oh my god I cant believe i have found such understandable youtube videos with regard to mechanic. I am a freshman in major of mechanical engineering. I have been so frustrated when it comes to dealing with this relative rigit body. now everything has turned. thanks a lot.
All your videos are truly amazing & and are getting me through Dynamics with an A. They're definitely short and concise. Thank you so much for all that you do! 👍🏽👍🏽
Thank you so much for your kind words, I really appreciate it. On a different note, if possible, please kindly share these videos with your friends/classmates. Maybe they too will find them helpful, and in turn, it will help the channel out too :) Again, thanks so much for your wonderful comment. Best of luck with your studies!
Really amazing explaination....i didn't get any videos of such great explaination on RUclips .....whenever i got video of yours on searching...a ray of hope generates...you have also too clear voice...cuz i am indian...so bit difficult to understand foriegner...but i get ur each point very clear.....lots of love sir ❤️☺️
Thank you very much Simran. I am glad these videos are helpful to you and I hope you do amazingly on your courses. Best wishes and thank you for taking the time to write your comment, I really appreciate it. ❤
Hi At 4:44 I'm sure most of you are confused at how Sir got the acceleration at the center.(a-not), Well think about it this way We looking for the horizontal acceleration at the center and well luckly for us that acceleration is parallel and equal to the tangential acceleration at circumference Which is calculated.... Mmm u getting it now... A =r×α where α is the angular acceleration, and r is the radius of the circle.( U can find that formula anyway)..... Sharp
Thanks for the valuable information that you provide. Without you, I was surely going to fail Dynamics (btw the mechanisms that you showed seem kinda sus ngl)
Hello, I have a question. At 8:53, α_AB was negative, so we knew it was clockwise. But why did α_C turn out clockwise even though it's output was positive? Shouldn't it be negative for it to be clockwise?
Hello. For the first problem, the equation you used to find the acceleration at point B was: a(b) = "a"(ab) x r(ab) - w(ab)^2r(ab), but then, to find the acceleration at C, the equation was this: a(c) = a(b) + "a"(bc) x r(c/b) - w(bc)^2r(c/b). I understand why you must add the extra a(b) term in the second equation, but shouldn't you add an extra a(a) term to the first equation because you are solving for the acceleration at point b using the values at point A? Please explain and ask questions if my question is not clear. Thank you.
For the acceleration at point B, it's "simpler" because we already have the angular acceleration of rod AE given, so we can use that. When we do point C, we don't know the acceleration of link BC, so we need to use the acceleration of one point and compare it to another, in this case, we used a_B to get a_C.
@@QuestionSolutions Hi. You said that we already know the acceleration at point A but you did not use that value at all when calculating for the angular acceleration at point B. Thank you.
So you don't know "exactly" where the vector would point, but you have a really good idea. What I mean is, you know that it has to point to the bottom-left, just from the way this contraption moves, along with the fact that we are given the angular acceleration and angular velocity of link AB. So it must be pulled towards the left, and seeing the relative locations of the links allows us to make a very good, educated guess. After that, doing the math gives us the actual vector -7 in the x-direction and -3.5 in the y-direction (imagine a coordinate system at point B with the acceleration vector starting at the origin).
No, no, it's not "a" cross "r". This is just the scalar version. So you multiply angular acceleration by the radius for a disk that's not slipping to get the acceleration (NOT angular acceleration) at the center.
First of all, thank you for saving my life in dynamics. Secondly i just have a small question, when you wrote down the equation for the acceleration at B why did we not not write down the angular velocity is in the K direction and just wrote down the number? Please explain. Thank you for your time!
You are very welcome and I am very glad to hear these videos helped you. To answer your question, can you give me a timestamp so I know where to look? Many thanks! :)
@@mohammadmerheb4597 So every part of that equation is written in cartesian form. Now let's look at our diagram, see how the acceleration is counter-clockwise? You can see it from the blue arrow on the left side of the diagram. Now if you curl your right hand fingers to match the rotation, meaning your fingers are now curled in a counter-clockwise direction, you will see that your thumb points outwards, so towards you. That's the positive k direction. If it was a clockwise rotation, then your thumb would point away from you into the screen. So that would be -k direction. When we use relative motion analysis, you should have everything in Cartesian form. Also on a side note, on RUclips, you can just type 3:00 or whatever time like 5:45 and it will become a hyperlink so you can click on it.
Hey, I really love your content, and I am so glad to have found your channel. I do have a question, I was working through a practice problem on my own and it had a crank, connected to a beam, connected to a 2nd crank. since the motion isnt constrained and doesnt inform my angular acceleration in any way, im confused as to how I should solve for angular acceleration of the beam since I have two equations and 4 unknowns (alpha X alpha Y Vcx and Vcy) since I am solving outward from the crank with given information. I am able to solve for velocities angular and linear, but I dont see how that helps me to solve further for angular acceleration.
I am glad you love the content :) As for your question, it's really hard to say without seeing it. It sounds like it might be a general motion problem? Hard to say without seeing it. Try watching these 3 videos and maybe it'll cover it: ruclips.net/video/lvP_ZcL0WEQ/видео.html ruclips.net/video/G4f0-MGtezc/видео.html ruclips.net/video/UNuRhIHthhY/видео.html
Question: The way the velocity of a crank is perpendicular and in the direction of motion,how do we draw that of acceleration? And how do we locate point I whichvwe use to find 'r' as we did with velocity?
You can find the instantaneous center of velocity whenever you have enough lengths/angles given. But if a question doesn't give you lengths and angles, which are necessary to use instantaneous center of velocity, then you can't use it. 👍
i really love your vids to help me go through aerospace engineering. but I have question at 3:24 when you cross the angular acceleration of ab (k) with negative notation behind it and crossing it with 0.5 sin(45) J, should it be equal to + 5,657 I, but you put in the video the notation is -5,657 I ?
So the 5.657 doesn't come from a cross product. It's just 4^2 x 0.5cos45. The cross product values are in the brackets. You end up with 2 negative terms because there is a negative in front of the 4.
Thank you for these super helpful videos! A question, how do you predict the direction of the acceleration vectors (at 3:41 of the video, for example)?
So you can make a very good assumption by looking to see how the object/rods would move. If by chance, your assumption is not right, you will get a negative answer, so you know it's opposite to your assumption.
at 1:00 and 2:40 , for the formula of acceleration, why did we subtract normal acceleration from tangential acceleration? as i know acceleration was addition of them.
That equation and proof should be in your textbook under the rotation about a fixed axis section. In a nutshell, the normal acceleration component can be found by "-ω²r". The tangential component is "α x r". Adding the two together gives us: α x r - ω²r
You know that when solving for velocity using Cartesian method, if the angular velocity is in the clockwise direction, then its (-wk) and positive when and clockwise. Does that affect angular velocity too? Or does velocity remain +ve throughout?
So in cartesian form, we use the right hand rule to figure out the direction of the vector. I am not sure if I understand the 2nd part of your question. "Does that affect angular velocity too?" What do you mean by this? Please let me know so I can help, thanks!
Thank you sir !! I do really appreciate your work :) but one question ... how do we guess the directions of velocities when we are given just one direction?? It’s not always that clear to figure out where our components are moving to
@@zarapiryaei5569 Usually, you can assume the direction. When you do this, make sure you follow through to the end of the calculations and if you get a negative value, then it's opposite to the direction you picked.
4:53 To find the acceleration at the center of the disk, why do you use a = angular acceleration * r and not a = angular velocity^2 * r? Since the acceleration is towards the left, I would think that it is a normal acceleration and not tangential, hence using the latter formula.
So the direction of acceleration doesn't matter, in other words, it doesn't matter if it's to the left or right. Normal acceleration points towards the center of the curve, but what we want is to find tangential acceleration.
@@QuestionSolutionsThank you for your reply! Although I don’t think I understand why we want to find tangential acceleration and not normal. How do we know which one we want to find?
@@szeloklau7380 You're very welcome. I think you should ask this, why do we want normal acceleration instead of tangential acceleration? In almost all cases, unless you have some object travelling along a curve, and the question specifically asks for this, you will end up using tangential acceleration. Also, you will notice that if a question just says "acceleration" it is usually referring to just tangential acceleration. If it says "magnitude of acceleration" then you have to find normal and tangential, square each term, add it together and then take the square-root. If a question says "normal acceleration" then you should specifically focus on the normal acceleration.
thank you for your quality content.. i have a doubt if i want to solve question 2 5:52 with IC method so firstly i will calucate distance from point of content to point C let say it x.. then i will multiply x with alpha to get tangential acc and then multiply x with omega² to get normal acceleration and then i will resultant to get Ac... is this a right approch
I am not exactly sure what you mean, but in general, we cannot use IC to figure out accelerations. If you draw a perpendicular line to the velocity vector at C, not the acceleration vector, I am unsure how that will help solve the question. Sorry I couldn't be more of help to you :(
So you have to think about these problems in a 3D sense. If the object is going counter-clockwise, then using the right hand rule, (so curl your fingers counter-clockwise), your thumb will point out of the screen, towards you. So that's the positive z-direction. That's the acceleration vector in the z-axis.
Great videos! I have a question. In your previous video(Rotation about a fixed axis) you showed that to get the total acceleration we have to sum tangential and normal acceleration. But here you are subtracting normal acc. from tangential acc. Could you kindly clarify it for me? Thank you.
@@samiulhassan9127 Okay, that makes it a lot easier. In the video about rotation about a fixed axis, we are talking about an acceleration vector, and how it has 2 components. It would have a normal component and a tangential component. This is true for any acceleration. So when you accelerate in a car, you also have those 2 components, but unless you're driving along a curve, normal acceleration is 0. Please see this video for more information: ruclips.net/video/hwHy4Dewc7g/видео.html In this video, at 0:15, we are talking about the relative acceleration equation. This equation is used to compare the acceleration of one point to the acceleration of another point.
So a way to think about it is to realize that if slider B is pulling to the right, then the acceleration will be along rod AB at that instant. It's easy for us to tell since point A is lying on the x-axis of the wheel. I guess you sort of get a feel for these things by doing a lot of questions :)
One thing I did not understand in the first example is how we determined the direction of the acceleration. I am asking because on the exam we will be asked to show the directions for each point. Also, how did we determine the direction of the angular acceleration for the second bar in the first ex. ?
Generally, you can make an educated guess. If the angular acceleration is positive, then your acceleration will be usually in that direction and also positive. So in the first example, slider C is locked onto the bar, so it's easy to know which way it can go, so velocity must be to the left. You really have to try and imagine these pieces moving in your mind, which is why I try to animate them as much as possible, so it's easy to see. When slide C moves to the left, that means that rod has to turn clockwise. So then we can show the direction of acceleration with that information. I hope that helps :)
Mechanics for engineers - dynamics by R. C. Hibbeler and K. B. Yap. They also have a book for statics which you should learn before doing dynamics, since it gets carried over.
4:54 why could we measure the acceleration of the centre of the wheel directly using (angular acceleration)*r ? Where is the reference of 0.5m? Is it the zero velocity at the bottom of the wheel? I thought that the acceleration equals (angular acceleration*r)-w^2r so how could we ignore the last part?
Please see: ruclips.net/video/zrmBObWEDuE/видео.html You can use the scalar method or the vector method to get the answer. Here, it's easier to use the scalar method since it's just "αr" instead of using the cross product.
Hi! Thanks a lot for your work, it has been incredibly helpful thus far! I have a question though: Why and how exactly you derived the tangential acceleration formula at 4:47 to find the acceleration on the origin? I thought about deriving it through the general formula (a[b] = a[a] + alpha x r[b/a] - omega^2.r[b/a]) with point b being the origin and point a being the contact point between the circle and the ground, but the ( - omega^2.r[b/a]) term wouldn't disappear to leave just (a = alpha.r). Am I missing something or is this not a viable way of doing it?
So this is the scalar version for a rotation about a fixed axis, which is simply angular acceleration times the distance from a fixed axis to where you are calculating the acceleration. Most of the time, this works with wheels and such. The derivation for this should be in your textbook. I don't really cover derivations of equations because it takes too long.
So if you look at the equation we use, the angular velocity itself is a scalar value in the equation itself. On the other hand, the angular acceleration in the equation has to be a vector. This is because the angular acceleration going through a cross product while the angular velocity goes though a simple multiplication. Note that bold letters indicate vectors.
Is it possible to solve this type of questions (acceleration) using only Scalar Method? My lecturer said that it's do-able but apparently I couldn't do it so that's why I'm here, to learn Vector Method. If it's possible to solve with Scalar Method, can you make another video with that?
I mean I guess you can, but in almost all the cases, it's so much easier to use the vector method. To find velocity, I think scalar is easier. But for acceleration, unless the question is very simple, it's really not efficient to do it using the scalar method. Does the textbook you use talk about a scalar method to solve relative motion acceleration problems?
@@QuestionSolutions Well in my lecture notes, they use both vector and scalar methods to solve (but like you said, they use vector when it comes to be tricky) And apparently I realize the Scalar Method solutions are based on certain assumptions (direction of acceleration), where I couldn't do the same thing when it comes to real problem-solving. Anyways I love your videos! keep it up!
@@shinjuku4850 Gotcha, that makes sense. A lot of students don't like to mess around with vectors, but I think vectors are so much easier than just using scalar methods for long problems. For one, you don't have to make any direction assumptions like you said since the vectors automatically give you positives and negatives. Two, the equations are smaller and simpler in the long run, whereas with scalars, you have to do more steps to get to the same answer. I am glad these videos help, I wish you the best with your studies! :)
@@QuestionSolutions Anyways I found something wrong (or maybe I'm incorrect), at 8:47 , why is the angular acceleration of C is clockwise? In early stage of the calculation we assumed that the angular acceleration is anticlockwise (+k), so when we calculated the answer as +126.67 m/s^2, it shows that our assumption is correct (anticlockwise), did you make an error on the direction of arrow?
Yes, but I don't remember which one. All of the videos go in a series, so one builds upon the other. It's really hard to cram all the info into a single video, or keep repeating the same stuff because it makes the videos too long 😅 Which question are you referring to that needed background info first?
if I find the angular accleartion or velocity to be a negative value. should I substitute the negative value or ablouste value to find the velocity or accleartion, Because sometimes subsisting for example W = 3 gave me the right answer but sometimes W = -3 gave me the right answer
Depending on which side you picked positive, a negative sign only implies that it's opposite to your assumption. Carry out the assumption until the very end of the question, and then you can figure out the directions by seeing if the final value is negative or positive. So don't change signs in the middle of the question.
@@mustardthe2ndkingoffastfoo570 You only need to pick positives and negatives if you are using a scalar method. If you use vectors, they automatically take care of it and you don't have to worry about signs since it'll be +/- ijk. 👍
@@QuestionSolutions the problem is I have answered two questions using vectors before. Once I used W = absolute(negative value) ans got the right answer the other time I used the Negative Omega I got to get the right answer which means I think that I either messed up the Vector multiplication or the radius but I should always use the Omega I get
Hi, thanks a lot for these videos, you do help to revise for finals. I am confused towards the negative sign before the normal acceleration ("-w^2*r") at 0:58 , the original equation has a plus instead of minus. In your video: Rigid Bodies: Rotation About a Fixed Axis Dynamics at 3:30 you show that the normal acceleration is positive (" + w cross x cross r "). It would be great if you could clarify the use of different signs. Thank you Edit: In our lectures it's called the two point theorem
There isn't a difference. On the Rotation About a Fixed Axis Dynamics video, at 3:30, we have w cross w cross r, not w multiplied by w multiplied by r. Here, it's shown in a different form, with w^2 multiplied by r. The w cross w cross r takes care of the signs.
I am not sure I understand. What is aC? Point C is fixed, so the wheel turns around it. If possible, please give more details so I can help you out better, thanks! :)
@@QuestionSolutions well I already found my answer. I was little confuse. In the third problem, the wheel is only rotating not "translating and rotating". Thank You.
So you have to think about these problems in a 3D sense. If the object is going counter-clockwise, then using the right hand rule, (so curl your fingers counter-clockwise), your thumb will point of the screen, towards you. So that's the positive z-direction. That's the acceleration vector in the z-axis.
Hello sir, at the first example when you calculate Vb we use the formula wxr but we dont split the r in i and j components, why should we split r in i and j components when we calculate voor ab?
Just by looking at the only way this rod can move with the given information. To understand it better, you can take a piece of cardboard, cut it into rough shapes and then pin them. Afterwards, look to see how the pieces move. 👍
It's not precise, just an educated guess as to the location and direction. Solving the formula gives us the exact value. So if you draw a coordinate system at C, then the acceleration vector would be -10 in x-direction and -0.35 in y-direction.
Usually, you have to imagine these objects moving in your mind, and then you can figure out the directions. Generally, however, if you get a negative value for acceleration, then it's actually opposite to the direction you assumed.
@@QuestionSolutions Thank you sir! But one more thing, at 8:46, when I substitute the answer to the J components, I get different answer for αC which is 406.67. Am I correct in pointing that out?
@@afiffazizz Okay, so I figured out how you got that value. You plugged in positive 70 for α_AB. So at the beginning of the question, we made an assumption, and it turned out, its actually the opposite, but our calculations must be carried out until the very end with our assumption, even if it's opposite. So you have to use negative 70 for α_AB. Then at the end, you can write them positive, and show the real direction.
@@afiffazizz You're very welcome. Always remember, carry out the calculations with your assumption until the very end, and at the very end, change the signs to show direction. This is true for any subject, any question, etc :)
@@QuestionSolutions by drawing the directions of the components of accelerations, like you have drawn for the velocities Vb and Vc creating an acceleration diagram(Ab, Ac, Ab/c) in order to solve analytically instead of using vector analysis. My lectures insist on doing that method ad well and is soo complicated. Can you please help?
@@Pastorella_COD_Mobile Why does your lecturer insist on that method? It's subpar and cannot be used in advanced problems. 😅 I will add it to my list to do, but I don't think it'll get done for some time. But I will try my best.
Could you tell me which time you are referring to? If it's the first example, in reality, while I stop the animation half way, point C is actually increasing in velocity. Because angular acceleration is not negative, it's increasing the angular velocity. So the velocity at C would increase.
@@darrylcarter3691 Yes. So when the wheel C turns, the angular acceleration of rod AB changes with it. At a certain time, t, the angular acceleration is now opposite to what it was before, thereby slowing down the slider at B. You should note that the angular acceleration for wheel C is 0, since its a constant angular velocity. It is this velocity, and the motion of the wheel, along with how the rod is attached that causes the angular acceleration of rod AB to change. This change then causes the acceleration at B to change, but the question is asking about this before the acceleration slows down, since we are given the acceleration at point B to be positive 2 m/s^2 to the right.
So the position vector would go from B to C. If you draw a line straight down from B, you can form the right angle triangle. BC would be the hypotenuse. The opposite side to the angle would give us the y-component, and adjacent side to the angle would give us the x-component. Since the position vector is going from B to C, we go down and then to the right. So down is negative while to the right is positive. In other words, j-component is negative while the i-component is positive. @@abdallahindimi3484
I guess you have to visualize it in your head, but another way is to see the animation. Look at which way BC rotates, and if you want to figure out the vector, then you'd use the right hand rule. If we are looking at the first example, notice how the bottom of link BC, so point C moves towards the left while point B moves to the right (from the perspective of just link BC, not link AB), so it's clockwise.
So the position vector is from B to C. In other words, if we break the position vector into components, the x-component will face right, and the y component will face down. I think you thought of it as it going from C to B, if it's from C to B, then you would be right. :)
So you can think of it as the link pulling on the wheel with the rod. The acceleration vector will point along where the pull is happening. This in turn causes the wheel the spin, but the velocity vector is perpendicular to edge of the wheel.
@@QuestionSolutions Wait, but at 8:03, we learn the exact x component of the vector aB, which is 60. But if the direction of the vector is known (towards the link AB), why can't we find the y component from there? If the x-component is 60, and we are in a 3 4 5 triangle, then the y component would have to be 80? However, we find a different answer later in the video?
Sorry but I am not understanding your question. Vector ab just has a single component, along the x-axis, nothing else. aB is the acceleration of the slider, not the wheel or the rod. @@itsyaboybinnybenis8011
So you look at member AB and you see what's given. We see that member AB has a counterclockwise angular velocity, with an angular acceleration in the same direction, so there is only one direction the vector can go, which is to the left. These questions do require a bit of imagination where you need to imagine the mechanism moving, which is why I animated them. That way you can see it easier :)
Sorry, could you kindly give me a timestamp as to where you think we should break the velocity into components? Then I can give an answer as to why we do or we don't. Thanks!
So the vector goes from B to C. Imagine you're at point B. To get to C, you have to walk down and to the right. The down (y-component) is given by sine, we chose up to be positive, so down would be negative. Going to the right (x-component) is given by the cosine component, which is positive since right is positive and left is assumed negative.
@@jenncycarranza3370 Everything is written in cartesian form except for the angular velocity, which is a scalar. So the 0.5m is the magnitude (length of the rod), which we break into components. Since r_AB is a position vector, it will always be in cartesian form. Also, the bold letters represent cartesian vectors, while non-bold letters represent scalar values. I hope that helps 👍👍
@@QuestionSolutions thank you! also, if angular velocity is a scalar, do we not use negative signs when it's going cw instead of ccw? (same problem but when you were finding ac)
@@jenncycarranza3370 No, if you notice, we multiply it by the positive vector on the rod, which is in cartesian form, so the answer you get will also be in cartesian form, which means positives and negatives are already taken care of. I hope that makes sense?👍
Good day sir!, I've got a question, at 2:51 , isnt the acceleration aB, found from the formula for acc. about a fixed point , so basically I mean is it not supposed to be aB=(alpha)AB X rAB + (omega) X [Omega X rAB]? Or does the double cross product just simplifies to a scalar squared so omega^2, I have tried doing it using the notation/expression i just mentioned yet I get a wrong answer so I am wondering if it actually is the same and I am just have a sign error/or a simple calcualtion mistake, or theyre actually 2 differnet formulas.
No, you have to use the equation I show at 1:00. Where did you get the equation "aB=(alpha)AB X rAB + (omega) X [Omega X rAB]" from? I am not sure where this one is from, maybe I am forgetting something 😅
@@QuestionSolutions I have got it from your video on Rigid bodies: Rotation About A fixed axis dynamics at 3:34, ruclips.net/video/zrmBObWEDuE/видео.html
@@radiatedbug Technically, this is the expanded form when you specifically find tangential and normal accelerations. Most of the time, you're not going to use that equation, but the one I show at 1:00 mark, which is the same, but "simplified."
@@QuestionSolutions So mathmatecially speaking, theyre the same equations just that the double cross product simplifies to a scalar? (Omega^2) So ultimately they have to give the same answer?
You can refer to your textbook, but here is another resource: eng.libretexts.org/Bookshelves/Mechanical_Engineering/Mechanics_Map_(Moore_et_al.)/11%3A_Rigid_Body_Kinematics/11.4%3A_Relative_Motion_Analysis Look at section 11.4.12
I am sorry for asking so much is it possible with same assumptions and same axises to get Wcb = negative value. ALPHAcb = postive value? this happend to me and I got correct values somehow
Yes, it's possible. Imagine you're driving a car and you have a velocity of 5m/s. However, your acceleration could be -2m/s^2. All it means is that while going forward, you are hitting the breaks, slowing the car down. Instead of linear velocity, we are just talking about angular velocity and angular acceleration, but its the same concept. They both don't have to be positive, or both negative, etc.
Hey, great videos I watch all of them. Just wanted to let you know that I think there's a typo at 1:25. For the angular acceleration, you have AE but it should be AB. Am I correct in pointing that out? Have a great day!
@@dwightbenedict6975 It's hard to tell without seeing the question. If a lot of angles and side lengths are given, maybe you can solve it using instantaneous center of zero velocity or this could be an absolute motion analysis question.
I show the arrows in color, all labeled. They point in the same direction, they are just at different angles. You can see them if you look at the acceleration vector drawn, and the velocity vector drawn. Acceleration vector for point B is drawn in purple. @@tarifulislam7934
I just want to make sure, did you see the arrows and did you understand what I meant by same direction but different angles? Because that's usually how it is, acceleration vector will be at a different angle, where as the velocity vector is perpendicular to the rod. See: ruclips.net/video/1aQ9EZGMdDk/видео.html at this time, I show the velocity vector, notice it's perpendicular to the rod. @@tarifulislam7934
So I am not sure how much fundamental knowledge you have with cartesian vectors. But in a nutshell, x-components are represented with i, y-components with j, and z-components with k. This should have been covered in your statics course, which comes before dynamics. I cover a bit of it on this video: ruclips.net/video/mz7gPpIL0Gk/видео.html
@@QuestionSolutions noo thats not my question bro, i mean i know cartesian products and all very well.... my question is when do i have to solve with vector methods and when with scalar methods...?
@@tahabinaamir2749 Okay, now I understand. So it boils down to the information you're given, and what you're looking for. For example, if you're going to find values using the instantaneous center of zero velocity, I would go with scalar. If the information given for the question can be found using scalar methods, so simple things like angular velocity can be easily found using just scalar methods. For everything else, I'd keep my eye out and use vector analysis. Using vector analysis keeps the errors low since directions, positives and negatives are accounted for automatically by the vectors.
@@QuestionSolutions thank you, so you mean becaue it it the center of a rotating body the center will not have normal acceleration however any other point will have?
That's the angular velocity of w_AB, it's given. It's written on the diagram, on the left side, please pay attention :) There is a typo on the diagram, but that's regarding angular acceleration, it should be AB not AE. But I think it's easy to see with the arrows already there.
![[2015] Dynamics 28: Relative Motion Analysis Using Rotating Axes [with closed caption]](http://i.ytimg.com/vi/1Y-7jtoHy4g/mqdefault.jpg)
oh my god I cant believe i have found such understandable youtube videos with regard to mechanic. I am a freshman in major of mechanical engineering. I have been so frustrated when it comes to dealing with this relative rigit body. now everything has turned.
thanks a lot.
Glad to hear these videos helped! I wish you the best with your studies.
These make everything so simple. Your detailed animation and vector analysis is top notch. Thank you for your time and hard work.
Thank you for taking the time to comment on these videos. I am really glad they help you :)
@@QuestionSolutions Yes sir. I hope they help others find them too!
@@camerongillespie870 ❤
I wish I’d found these videos earlier. They’re very helpful, thank you!
Thank you so much. I wish you the best with your studies!
@@QuestionSolutions I have a problem question can you solve it ..Quickly please 😊😭
All your videos are truly amazing & and are getting me through Dynamics with an A. They're definitely short and concise. Thank you so much for all that you do! 👍🏽👍🏽
Thank you so much for your kind words, I really appreciate it.
On a different note, if possible, please kindly share these videos with your friends/classmates. Maybe they too will find them helpful, and in turn, it will help the channel out too :)
Again, thanks so much for your wonderful comment. Best of luck with your studies!
@@QuestionSolutions I can absolutely do that for you! Thanks again
@@adrianalopez9282 Thank you! I am very grateful.
An A?!?? Girl I'm just tryna pass
Omg only when the figures are moving I eventually understand what’s going on. Thank you very much!
I am glad the animations help! Best of luck with your studies.
Bru you saved my life. your work is appreciated
Glad to hear! Best of luck with your studies.
Hi this is vikram from India. Finally I got rigid body dynamics from you...
Yup. One step at a time Vikram, one step at a time.
You rock!!! Your explanations are clear as vodka
Glad to hear they are helpful 👍
Your voice sounds so calm 💙
Thank you for the TUT. Very informative and straight foward.
Thank you very much :)
This man is a national hero
🤩
Really amazing explaination....i didn't get any videos of such great explaination on RUclips .....whenever i got video of yours on searching...a ray of hope generates...you have also too clear voice...cuz i am indian...so bit difficult to understand foriegner...but i get ur each point very clear.....lots of love sir ❤️☺️
Thank you very much Simran. I am glad these videos are helpful to you and I hope you do amazingly on your courses. Best wishes and thank you for taking the time to write your comment, I really appreciate it. ❤
Mind blowing visualization 👌..superb
Thank you very much!
These contents are masterpieces❤❤❤
Thank you so much! ❤❤❤
Hi
At 4:44
I'm sure most of you are confused at how
Sir got the acceleration at the center.(a-not),
Well think about it this way
We looking for the horizontal acceleration at the center and well luckly for us that acceleration is parallel and equal to the tangential acceleration at circumference
Which is calculated.... Mmm u getting it now...
A =r×α
where α is the angular acceleration, and r is the radius of the circle.( U can find that formula anyway).....
Sharp
Thank you for your tips!
I love this so much, the best educational videos on yt I came across. Thank you!
Thank you very much :)
damnit I wish I found this channel a week ago not a day before the exam...
I hope you did well on your exam! Best wishes with your studies.
Too much good video ang good concepts I had never comment on any video on utube before but yours videos are amazing
Thank you very much! I appreciate it :)
Thanks bro.. short and simple
You're welcome!
thank you so much . What a great explanation of high quality and in a short time
You're very welcome! Thank you for the nice comment.
Amazing video!
Thank you very much!
So helpful! Thank you!
Glad it was helpful and you're very welcome!
Thank you lots. God bless.
You're very welcome.
Seriously thank you for these amazing videos. You are amazing
You're very welcome!
Thanks for the valuable information that you provide. Without you, I was surely going to fail Dynamics (btw the mechanisms that you showed seem kinda sus ngl)
You're very welcome. I hope you did great on all your courses, keep up the great work!
I can't thank you enough!! such helpful videos !
You're very welcome! :)
Hello, I have a question. At 8:53, α_AB was negative, so we knew it was clockwise. But why did α_C turn out clockwise even though it's output was positive? Shouldn't it be negative for it to be clockwise?
The direction of the circular arrow is wrong, it's a typo 😅
@@QuestionSolutions So, α_C is counterclockwise, right?
@@studyhard6214 Yes 👍
@@QuestionSolutions Thank you very much for answer and these helpful videos! 😇🙏
@@studyhard6214 You're very welcome! :)
Hello. For the first problem, the equation you used to find the acceleration at point B was: a(b) = "a"(ab) x r(ab) - w(ab)^2r(ab), but then, to find the acceleration at C, the equation was this: a(c) = a(b) + "a"(bc) x r(c/b) - w(bc)^2r(c/b). I understand why you must add the extra a(b) term in the second equation, but shouldn't you add an extra a(a) term to the first equation because you are solving for the acceleration at point b using the values at point A? Please explain and ask questions if my question is not clear. Thank you.
For the acceleration at point B, it's "simpler" because we already have the angular acceleration of rod AE given, so we can use that. When we do point C, we don't know the acceleration of link BC, so we need to use the acceleration of one point and compare it to another, in this case, we used a_B to get a_C.
@@QuestionSolutions Hi. You said that we already know the acceleration at point A but you did not use that value at all when calculating for the angular acceleration at point B. Thank you.
@@byungshin8773 But we did, maybe you missed it, but at 2:58, you can see that we use the 6 rad/s^2 to get a_B. Then we used a_B to find a_C.
@@QuestionSolutions wow you are amazing. thank you
Thanks for the video! Quick question at 2:48 how do you know that the acceleration vector of B is going in that direction.
So you don't know "exactly" where the vector would point, but you have a really good idea. What I mean is, you know that it has to point to the bottom-left, just from the way this contraption moves, along with the fact that we are given the angular acceleration and angular velocity of link AB. So it must be pulled towards the left, and seeing the relative locations of the links allows us to make a very good, educated guess. After that, doing the math gives us the actual vector -7 in the x-direction and -3.5 in the y-direction (imagine a coordinate system at point B with the acceleration vector starting at the origin).
Hi,
Why in 4:50, a0 = α x r (only). I was solving on my own with ao = α x ro/point of contact -ω2 x r0/point of contact.
No, no, it's not "a" cross "r". This is just the scalar version. So you multiply angular acceleration by the radius for a disk that's not slipping to get the acceleration (NOT angular acceleration) at the center.
First of all, thank you for saving my life in dynamics. Secondly i just have a small question, when you wrote down the equation for the acceleration at B why did we not not write down the angular velocity is in the K direction and just wrote down the number? Please explain.
Thank you for your time!
You are very welcome and I am very glad to hear these videos helped you. To answer your question, can you give me a timestamp so I know where to look? Many thanks! :)
@@QuestionSolutions @ the time 3 mins please
@@mohammadmerheb4597 So every part of that equation is written in cartesian form. Now let's look at our diagram, see how the acceleration is counter-clockwise? You can see it from the blue arrow on the left side of the diagram. Now if you curl your right hand fingers to match the rotation, meaning your fingers are now curled in a counter-clockwise direction, you will see that your thumb points outwards, so towards you. That's the positive k direction. If it was a clockwise rotation, then your thumb would point away from you into the screen. So that would be -k direction. When we use relative motion analysis, you should have everything in Cartesian form.
Also on a side note, on RUclips, you can just type 3:00 or whatever time like 5:45 and it will become a hyperlink so you can click on it.
@ 8:47 you have -10^2(.3i-.4j) then the next line you lose your negative value for the .3 ( should be - 30i+40j)
60i - 30i = 30i. You forgot the positive 60i.
Hey, I really love your content, and I am so glad to have found your channel. I do have a question, I was working through a practice problem on my own and it had a crank, connected to a beam, connected to a 2nd crank. since the motion isnt constrained and doesnt inform my angular acceleration in any way, im confused as to how I should solve for angular acceleration of the beam since I have two equations and 4 unknowns (alpha X alpha Y Vcx and Vcy) since I am solving outward from the crank with given information. I am able to solve for velocities angular and linear, but I dont see how that helps me to solve further for angular acceleration.
I am glad you love the content :)
As for your question, it's really hard to say without seeing it. It sounds like it might be a general motion problem? Hard to say without seeing it. Try watching these 3 videos and maybe it'll cover it:
ruclips.net/video/lvP_ZcL0WEQ/видео.html
ruclips.net/video/G4f0-MGtezc/видео.html
ruclips.net/video/UNuRhIHthhY/видео.html
Question: The way the velocity of a crank is perpendicular and in the direction of motion,how do we draw that of acceleration? And how do we locate point I whichvwe use to find 'r' as we did with velocity?
Can you give me a timestamp? I am not sure where you're referring to. Thanks!
In what situations should I know to apply instantaneous center on a test bc it doesn't always seem so obvious
You can find the instantaneous center of velocity whenever you have enough lengths/angles given. But if a question doesn't give you lengths and angles, which are necessary to use instantaneous center of velocity, then you can't use it. 👍
i really love your vids to help me go through aerospace engineering. but I have question at 3:24 when you cross the angular acceleration of ab (k) with negative notation behind it and crossing it with 0.5 sin(45) J, should it be equal to + 5,657 I, but you put in the video the notation is -5,657 I ?
So the 5.657 doesn't come from a cross product. It's just 4^2 x 0.5cos45. The cross product values are in the brackets. You end up with 2 negative terms because there is a negative in front of the 4.
You never disappoint :)
Thank you :)
Thank you for these super helpful videos! A question, how do you predict the direction of the acceleration vectors (at 3:41 of the video, for example)?
So you can make a very good assumption by looking to see how the object/rods would move. If by chance, your assumption is not right, you will get a negative answer, so you know it's opposite to your assumption.
@@QuestionSolutions I see, thank you for the explanation! I really enjoy your videos.
@@noalily6922 Glad to hear :)
perfect
keep going
Thank you 👍👍
Great video! Can you please also do the chapter Virtual work in statics?
Thanks! I'll add it to my list but I probably won't get back to statics for a while.
Thanks a lot :)!! And if possible, can you also add thin-walled assumption when doing area moment of inertia to the list?@@QuestionSolutions
at 1:00 and 2:40 , for the formula of acceleration, why did we subtract normal acceleration from tangential acceleration? as i know acceleration was addition of them.
That equation and proof should be in your textbook under the rotation about a fixed axis section. In a nutshell, the normal acceleration component can be found by "-ω²r". The tangential component is "α x r". Adding the two together gives us: α x r - ω²r
@@QuestionSolutions thanks a lot but why negative sign?
You know that when solving for velocity using Cartesian method, if the angular velocity is in the clockwise direction, then its (-wk) and positive when and clockwise.
Does that affect angular velocity too?
Or does velocity remain +ve throughout?
So in cartesian form, we use the right hand rule to figure out the direction of the vector. I am not sure if I understand the 2nd part of your question. "Does that affect angular velocity too?" What do you mean by this? Please let me know so I can help, thanks!
@@QuestionSolutions ohh.
Sorry, I meant angular acceleration.
Good day sir, do you not have a video on motion relative to rotating axes ? If not can you please make one 🙏
If it's not in the dynamics playlist, it probably isn't. I'll look to see and if there isn't one, I will put it on my to do list :)
Thank you sir 🙏
Thank you sir !! I do really appreciate your work :) but one question ... how do we guess the directions of velocities when we are given just one direction?? It’s not always that clear to figure out where our components are moving to
You're very welcome. For your question, are you specifically asking about a question I used as an example, or as a general case?
@@QuestionSolutions In general case , when our components are not moving along a specific rod or sth like that and they can move freely ...
@@zarapiryaei5569 Usually, you can assume the direction. When you do this, make sure you follow through to the end of the calculations and if you get a negative value, then it's opposite to the direction you picked.
@@QuestionSolutions Thanks for your help !
4:53 To find the acceleration at the center of the disk, why do you use a = angular acceleration * r and not a = angular velocity^2 * r? Since the acceleration is towards the left, I would think that it is a normal acceleration and not tangential, hence using the latter formula.
So the direction of acceleration doesn't matter, in other words, it doesn't matter if it's to the left or right. Normal acceleration points towards the center of the curve, but what we want is to find tangential acceleration.
@@QuestionSolutionsThank you for your reply! Although I don’t think I understand why we want to find tangential acceleration and not normal. How do we know which one we want to find?
@@szeloklau7380 You're very welcome. I think you should ask this, why do we want normal acceleration instead of tangential acceleration? In almost all cases, unless you have some object travelling along a curve, and the question specifically asks for this, you will end up using tangential acceleration. Also, you will notice that if a question just says "acceleration" it is usually referring to just tangential acceleration. If it says "magnitude of acceleration" then you have to find normal and tangential, square each term, add it together and then take the square-root. If a question says "normal acceleration" then you should specifically focus on the normal acceleration.
@@QuestionSolutions This makes sense! Thank you!
Thanks for the video, very helpful indeed.
But at 2:40 , shouldn't position vectors to be r_BA, not r_AB?
Do you mean the vector should go from B to A, or do you mean the notation should be flipped but the vector still goes from A to B?
thank you for your quality content.. i have a doubt if i want to solve question 2 5:52 with IC method so firstly i will calucate distance from point of content to point C let say it x.. then i will multiply x with alpha to get tangential acc and then multiply x with omega² to get normal acceleration and then i will resultant to get Ac... is this a right approch
I am not exactly sure what you mean, but in general, we cannot use IC to figure out accelerations. If you draw a perpendicular line to the velocity vector at C, not the acceleration vector, I am unsure how that will help solve the question. Sorry I couldn't be more of help to you :(
How is the angular acceleration for aB in the K direction? You plugged in 6 rad/s^2 in the k direction around 3:15
So you have to think about these problems in a 3D sense. If the object is going counter-clockwise, then using the right hand rule, (so curl your fingers counter-clockwise), your thumb will point out of the screen, towards you. So that's the positive z-direction. That's the acceleration vector in the z-axis.
Great videos!
I have a question. In your previous video(Rotation about a fixed axis) you showed that to get the total acceleration we have to sum tangential and normal acceleration.
But here you are subtracting normal acc. from tangential acc.
Could you kindly clarify it for me?
Thank you.
Could you give me a timestamp to the location you're referring to? It'd be easier to answer if I know where. Thanks!
Thank you for your reply.
For Rotation about a fixed axis video: 3.30
For This video you could see at 0.15
@@samiulhassan9127 Okay, that makes it a lot easier. In the video about rotation about a fixed axis, we are talking about an acceleration vector, and how it has 2 components. It would have a normal component and a tangential component. This is true for any acceleration. So when you accelerate in a car, you also have those 2 components, but unless you're driving along a curve, normal acceleration is 0. Please see this video for more information: ruclips.net/video/hwHy4Dewc7g/видео.html
In this video, at 0:15, we are talking about the relative acceleration equation. This equation is used to compare the acceleration of one point to the acceleration of another point.
Thank you very much.
Learning a lot from you playlists.
(Please excuse my late reply)
Thank you so much for this. Can you explain it to me at 6:25 why is the direction of aA like that ? How do you know it lay on the rod AB ?
So a way to think about it is to realize that if slider B is pulling to the right, then the acceleration will be along rod AB at that instant. It's easy for us to tell since point A is lying on the x-axis of the wheel. I guess you sort of get a feel for these things by doing a lot of questions :)
One thing I did not understand in the first example is how we determined the direction of the acceleration. I am asking because on the exam we will be asked to show the directions for each point. Also, how did we determine the direction of the angular acceleration for the second bar in the first ex. ?
Generally, you can make an educated guess. If the angular acceleration is positive, then your acceleration will be usually in that direction and also positive. So in the first example, slider C is locked onto the bar, so it's easy to know which way it can go, so velocity must be to the left. You really have to try and imagine these pieces moving in your mind, which is why I try to animate them as much as possible, so it's easy to see. When slide C moves to the left, that means that rod has to turn clockwise. So then we can show the direction of acceleration with that information. I hope that helps :)
Thanks for the video! Quick question, when its rC/O is it the vector in direction C to O or the other way around? I always get the signs mixed up 😅
So if it's rA/B then it's B to A.
what is a good book to buy that teaches you all this including forces?
Mechanics for engineers - dynamics by R. C. Hibbeler and K. B. Yap. They also have a book for statics which you should learn before doing dynamics, since it gets carried over.
Thank you very much! Love your videos BTW!
4.13 how is cos positive as it's going towards negative x axis it should be -1cos60i+1sin60j?
The position vector goes from B to C. Listen again at 4:07, which means to goes left to right. So that's positive 👍
4:54 why could we measure the acceleration of the centre of the wheel directly using (angular acceleration)*r ? Where is the reference of 0.5m? Is it the zero velocity at the bottom of the wheel? I thought that the acceleration equals (angular acceleration*r)-w^2r so how could we ignore the last part?
Please see: ruclips.net/video/zrmBObWEDuE/видео.html
You can use the scalar method or the vector method to get the answer. Here, it's easier to use the scalar method since it's just "αr" instead of using the cross product.
@@QuestionSolutions thank you so much
@@迪安-z6i You're very welcome!
same question , thank you!
Hi! Thanks a lot for your work, it has been incredibly helpful thus far!
I have a question though: Why and how exactly you derived the tangential acceleration formula at 4:47 to find the acceleration on the origin? I thought about deriving it through the general formula (a[b] = a[a] + alpha x r[b/a] - omega^2.r[b/a]) with point b being the origin and point a being the contact point between the circle and the ground, but the ( - omega^2.r[b/a]) term wouldn't disappear to leave just (a = alpha.r). Am I missing something or is this not a viable way of doing it?
So this is the scalar version for a rotation about a fixed axis, which is simply angular acceleration times the distance from a fixed axis to where you are calculating the acceleration. Most of the time, this works with wheels and such. The derivation for this should be in your textbook. I don't really cover derivations of equations because it takes too long.
@@QuestionSolutions oooh, I see. Thank you very much for the quick response!
@@Wolfgandag You're very welcome!
at 3:19, why is the angular acceleration of AB in vector form, but the angular velocity is just a scalar value?
So if you look at the equation we use, the angular velocity itself is a scalar value in the equation itself. On the other hand, the angular acceleration in the equation has to be a vector. This is because the angular acceleration going through a cross product while the angular velocity goes though a simple multiplication. Note that bold letters indicate vectors.
how will the disk slipping or not slipping change our solution to any problem? will there be additional requirements if it did slip?
Usually, if it slips, you'll be given the coefficient of friction to consider. But this is usually rare.
May I ask why (omega_C)^2(r_c) is negative at 7:52? Thank you
That's the equation to find acceleration, see: 1:01 if you need the proof, please refer to your textbook or course notes. :)
Is it possible to solve this type of questions (acceleration) using only Scalar Method? My lecturer said that it's do-able but apparently I couldn't do it so that's why I'm here, to learn Vector Method.
If it's possible to solve with Scalar Method, can you make another video with that?
I mean I guess you can, but in almost all the cases, it's so much easier to use the vector method. To find velocity, I think scalar is easier. But for acceleration, unless the question is very simple, it's really not efficient to do it using the scalar method. Does the textbook you use talk about a scalar method to solve relative motion acceleration problems?
@@QuestionSolutions Well in my lecture notes, they use both vector and scalar methods to solve (but like you said, they use vector when it comes to be tricky)
And apparently I realize the Scalar Method solutions are based on certain assumptions (direction of acceleration), where I couldn't do the same thing when it comes to real problem-solving. Anyways I love your videos! keep it up!
@@shinjuku4850 Gotcha, that makes sense. A lot of students don't like to mess around with vectors, but I think vectors are so much easier than just using scalar methods for long problems. For one, you don't have to make any direction assumptions like you said since the vectors automatically give you positives and negatives. Two, the equations are smaller and simpler in the long run, whereas with scalars, you have to do more steps to get to the same answer. I am glad these videos help, I wish you the best with your studies! :)
@@QuestionSolutions Anyways I found something wrong (or maybe I'm incorrect), at 8:47 , why is the angular acceleration of C is clockwise? In early stage of the calculation we assumed that the angular acceleration is anticlockwise (+k), so when we calculated the answer as +126.67 m/s^2, it shows that our assumption is correct (anticlockwise), did you make an error on the direction of arrow?
@@shinjuku4850 Yes, I believe there is a typo with the arrow. 😅
Hello, did you cover ROLLING MOTION in one of your videos? I mean with formulas and generalities just for this topic.
Yes, but I don't remember which one. All of the videos go in a series, so one builds upon the other. It's really hard to cram all the info into a single video, or keep repeating the same stuff because it makes the videos too long 😅 Which question are you referring to that needed background info first?
@@QuestionSolutions It is ok, I think I got the idea it was very basic thing I saw in my lecture slides.
Thank you anw for your quick response!
@@alexboudakian2148 You're very welcome :)
Hello, in the last question ac shouldn't it be counter-clockwise?
Please give me a timestamp so I can take a look. Thanks!
if I find the angular accleartion or velocity to be a negative value. should I substitute the negative value or ablouste value to find the velocity or accleartion, Because sometimes subsisting for example W = 3 gave me the right answer but sometimes W = -3 gave me the right answer
Depending on which side you picked positive, a negative sign only implies that it's opposite to your assumption. Carry out the assumption until the very end of the question, and then you can figure out the directions by seeing if the final value is negative or positive. So don't change signs in the middle of the question.
@@QuestionSolutions so if I for example picked the right to be postive and I have gotten W = - sqrt 2
Vcj = 3sqrt2 + Wsqrt 2 j
Vc = 2sqrt2?
@@mustardthe2ndkingoffastfoo570 You only need to pick positives and negatives if you are using a scalar method. If you use vectors, they automatically take care of it and you don't have to worry about signs since it'll be +/- ijk. 👍
@@QuestionSolutions the problem is I have answered two questions using vectors before.
Once I used W = absolute(negative value) ans got the right answer
the other time I used the Negative Omega I got to get the right answer
which means I think that I either messed up the Vector multiplication or the radius
but I should always use the Omega I get
@@mustardthe2ndkingoffastfoo570 Yes, you have to carry out the value you get until the very end.
Hi, thanks a lot for these videos, you do help to revise for finals.
I am confused towards the negative sign before the normal acceleration ("-w^2*r") at 0:58 , the original equation has a plus instead of minus.
In your video: Rigid Bodies: Rotation About a Fixed Axis Dynamics at 3:30 you show that the normal acceleration is positive (" + w cross x cross r ").
It would be great if you could clarify the use of different signs. Thank you
Edit: In our lectures it's called the two point theorem
There isn't a difference. On the Rotation About a Fixed Axis Dynamics video, at 3:30, we have w cross w cross r, not w multiplied by w multiplied by r. Here, it's shown in a different form, with w^2 multiplied by r. The w cross w cross r takes care of the signs.
Thank you for the video. I have question at 7:53 "Why are you not adding center acceleration (aC) into aA?"
I am not sure I understand. What is aC? Point C is fixed, so the wheel turns around it. If possible, please give more details so I can help you out better, thanks! :)
@@QuestionSolutions well I already found my answer. I was little confuse. In the third problem, the wheel is only rotating not "translating and rotating". Thank You.
@@niketshah1000 Glad you got it :)
How do you know the angular acceleration is about the z axis in the first question 2:52
So you have to think about these problems in a 3D sense. If the object is going counter-clockwise, then using the right hand rule, (so curl your fingers counter-clockwise), your thumb will point of the screen, towards you. So that's the positive z-direction. That's the acceleration vector in the z-axis.
Hello sir, at the first example when you calculate Vb we use the formula wxr but we dont split the r in i and j components, why should we split r in i and j components when we calculate voor ab?
Nvm, because the velocity is perpendicular to the arm you dont need to divide into components I think.
@@ABCD-yw9fw Sorry, could you give me the timestamps to the locations where you're referring to? I can help you better that way. :)
for the 1st example, how did you figure out that omega BC is clockwise?
Just by looking at the only way this rod can move with the given information. To understand it better, you can take a piece of cardboard, cut it into rough shapes and then pin them. Afterwards, look to see how the pieces move. 👍
@@QuestionSolutions thank you for the explanation!
@@masanobunishimura2637 You're very welcome :)
for the disc how do you know the acceleration on point c is in that direction? and in c does it only have tangential and normal accelration?
It's not precise, just an educated guess as to the location and direction. Solving the formula gives us the exact value. So if you draw a coordinate system at C, then the acceleration vector would be -10 in x-direction and -0.35 in y-direction.
Hi sir, how do I know the direction of acceleration and where to place my arrow
Usually, you have to imagine these objects moving in your mind, and then you can figure out the directions. Generally, however, if you get a negative value for acceleration, then it's actually opposite to the direction you assumed.
Hi, may I know why the acceleration of O in question 2 is towards left side but not right?
Because the wheel is going left and it's accelerating while it's travelling.
@4:15 , why isnt the cos component negative since it's going from right to left?
The position vector is going from B to C. So Left to right.
@8:56 Why is the angular acc. of the wheel C clockwise while the angular velocity of the wheel c counter clockwise?
The direction of the arrow is wrong, it's a typo. 😅
If required by the professor i should add Coriolis, right?
Yes, you should use whatever is required by your professor.
Great video. But I have a question at 8:02, why rC is negative?
The position vector is going from C to A, which is towards the left, so negative. I hope that helps!
@@QuestionSolutions Thank you sir! But one more thing, at 8:46, when I substitute the answer to the J components, I get different answer for αC which is 406.67. Am I correct in pointing that out?
@@afiffazizz Okay, so I figured out how you got that value. You plugged in positive 70 for α_AB. So at the beginning of the question, we made an assumption, and it turned out, its actually the opposite, but our calculations must be carried out until the very end with our assumption, even if it's opposite. So you have to use negative 70 for α_AB. Then at the end, you can write them positive, and show the real direction.
@@QuestionSolutions Okay, I understand now, thank you !
@@afiffazizz You're very welcome. Always remember, carry out the calculations with your assumption until the very end, and at the very end, change the signs to show direction. This is true for any subject, any question, etc :)
nice! what program do u use for animations? thnks
After effects for animations 👍
Please can u solve these questions by acceleration diagram ???
Sorry, what do you mean by an acceleration diagram?
@@QuestionSolutions by drawing the directions of the components of accelerations, like you have drawn for the velocities Vb and Vc creating an acceleration diagram(Ab, Ac, Ab/c) in order to solve analytically instead of using vector analysis. My lectures insist on doing that method ad well and is soo complicated. Can you please help?
@@Pastorella_COD_Mobile Why does your lecturer insist on that method? It's subpar and cannot be used in advanced problems. 😅 I will add it to my list to do, but I don't think it'll get done for some time. But I will try my best.
@@QuestionSolutions thank you very much for replying. Yes we really find it unnecessary, as if solving the vector method wasn't hard enough.
Is it true that the Angular Acceleration at point AB an Angular Acceleration at point C is causing the motion to Slow Down?
Could you tell me which time you are referring to? If it's the first example, in reality, while I stop the animation half way, point C is actually increasing in velocity. Because angular acceleration is not negative, it's increasing the angular velocity. So the velocity at C would increase.
@@QuestionSolutions I am talking about the third problem
@@darrylcarter3691 Yes. So when the wheel C turns, the angular acceleration of rod AB changes with it. At a certain time, t, the angular acceleration is now opposite to what it was before, thereby slowing down the slider at B. You should note that the angular acceleration for wheel C is 0, since its a constant angular velocity. It is this velocity, and the motion of the wheel, along with how the rod is attached that causes the angular acceleration of rod AB to change. This change then causes the acceleration at B to change, but the question is asking about this before the acceleration slows down, since we are given the acceleration at point B to be positive 2 m/s^2 to the right.
how do we draw the triangle for the position coordinates using the angle?
Please give me a timestamp so I know where you're referring to. Thanks!
4:07 and sorry I meant position vector@@QuestionSolutions
So the position vector would go from B to C. If you draw a line straight down from B, you can form the right angle triangle. BC would be the hypotenuse. The opposite side to the angle would give us the y-component, and adjacent side to the angle would give us the x-component. Since the position vector is going from B to C, we go down and then to the right. So down is negative while to the right is positive. In other words, j-component is negative while the i-component is positive. @@abdallahindimi3484
wait wait nvm i got , thank you so much😅
@@abdallahindimi3484 Awesome :) keep up the good work!
how do you know if W(BC) is clockwise or anticlockwise?
I guess you have to visualize it in your head, but another way is to see the animation. Look at which way BC rotates, and if you want to figure out the vector, then you'd use the right hand rule. If we are looking at the first example, notice how the bottom of link BC, so point C moves towards the left while point B moves to the right (from the perspective of just link BC, not link AB), so it's clockwise.
Wait shouldn’t be -1cos60 + 1sin60? X is in the negative direction and Y is in the positive.
So the position vector is from B to C. In other words, if we break the position vector into components, the x-component will face right, and the y component will face down. I think you thought of it as it going from C to B, if it's from C to B, then you would be right. :)
At 6:32, why is the acceleration in the direction of the link? I thought it would also be in the same direction as the velocity of A?
So you can think of it as the link pulling on the wheel with the rod. The acceleration vector will point along where the pull is happening. This in turn causes the wheel the spin, but the velocity vector is perpendicular to edge of the wheel.
@@QuestionSolutions Ahh that is very clear, thank you so much for the response!
@@itsyaboybinnybenis8011 You're very welcome!
@@QuestionSolutions Wait, but at 8:03, we learn the exact x component of the vector aB, which is 60. But if the direction of the vector is known (towards the link AB), why can't we find the y component from there? If the x-component is 60, and we are in a 3 4 5 triangle, then the y component would have to be 80? However, we find a different answer later in the video?
Sorry but I am not understanding your question. Vector ab just has a single component, along the x-axis, nothing else. aB is the acceleration of the slider, not the wheel or the rod. @@itsyaboybinnybenis8011
2:11 How do you know the direction of the V_B vector?
So you look at member AB and you see what's given. We see that member AB has a counterclockwise angular velocity, with an angular acceleration in the same direction, so there is only one direction the vector can go, which is to the left. These questions do require a bit of imagination where you need to imagine the mechanism moving, which is why I animated them. That way you can see it easier :)
in the first example why don't you break up up velocity into components. When do you know when you need to break it up or not.
Sorry, could you kindly give me a timestamp as to where you think we should break the velocity into components? Then I can give an answer as to why we do or we don't. Thanks!
Hello! Why at 4:07 the cos is positive and the sin is negative?
So the vector goes from B to C. Imagine you're at point B. To get to C, you have to walk down and to the right. The down (y-component) is given by sine, we chose up to be positive, so down would be negative. Going to the right (x-component) is given by the cosine component, which is positive since right is positive and left is assumed negative.
@@QuestionSolutions Thank you for your reply!
@@AgnesaSopaj-fj7os You're very welcome!
8:15 is the angular acceleration for c not meant to be 406.67
Nvm😅
@@yusuffabdulmusawwir6807 👍
hi! at 3:22, how do you know to use the i and j components of r(ab) instead of 0.2? is it because its a cross product?
sorry *0.5
@@jenncycarranza3370 Everything is written in cartesian form except for the angular velocity, which is a scalar. So the 0.5m is the magnitude (length of the rod), which we break into components. Since r_AB is a position vector, it will always be in cartesian form. Also, the bold letters represent cartesian vectors, while non-bold letters represent scalar values. I hope that helps 👍👍
@@QuestionSolutions thank you! also, if angular velocity is a scalar, do we not use negative signs when it's going cw instead of ccw? (same problem but when you were finding ac)
@@jenncycarranza3370 No, if you notice, we multiply it by the positive vector on the rod, which is in cartesian form, so the answer you get will also be in cartesian form, which means positives and negatives are already taken care of. I hope that makes sense?👍
@@QuestionSolutions yes thank you so much!
Good day sir!, I've got a question, at 2:51 , isnt the acceleration aB, found from the formula for acc. about a fixed point , so basically I mean is it not supposed to be aB=(alpha)AB X rAB + (omega) X [Omega X rAB]? Or does the double cross product just simplifies to a scalar squared so omega^2, I have tried doing it using the notation/expression i just mentioned yet I get a wrong answer so I am wondering if it actually is the same and I am just have a sign error/or a simple calcualtion mistake, or theyre actually 2 differnet formulas.
No, you have to use the equation I show at 1:00. Where did you get the equation "aB=(alpha)AB X rAB + (omega) X [Omega X rAB]" from? I am not sure where this one is from, maybe I am forgetting something 😅
@@QuestionSolutions I have got it from your video on Rigid bodies: Rotation About A fixed axis dynamics at 3:34, ruclips.net/video/zrmBObWEDuE/видео.html
@@radiatedbug Technically, this is the expanded form when you specifically find tangential and normal accelerations. Most of the time, you're not going to use that equation, but the one I show at 1:00 mark, which is the same, but "simplified."
@@QuestionSolutions So mathmatecially speaking, theyre the same equations just that the double cross product simplifies to a scalar? (Omega^2) So ultimately they have to give the same answer?
@@radiatedbug Just out of curiosity, what textbook does your course use?
i have a simple confusion , in the formula of accleration isn't there +ve omega square r ? why is there a minus
You can refer to your textbook, but here is another resource: eng.libretexts.org/Bookshelves/Mechanical_Engineering/Mechanics_Map_(Moore_et_al.)/11%3A_Rigid_Body_Kinematics/11.4%3A_Relative_Motion_Analysis
Look at section 11.4.12
I am sorry for asking so much
is it possible with same assumptions and same axises to get
Wcb = negative value.
ALPHAcb = postive value?
this happend to me and I got correct values somehow
Yes, it's possible. Imagine you're driving a car and you have a velocity of 5m/s. However, your acceleration could be -2m/s^2. All it means is that while going forward, you are hitting the breaks, slowing the car down. Instead of linear velocity, we are just talking about angular velocity and angular acceleration, but its the same concept. They both don't have to be positive, or both negative, etc.
i'm confused, does the angular velocity should have a k-hat subscript?
Please let me know where you're referring to with a timestamp. Thanks!
Hey, great videos I watch all of them. Just wanted to let you know that I think there's a typo at 1:25. For the angular acceleration, you have AE but it should be AB. Am I correct in pointing that out? Have a great day!
Yes, you are absolutely right, this is a typo on my part. Thank you for pointing it out, I appreciate it. :)
How can i solve for acceleration without the angular acceleration?
The only given are angles, side lengths, and velocity at a point. I'm tasked to find the angular velocity and acceleration.
@@dwightbenedict6975 It's hard to tell without seeing the question. If a lot of angles and side lengths are given, maybe you can solve it using instantaneous center of zero velocity or this could be an absolute motion analysis question.
In the first problem, why did the direction of the acceleration of B not on the direction of the velocity of B?
They are in the same direction, just at different angles.
@@QuestionSolutions Will you please elaborate?
I show the arrows in color, all labeled. They point in the same direction, they are just at different angles. You can see them if you look at the acceleration vector drawn, and the velocity vector drawn. Acceleration vector for point B is drawn in purple. @@tarifulislam7934
@@QuestionSolutions thanks a lot. Will you give more videos on different topics like Heat Transfer, Fluid Mechanics etc?
I just want to make sure, did you see the arrows and did you understand what I meant by same direction but different angles? Because that's usually how it is, acceleration vector will be at a different angle, where as the velocity vector is perpendicular to the rod. See: ruclips.net/video/1aQ9EZGMdDk/видео.html at this time, I show the velocity vector, notice it's perpendicular to the rod. @@tarifulislam7934
how do i know when to use i j k components, like questions have same data and diagrams so i cant figure it out.... please tell me
So I am not sure how much fundamental knowledge you have with cartesian vectors. But in a nutshell, x-components are represented with i, y-components with j, and z-components with k. This should have been covered in your statics course, which comes before dynamics. I cover a bit of it on this video: ruclips.net/video/mz7gPpIL0Gk/видео.html
@@QuestionSolutions noo thats not my question bro, i mean i know cartesian products and all very well.... my question is when do i have to solve with vector methods and when with scalar methods...?
@@tahabinaamir2749 Okay, now I understand. So it boils down to the information you're given, and what you're looking for. For example, if you're going to find values using the instantaneous center of zero velocity, I would go with scalar. If the information given for the question can be found using scalar methods, so simple things like angular velocity can be easily found using just scalar methods. For everything else, I'd keep my eye out and use vector analysis. Using vector analysis keeps the errors low since directions, positives and negatives are accounted for automatically by the vectors.
@@QuestionSolutions yes thats true, signs are easy to figure out in vectors... BTW big thanks bruh!! May god bless you! ❤
@@tahabinaamir2749 You're very welcome! Best wishes with your studies! ❤
Nyc
🤔
at 4:51 why a0 is just aplha*r
See: ruclips.net/video/zrmBObWEDuE/видео.html
@@QuestionSolutions thank you, so you mean becaue it it the center of a rotating body the center will not have normal acceleration however any other point will have?
i love you
❤
3.13 minute 4^2k instead of Wab^2 ? why not?
ı found it , it is not cross product :ddd
That's the angular velocity of w_AB, it's given. It's written on the diagram, on the left side, please pay attention :) There is a typo on the diagram, but that's regarding angular acceleration, it should be AB not AE. But I think it's easy to see with the arrows already there.
@@QuestionSolutions hi teacher ı love you after 10 months . It's good to listen to you even though I passed the this lesson. Have you social media ?
No coriolis?
Not covered.
Can I contact you personally please?
yes, email is contact @ questionsolutions.com