Damn man I'm so glad I found this channel because ever since I bombed my last exam I started watching your videos and now I understand the material way more. I thought I was going to have to retake Dynamics, but you saved my grade. Keep up the good videos man!
Oh my goodness you are so helpful! Thank you and please keep the videos coming! Perhaps consider doing other topics too such as solids, materials, etc.?
You're very welcome. I hope to branch out to many different subjects, though time has been my enemy so far. Thank you very much for your comment, best of luck with your studies!
@Elian Camilo thanks so much for your reply. I found the site thru google and I'm trying it out atm. Takes quite some time so I will reply here later with my results.
You’re so under rated!! More people should know this channel. Thank you so much for helping us understand these lessons. Definitely will share this with everyone!
I literally love you. The visuals are so amzing and easy to understand! I love how you provide examples as well and acknowlege that examples help clarify concepts/ help learners see the concept in action.
I’m so glad to hear that you found the visuals and examples helpful! Your feedback is greatly appreciated and I’m thrilled to hear that the examples are helping. Thank you for your kind words! 😊
Thank you so much for everything you do for us, you're really a kind soul, i can't thank you enough for what you do, i have my mechanics exam in 2 days and you've taught me everything i needed to know, thank you ❤️
You are so welcome! What a nice comment, thank you for taking the time to write it. I wish you the best with your exam, you'll do great! Keep up the awesome work ❤
Youre the goat 🐐 Thanks so much for making it so much more easier to understand. You explained it in 5minutes compared to my lectures that takes 2 hours 😂
You're just assuming. If you get a negative sign for your answer, then it's opposite to your assumption. But you can usually make a very good guess as to the direction just by visualizing how something moves.
Great video as always, quick question though, at 5:05 , why did we use r(B/IC) to solve for W(BC) in the equation VB = W(BC)*r(B/IC), why didnt we use the bar BC lengh of .5m instead of the r(B/IC) lengh? Also in this case when we use the equation relating the velocities at the instanteous center, VB = VA + W(BC)x(r(IC)) what would VA = 0 be in this moment? Like which linear velocity is that?
You canuse the 0.5 m if it's a simple rotation about a fixed axis. Here though, that's not the case. We have a fixed slot "C" and point B which is dependent on bar AB. When we calculate VB, you can think of it as bar BC disappearing, and you will see that bar AB going back and forth about point A. If we do the same for bar AB, and think of it disappearing, where does bar BC rotate about? It's dependent on the other bar so you have to use relative velocity or instantaneous center of zero velocity to figure it out. For this problem, you don't use VA, you would compare VB to VC to use relative velocity.
so for the first example, when we get the answer at 3:41, is this the angular velocity of the link about the IC or is it the angular velocity about either B or C?
This is the angular velocity of link BC. They are not separate entities, it's a single link. So we just name it link BC. Also, the IC point doesn't change the angular velocity of link BC (so we aren't saying it's about the IC point), it's just a different method to get the angular velocity instead of using relative velocity. I think, at least from your previous questions as well, you might be confusing linear velocity and angular velocity. You can calculate linear velocity at point B or C. When we say angular velocity of link BC, that is for the whole member. That is the angular velocity of the metal rod. 👍
Love the videos, saving my grade for real. I just don’t understand what the point of this method is though. Why couldnt we just use the relative velocity method for these problems?
I’m confused because when I’m doing other examples where the point of interest isn’t at the center of the circle, I don’t know which direction I should draw the radial IC line to, I hope you understand my question...
@@kanhchanaly6445 Yes, the IC point would be where the wheel makes contact. For example, if a wheel was rolling on the floor, the IC point would be straight down at the point where it touches the ground.
It depends on the question. If all your values are given in cartesian form, use a vector formulation, if you can easily solve it using scalar, use scalar. Both methods will give the same answer in the end.
You can use whatever method you like to solve these problems. It's completely up to you. Sometimes, you won't have the givens to use the instantons center of zero velocity, in which case, you should use the relative motion analysis.
@@QuestionSolutions I did seek for help. Could't get it through though. I feel ashamed for asking same question again so I'm looking for solution online. The picture can be found in google. The question goes like this if you could help. Gear A rotates counterclockwise with a constant angular velocity of Omega A=10rad/s, while arm DE rotates clockwise with an angular velocity of Omega DE =6rad/s and an angular acceleration of alpha DE =3rad/s2 . Determine the angular acceleration of gear B at the instant shown. I found the solution to this question online BUT I'm following your technique of IC and couldn't get the same answer. I'm wondering if both rotates then the way of finding IC changes.
Thank you so much❤️❤️btw I've got a question, I have a problem comprehending the purport of this method. When we say we assume point A has no velocity, does it mean we assume it's fixed? If that's the case , in the previous method , we also assumed(as an example) point A is fixed but it was moving? And another thing is that: When is it allowable for us to use this method?
So you can use this method whenever enough information about the geometry of the problem is given. If you have a question where you need to find the velocity of a point and you're given distances and angles, use this method, it'll get you to an answer faster than using relative velocity. Also, keep in mind, this only works with velocity, not with acceleration. We aren't saying point A has no velocity, we are just ignoring it because we are using the instantaneous center of zero velocity point. So if you saw the previous video about relative velocity, we compared one point (point A), to another unknown point (point B), and figured out the velocity using a position vector from A to B. Here, instead of that, we use the IC point, so no position vector from A to B, and we can just forget about the velocity at point A. You can imagine a single object rotating about this IC point and that's really what's happening behind the scenes. Your textbook should give a more detailed analysis of this method along with the proof. 👍
Does instantaneous centre of velocity help to determine the direction of the angular velocity as well? because I solved one problem with both velocity dynamics and the ic method only to find that the velocity dynamics method gives a negative when the angular velocity is clockwise, but the ic method does not
So generally speaking, it does give the direction since you can see how the object would move. For example, looking at 3:39, we established that the angular velocity of BC would be counter-clockwise. You don't need the answer to tell you that since you can see and imagine the object moving in your head. You know the direction it has to go since these involve fixed axes. Also remember, we with this method, you're getting scalar values where as with, for example, relative velocity, you can get vector answers.
For example 2. wouldn't vc=1.64. You said vc=wBC*rC/IC but wouldn't vC=wCB*rC/IC because it comes from vC=vB+wCB*rCB. Or does the order not matter is wCB=wBC
Order doesn't matter since it's a scalar value. So you can think of it in simple terms, the rod BC or CB, (same rod regardless of how you name it), has an angular velocity of 1.959 rad/s.
It is the same equation shown at 0:41. It is the scalar method of finding velocity using angular velocity and a length. So in a rotating circle, the velocity at the very edge is equal to the angular velocity multiplied by the radius.
Can you explain why in case of rolling wheel on ground without slipping, it's linear velocity on it's circumference equals zero at the instant it contacts with the ground. I know in case of gears their linear velocities are equal when they are meshed, make sense but here it seems the velocity can't be zero when contact with ground it should be same as when not in contact with the ground. Kindly elaborate.
So notice how we used the segment BC to calculate the IC point. Then we use the angular velocity of that piece to figure out the linear velocity. If you use AB to figure out the IC point, you'd use the angular velocity of AB.
@@readbhagwatgeeta3810 Where do you see wB/IC? You're just multiplying a distance labeled rB/IC by a variable labeled wBC. When you plug your values in, you get a solution for the variable wBC. I don't understand what you mean by "How wBC = wB/IC" where did we write that on the solution?
3:35 . I don't understand how Wbc can be used in the formula Vb = Wbc * rb/ic. In other words how can Wbc be used to to calculate Vb, because point c is also moving right. So it doesn't really have a angular velocity. I am confused. Great vids btw !!
So what do you mean by "it doesn't really have a angular velocity?" Rod BC definitely has an angular velocity, otherwise, it wouldn't be moving. It fact, we even show it to have an angular velocity of 6.787 rad/s. So maybe I am not understanding what you are asking, or maybe you can rephrase the question? Point C will have a linear velocity, point B will have a linear velocity, and they will not be equal either. Let me know if that helps. 👍
@@QuestionSolutions Sorry for the confusement. Someone else asked a similar question which said, "in the equation, V_B = W_BC(r_B/IC), why are we using the angular velocity of BC (W_BC)"
@@icu3545 We use ω_BC because that's what we used for the perpendicular lines. We found the IC point using rod BC, so that's what we need to use. That's the whole point of this idea, it's to use the IC point related to a rod, and use that to figure out the velocity at any given point on the said rod. We can't use AB since we didn't find the IC point with respect to rod AB. Another way to think about this is to imagine rod BC rotating about the IC point. It might make it more clear that all we are doing is multiplying the distance from the IC point by the angular velocity of BC. Remember that in the scalar form, we can find the linear velocity by simply multiplying the distance from the axis of rotation to the point where we are finding the linear velocity, it's v=ωr. So all we do is just figure out this "rotation point" which is the IC point, then figure out the distance and multiply it by the angular velocity. Does that make sense? :)
3:25 i don't get this step at all. It's using angular velocity of BC but the r_(B/IC), can someone explain this to me? doesn't solving this equation give the angular velocity about the instantaneous center? Namely (omega_IC). Unless omega_IC=omega_BC, but I don't see how that's the case.
So the cool thing about this method is that by multiplying the angular velocity by the instantaneous center distance, you actually end up getting the velocity of the link. Remember that velocity is just angular velocity multiplied by the distance from the axis of rotation. Here, we are pretty much doing the same thing, but now, we are just imagining the rod rotating about the IC point. The proof for this should be in your textbook, (if not, a quick search should allow you to find it). Also, I think doing a few questions will allow you to visualize what's actually happening, and this technique will become really helpful in the future for some problems where going through the traditional method of relative velocity can be tedious.
