me wonder why I am paying $20k/yr for college 🙃
Glad to hear you found it concise and easy to follow! Keep up the good work and best wishes with your studies :)
Thank you so much for this video mate, great explanation and great video quality too. Btw at my university our prof. talked about something called "Stübler Theorem" or Stübler's Theorem" when she was explaining relative acceleration on rigit bodies. I couldn't find anything about it online, have you ever heard of such a thorem?
Thanks! Hmm, I don't think I heard of Stübler Theorem. If you do ask your professor again on what it is, please share with me also, I am curious to know what it is. I searched for it but couldn't get any hits.
You saved me from failing. I have an exam tomorrow morning.
Thank You my unknown friend 🙂
You're very welcome! I hope your exam went okay. Best wishes with your studies :)
Bravo sir. This was very helpful, thank you
Glad it was helpful and you're very welcome!
For the example at 4:20, why do you use the cross product to solve for Vb and why not just Vb=wr, whereas at 6:20 you just used Ve=wr?
You can solve these questions in many different ways. I showcased relative motion analysis, but it's up to you what method to use.
Great video, really appreciate it!
I am glad to hear that! Best of luck with your studies.
you are perfect man, thanks a lot
Thank you for the kind comment :)
Thank you 🙏🙏🙏
You're very welcome!
Im confused, if the problem is in 2D why are we using K if that's technically in the Z direction? Wouldn't we only need I and J?
For the future, please kindly use timestamps so I know where you are referring to. I assume you're asking about the k component of the angular velocity? If so, the vector actually points into the screen. You have to use the right hand rule and see where the thumb points. So here (2:24), the angular velocity is clockwise, so if you curl your fingers so that they are clockwise, the thumb will point into the screen, which is the negative k direction. Here is a diagram illustrating the vector directions. scripts.mit.edu/~srayyan/PERwiki/index.php?title=File%3AAngularKinematics02.png
@@QuestionSolutions Thank you that explained it!
2:54 can u pls explain how do u get the Vc?
Vc in the question is given to us as 8m/s. All we did there was just break that into components. Since the 8m/s is at an angle of 45°, we can use cosine and sine to break the velocity into x and y components.
Thank you so much for these videos!
You're very welcome!
Very helpful! Thnx a lot!
You're very welcome!
Nice work! How you do this animation.
Thank you. I use illustrator for the diagrams and after effects for animations.
at 5:02, why do we write the relative velocity as Va=Vb + (Vab) and not Vb= Va+ (Vab)?
So you're trying to solve for va, which is why it's isolated on the left side. You can do it the other way, make sure your position coordinates and values are properly set.
Very good video! May I please ask how do u know where to start please? For example, U start with Velocity of B in the first example, but in the last one, u start with velocity of e. Thanks in advance!
Thank you. To answer your question, I think it comes from intuition. Just a hunch that going this way will work our better and give an answer faster. Sometimes, it's wrong but usually, it's right. As you do more questions, your intuition becomes better and you can just visualize the flow of the question and answer. :)
All of ur videos are very useful & very easy to understand...☺️🥰
Thanks a lot 🌝
You are very welcome! :)
could you please explain where you get vc = 8 sin theta - 8 cos theta? Appreciated your video so much
Please see 2:46. We are breaking the velocity into x and y components using the 45 degree angle given to us. So the x-component can be found using sine and y-component using cosine. Y-component is negative because we chose up to be positive, but the component is downwards. I hope that helps!
@@QuestionSolutions Thank you!
why do you use the 30rad/s for Ve and 15 rad/s for Vc @6:26 ?
So Ve is found by multiplying the angular velocity of gear A by the radius, where as Vc is found by multiplying the angular velocity of the rod BC by the length. So each have their own corresponding angular velocities.
ı owe you a comment because of your great content
Thank you very much :)
Amazing thanks alot
You're very welcome!
These are great videos, thank you so much! at 2:24 how did you determine that the angular velocity was in the k direction?
So the vector would point into the screen. Curl your right hand fingers in the direction of the rotation and look to see where your thumb points. Since it points into the screen, that's the negative k-direction, or negative z-axis.
Really helpful video thank you. How do you know whether a rotation is positive or negative?
Thanks!
Can you give me a timestamp where you're confused about which direction is positive or negative?
Hi, how do u determine the direction of velocity? like for the very first example, we know Vb from w but how do we know Va from Vb?
You can usually make an educated guess as to the the direction of velocity, you just have to imagine the system moving in your mind. If not, that's alright, in the end, if you end up with a negative value, that means the assumption you made for the direction of velocity is incorrect, in other words, it's opposite to your assumption.
I wish you did more questions for this subtopic
I will create more in the future :)
@@QuestionSolutions Thank you.
At 4:30 why does the wheel spinning clockwise dictate that the value should be negative?
So clockwise means that if you use the right hand rule, where you curl your fingers to match the direction of rotation, you will see that your thumb points away from you. If it's away from you, it's in the negative z direction so it's a negative k component. Keep in mind that while the rotation is clockwise/counter-clockwise, the vector for angular velocity is found using the right hand rule. See: hyperphysics.phy-astr.gsu.edu/hbase/rotv.html
@@QuestionSolutions Wow... I don't know how to thank you enough. I appreciate you and your content so much 🙏🙏
You're very welcome! Best wishes with your studies :)
Got my 2nd test tomorrow, wish me luck guys!
Best of luck with your test!
well-done, this is great
Many thanks :)
You are the best thank you
You're very welcome 🙂
Hii there love your videos, I just want to clear up some of my confusion, at the third example (6:19) how do we know that we should calculate the VE and VC, like what would be the reasoning ? I understand what you are doing I just struggle to understand how to know in an exam that, that is what i should be doing. I hope this makes sense 🙈
I am not too sure how to answer that question. :( It should sort of "appear" in your head what steps to take next. If it doesn't happen, that's okay, but it just means you need to solve more problems and visualize the method to solve that problem in your head. Everyone does it at a different pace, but the more questions you solve, the easier it is to know how to tackle that problem.
Hi there :) in the example at 3:24 , from the equations of the i and j components, don't Wab and Wbc have the same sign and therefore direction of rotation?
Yes, there is a typo there :)
@@QuestionSolutions okay, thank you very much for the prompt reply. I really appreciate your videos! Many thanks.
@@noalily6922 You're very welcome!
God bless you...
Thanks!
i have a question, why did the i component in 3:26 changed to wab for the final solution but it uses wbc for the calculation
It's just a typo, you can see I mixed up WAB and WBC 😅
@@QuestionSolutions i see i see was figuring hard why until i looked into the next question, thanks for the video, much appreciated
@@changjunyang7471 You're welcome. I apologize for the typo.
Doing a very great job yar.
Many thanks!
I have a question at 5:18 why is it -0.5cos60? and not +0.5cos60?
So the position vector goes from B to A. In other words, to the left and up. We chose up and to the right to be positive. Since the x-component goes to the left, it's negative.
Ahhhh make sense, you are a legend subscribed.
@@sinahafezi425 Thank you! :)
at 5:18, can i ask why is "-0.5cos60i", but not "0.5cos60i" ?
So the position vector goes from B to A. In other words, to the left and up. We chose up and to the right to be positive. Since the x-component goes to the left, it's negative.
@@QuestionSolutions ohhh, thank you so much sir :>>
At 5:12, why the angular velocity is in k ?
So it's rotating clockwise, right? When we look for the vector, we have to use the right hand rule. So you curl your right hand fingers in the clockwise direction and see where your thumb points. It points towards you, in other words, out of the screen, so that's the z-axis, (k component).
Greatly explained as usual! I've got a question though, a simple one rather. At 6:51 , for the relative velocity equation. What if we wrote it for C with respect to E, so Vc/e instad. So [ Vc=Ve+Vc/e] I've done that and my answer came out to be negative (-105), what is this supposed to mean? Becaus clearly it cant go the other direction. The gear does infact rotate and has to rotate in a clockwise manner I believe haha for the system to work in the first place, atleast in this question per se.
Most likely a positive/negative error. For your r_C/E, was it -0.05? Because you switched it around, so it'll now be negative. If you get 105, but the sign is wrong, it's almost always just a positive/negative error, maybe when you did the cross multiplication, you forgot a positive or negative, etc. It's hard to say 😅
@@QuestionSolutions Ah yes, it was a simple sign error haha. To my luck, my errors are mostly always related to the signs. It was the R_c/e as u said. Thank u once again!! Have a graat day
thanks so much bro
You're welcome!
hi i have a question, i will really appreciated if you can responde quickly. the thing i wawnt to ask why in the 2:32 we call it -k. its spoused to be -i
So to figure out the direction of the vector, you have to use the right hand rule. While the angular velocity is clockwise, this is a 3D vector. If we curl our right hand fingers the same direction as our angular velocity, we see that our thumb points into the screen. So that's the negative z axis, or -k axis.
Thanks for the videos they are very helpful. One mistake you have is you have labelled the angular velocities in rad/s but your values have been calculated in degree mode. This means some of your values have come out incorrect w.r.t the units you have used
You are incorrect. The answers shown are correct. So I can see where the confusion comes from, but the degrees were used to break velocity into components. What you end up as units is still m/s. So the components of 8m/s can be 8sin45 degrees, but the answer that comes out is 5.66 m/s. So there is no correlation between the degrees and rads for the final answer. The units shown are correct :)
@@QuestionSolutions hm interesting. I never knew you could use rads and metres interchangeably. I was always taught to convert the units, when I presented this problem to others they also said the units were mixed up. Maybe this is a difference in method of teaching
@@truewassabian2491 Here is a solved example directly from a textbook. Please see: bit.ly/3y7xE0o
why is the cross product not used in the problem at 5:45
Cross product is only required if you're using vectors, otherwise, it's easier to use the scalar method.
Can someone please explain to me the following. In my class in a similar exercise to 1:50 when we defined Vb=Va+w*r(b/a) we also include Vb= V (relative) +Va +w*r(b/a) , and as for r(b/a) we also changed and make it relative. When are the cases that need to add relative velocity and make r(a/b) also relative ??
It's really hard to answer that question without actually seeing what example you took up in class. At 1:50, at A, there is no velocity since it's just pinned there, so it would be zero. And since VB is a rotation about a fixed axis, (here the fixed axis is pin A), we can figure it out by multiplying angular velocity by a vector from A to B. To use the relative equation, you need 2 velocities to compare. So at 2:43, we are comparing the velocity at B to velocity at C. So in simple terms, if we need a velocity and its a rotation about a fixed axis, we can find it by multiplying the angular velocity by "r" and if we are comparing one velocity to another, then we need to use the relative velocity equation. Vb = ωr is not a relative velocity equation, its just an equation to find the velocity at a point if it's a rotation about a fixed axis. Also, you mentioned r(b/a). So again, if we are just using V = ωr, then its the distance from the axis to the location where we are finding the velocity. If it's the relative velocity equation, then its a vector from the location of the second velocity to the first velocity. I think it'll help a lot if you do the question first, and then if you get stuck, stop and view what step I did next. Sometimes, doing it yourself is super helpful. 👍
@@QuestionSolutions So If I am getting it right it there was an external velocity acting at point b (for example a motor pushing b) then we would use the relative velocity equation?
@@melekmnif1933 Yes, you definitely can use it like that, but its not limited to just those cases. So let's look at just example 1 in the video. In that example, the block at C moves down because of gravity. When it moves down, it has a velocity, which is Vc. Now that block is connected to rod CB. So what that means is, velocity at point C is the same velocity as the block. The goal is to use that to figure out the velocity of the other parts that's connected. You can see that to rod CB, there is another rod connected, which is AB. You can also see that because block C slides down, rods CB and AB both rotate. So whenever you have a velocity with connected pieces, we can use relative velocity to figure out the missing angular velocities and linear velocities. Since in that example we know the velocity at C, we can use that to figure out the other parts. So instead of thinking about an external velocity, think of it like this: If we have connected parts, and we know the velocity of one of those parts (or even angular velocity of one of those parts), we can use that velocity to figure out the velocity of another part. That's why it's called relative velocity, the velocity with respect to another point. Does that make sense?
@@QuestionSolutions I got your point but I think I asked a wrong question. I hope I am not asking for too much but here is a link to the exercise :
To answer the velocity part of the question we first used Vb=Va+w*r(b/a)= -300j and then Vb= V(relative) +Vc +w*r(b/c) to find w
So my question is why did they add V(relative) to the formula? and how to find it in general?
www.google.com/search?q=The+gear+has+the+angular+motion+shown.+The+peg+at+A+is+fixed+to+the+gear+and+the+vertical+distance+from+D+to+A+is+1.2ft.+A.+Determine+the+angular+velocity+and+angular+acceleration+of+the+slotted+link+BC+at+this+instant.+B.+Don%E2%80%99t+forget+to+draw+your+FULL+kinetic+diagram+of+all+non-zero+force+and+velocity+vectors&rlz=1C5CHFA_enUS810US810&sxsrf=ALeKk03arZ5Cjk2Ex4ovEQ5o21y8CulZbA:1605637945859&source=lnms&tbm=isch&sa=X&ved=2ahUKEwj6h5q7m4rtAhWNg-AKHZV4AYIQ_AUoAXoECAcQAw&biw=1920&bih=1089#imgrc=1PebBbBekr9sEM
Thank you for the effort !!
@@melekmnif1933 So they wrote "v(relative)"? That was part of the equation? I've never come across it. What was the value of "v(relative)"?
(02:50) Shouldn't Vc = 8cos(45)i + 8sin(45)j ?
Instead of Vc = 8sin(45)i + 8cos(45)j
No, the opposite side to the angle gives us the x-component, so sine, and the adjacent side gives us the y-component, so cosine.
How do you know that this is not a general plane motion 6:21?
I have a separate video on general plane motion. Were you able to take a look at that one? You will see that general plane motion questions can't be easily solved like these ones, you need a gazillion equations to get an answer.
At the end of solution of question 1, why are the rotation directions opposite ? both the values of omega are taken to be positive and must be anticlockwise according to the convention
They are positive, meaning the directions we assumed were correct. Notice how we assumed a clockwise rotation for AB and a counter-clockwise rotation for BC. Getting positive values indicates that those assumptions were correct. But regardless of sign conventions, with these problems, it's easy to figure out directions just by imagining how these contraptions would move. Here, block C moves down, and there is no way for that to happen if link BC was going clockwise with AB also going clockwise. The same is true for the opposite, if link AB was going counter-clockwise, then block C can't move down, it has to move up.
my fav video
:) Thanks!
In first example, you have angular velocity of i component as wab and j component as wbc? Should it be wbci and wabj?
Could you provide me with the timestamps you're referring to? I am having a hard time understanding your question. I'll try my best to help. Also, the steps shown are correct, there are no mistakes in the solution.
In the first exercise (with the slider block C at 8m/s); when filling in the general equation of relative velocity it is confusing to have 8sin45.i . Shouldn't the i component be 8COS45 i and the j - 8 sin45° ? For this exercise it doesn't matter because cos and sin of 45° is the same, but I think the sin and cos should we switched, unless I'm wrong ofc.
So here, I can't give you a definite answer because I don't know whether you took the top angle or the bottom angle and both will be 45 degrees. If you took the top angle, then opposite to the angle give you the y-component, and adjacent gives you the x-component. If you took the bottom angle, opposite gives you the x-component and adjacent gives you the y-component. If you, for whatever reason have an idea that says x-components are always cosine and y-components are sine, you need to get rid of that before you fall into a lot of pitfalls and lose marks on your exams. Here is a video < 60 seconds that will really help you understand how it works: ua-cam.com/users/shortsvynnKlJD_Jo?feature=share
why is it it the negative k direction 2:25 please?
So you have to use the right hand rule to figure out the direction of the vector. Here, we have a clockwise rotation, so curl your right hand fingers clockwise and see where the thumb points. Notice that it will point into your screen, which is the negative z-axis.
For a given instant ICR has zero velocity and for the problem at 3:29 ICR is the point of contact between the surface i.e. poin C
If this statement is true why have you used V,C as 5i during calculation of V,B at 4:17
The gear rack has a velocity of 5 m/s, this is a given in the problem. So all we are doing, in the simplest sense, is actually just comparing the velocity of one point to the velocity of another. We use -5i because that is the velocity of the rack. If it's zero, then the gear above will go on top, carrying the whole system with it. I think for these problems, you really have to use your imagination to see how this device would move in real life. What you're thinking of is a system that's making contact with a surface that's not moving, but here, that is not the case. I hope that helps!
Thank you! :)
güzel özetlemişsiniz hocam, teşekkürler.
Thank you! I used google translate so I have an idea of what you said :)
@@QuestionSolutions I thought you wouldn't answer :D I m from Turkey, I m glad sir
@@e-nes4042 an itu student? right?
for the second question wouldnt the j component of r_a/b be negative and the i component be positive?
nvm numbers work out to be the same
@@ShangWeiorNoWay 👍
Great video as always, but why did we not find VB in the second problem the way we found VE in the 3rd problem which all we did was multiply the angular velocity by the radius?
Thank you
We can't do that because there is a gear rack on the bottom that's affecting the speed of the wheel. It's not an independent object, it's connected to something else. In the case of VE, you can see that it rotates about a fixed axis, in this case, point B, and point B doesn't move. In the VB question, the center moves along the x-axis. 👍
@@QuestionSolutions Thank you
when we say relative motion for the second point on the rod how can we assume if it is only on x axis or both for example and when to do this assumption
Sorry, but I don't understand your question. Where are you referring to? Is there a specific part on the video? Let me know and I will try to help you out. Thanks!
at 6:50 was the radius of the gear given?
300mm - 250mm = 50 mm. It's not given per say, but it's very easy to calculate with the given info on the diagram :)
@@QuestionSolutions lol sorry for my stupid question! thank you
@@Caroline-ot8jn No need to apologize. Small things are usually the easiest to miss. Best wishes with your studies!
Amazing!!!
Thank you!
Nice video . Please also solve problems from merriam and kraige
Thank you. I think if students understand how to apply the new equations they learn, regardless of where the questions come from, they should be solvable. But I will keep in mind the Merriam and Kraige books as well. Thanks again for your feedback, I appreciate it.
What does rb/c mean?? do we take components of r from b to c or from c to b??
it's c to b.
the formula is only applicable for rigid bodies right?
Yes, for relative velocity of particles, see: ua-cam.com/video/opVKNCedkRo/v-deo.html