This really short and straight to the point, i have been confused for 3 months until today when the confusion is cleared in just 6 minutes. thank you for this helpful content
Wonderful technique. You explain vast amount of things is few minutes. I need Centroid lectures because my semester around the corner. My sem was on 27 th feb. Please upload the lectures of centroid as quick as possible. 🙏🏻
I am really glad it was helpful. Unfortunately, I am unsure when I can upload videos about centroids as I am currently working on thermodynamic videos. Best wishes with your studies!
This is amazing. I’m taking dynamics over the summer during a shortened semester and hardly have time to study. This explained N-T component problems better than my professor could and it was only 5 minutes💀. Fkn legendary
Soooo muchhhh happy to find this channel 😃 I hv exams after 4 days and the syllabus is too much These 7-8 min vdos are helping me a lot Thank you so much 😊
Average high schooler taking engineering course and having problems doing this work that takes me 3 hours. Thank u for making video regarding this topic
Nothing is beyond your capacity of understanding. I am glad it was helpful. Try to do as many questions as you can and you will have an excellent foundation for future topics :)
@@QuestionSolutions Yess i will and in fact im doing it now, thank you for your encouraging words. Its such a light in this dark academic years of mine
Don't be too hard on yourself. Dynamics can be a bit difficult but the more questions you do, the more confident you'll be. Just because it's tough now does not mean it will be throughout your whole course. Be confident and give it all you got. YOU GOT THIS!@@cosmosatrosanguineus3171
For real, these words are what i really needed to hear. This year had been too dark and im too im overwhelmed to grasp any of these concepts. Perhaps ive been way too hard on myself and i didnt noticed that. Thanks for shedding light, and restoring my hope. ❤
Are you a first year university student? Even if you aren't, it's actually fairly common to feel overwhelmed especially because university courses move so quickly. Create a nice time table with time allocated for each course and take it one step at a time. Seriously, doing things one step at a time is really good for students so they don't get overwhelmed. Also, when you need help, PLEASE use office hours, talk to your professors and TAs. They are usually sitting in an empty office waiting for students to come and ask questions (few comes rarely). Best wishes, and remember, you really can do this!@@cosmosatrosanguineus3171
In the final plane question, if you convert the X value in the P equation into meters instead of kilometres it gives you an answer of something like 8x10^10m which is obviously way larger than your answer (even after you convert your answer into meters) , so in the calculation even though speed and everything else is in m/s and not km/s for some reason the only correct way to get P is by using km, is there a specific reason for this? previously we could easily work out p by using meters? why is it that this has changed for this specific question? i was always under the impression that you are meant to convert your units within the actual calculation and not once you already have a final answer... could you please assist xd
Great question! There is a small misunderstanding on your part :) The whole P equation is in kilometres, not meters. I think maybe you thought the 0.8 on the bottom is m/s, but that is not it. It is the second derivative of our y=0.4x^2 equation. So to understand, notice that our original equation for the path of the jet is actually an equation given to us in kilometres, not meters. What I mean is, if I substitute 1 into x, we are saying that's 1 km, not 1 m because if you look at the diagram, we see that our distances are given in km. I'll go a bit deeper by giving an example. Let's say I substitute x=2 into the equation. We get y = 1.6 km. That means at x=2 km, y=1.6 km, not 1.6 m. So when we take the derivative of this equation, every equation we get is also in kilometres. This is why you cannot substitute a meter value into the p equation. After everything is done, you can convert your kilometre value into meters. I think it would have been good if the units for the equation are shown, which is usually the case, but since its not, it might be a bit of a pitfall. Regardless, excellent question and I hope this answer helped.
I couldn't find videos of this topic in spanish, I'm so happy I found your channel, it's exactly what I was looking for!!! Do you have videos of this topic combined with springs??, sorry for bad English
4:48 if i plug in 5000 m instead of 5km, i get a different radius of curvature. why can't i plug it in there so i don't have to convert it back to meters later
Great video . Could you help me with a similar problem. I need to calculate the angle between the tangential component of the acceleration and the vertical line (y-axis) . Is there a way to relate the y(x) equation such that i can calculate the angle at any given point on the curve?
I am not too sure I understand your question, sorry about that. But at 2:13, I show a purple arrow and green arrow. Using those, can you tell me what angle you're trying to find? From what I understand, you want the angle between the green arrow and the y-axis? Are you looking for where it intercepts and the angle there?
Thank you for this video! But how can you solve a problem requiring the normal acceleration, wherein only the constant speed and diameter of the circle are given?
I show the equation at 0:20. It's velocity divided by the radius. Since you have the diameter, you just divide the diameter by 2 to get the radius. Btw, speed is the velocity (without a direction). 👍
how did the radius of curvature in the last example became 87.62? when i plug in the same equation in my calculator it shows 113.05 is this some kind of error? please clarify thanks! 4:50
I hope you will see this in time, i have a question about the last excercise (with the jet). When you search the radius of curvature, I used as my x value 5000 m instead of using 5 km. This gives a whole other outcome that is impossibly great. Why doesn't it work if you convert the distance to meters. I love your videos btw they help me so much
Thanks! This helps a lot! However how do I solve a problem like the one at 2:50 when the acceleration is a function of time instead of a function of distance, and that I need to know the velocity at a specific distance without knowing the time it takes?
Glad to hear it helped! It's hard to answer your question without knowing the givens in your questions, but there should be enough values for you to use your kinematics equations. The overall process remains the same.
Glad it was helpful. There really isn't any more to upload on curvilinear motion, I think there are some solved problems already uploaded as well. I will keep what you said in mind for the future 👍
bro is there any chance you graduated somewhere in canada? Because all these examples and the format of the videos matches some of my courses for statics, thermo and dynamics. Which makes it even more helpful for me but I am just curious.
Thank you sir But i think there is a mistake In the calculation of an i think you can't divide m/s With radius in km but all in all I sincerely thank you sire I've watch your video in dynamic from A-Z and they really helped me
You're welcome. Can you give me a timestamp to the location where the mistake is? If there is an error, I'd like to make a pinned post about it. Thanks!
This is freaking awesome stuff. I am studying this for my back log exam and this video seriously had me questioning what the fuck was wrong with my teacher.
It's given in the question, so when questions say a car accelerated at 2 m/s^2, that's tangential acceleration. Normal acceleration only occurs when an object is travelling along a curve. So there is nothing to relate accelerating to tangential acceleration since that's what's given in the question.
hello, I am having some confusion with P. Do you know why I might be getting 1202.93?? I'm sure I might be miscalculating something, but I have checked me work so many ways. Let me know if you can figure out my problem. Thank you!! Amazinggg videos!
😅 Hopefully, anyone anywhere can watch these and gain a good understanding of the concepts, even if I can't be there physically. Many thanks for your comment!
The first example I already knew the tangential acceleration was going to be 3m/s^2 because the sled is sliding down the hill therefore it is speeding up. That means it is accelerating. And it is tangential to the curve. Though you said vA dot is 3 m/s^2, we can assume that speed at point A with a dot on top is the same as acceleration. I even know that the radius of Curvature was going to be in use when I saw the y = 0.01x^2. So I didn’t need much help with that one.
@@QuestionSolutions second one I know what had to do and what equations to use. So I did it myself than when I played the video to see how you did it. I realized that I made little booboo. 😅 I just forgot to make v one half v squared. But again I had the right idea. I just needed to slow down a bit. 😅
Hello! I used the formula that you taught in this video. But I'm not sure if I got the correct answer, here is the problem: A particle having a constant speed of 15 ft/s moves around a circle of 10 ft. in diameter. What is the normal acceleration? And my answer is 45 ft/s², may I know if it's correct? Thank you!
AWESOME VIDEO! Quick question, why is the tangential acceleration equal to the value of the sentence "is increasing at a rate"? Is the tangential acceleration related to the magnitude and the normal acceleration related to the direction?
I guess the easiest answer I can give is this. In almost all the cases, the acceleration you're given in a question is tangential. Normal acceleration is zero unless the object is traveling along a curve. So imagine there is a question and it said a car is accelerating at 2 m/s^2, or the velocity is increasing at a rate of 2 m/s^2. That's tangential acceleration. If the car was on a curve, then you will end up finding normal acceleration. Normal acceleration is related with the radius of curvature and velocity. It will ALWAYS point towards the center of the curve. Tangential acceleration is acceleration along a tangent at an instantaneous point. I hope that helps. Best of luck with your studies :)
Thank you so much. Your explanations are crystal clear. Would you please teach us how to find the angle between posetive x-axis and at (tangential acceleration? Thank you in advance. PS. Please upload more videos for dynamic.
Hi, thank you so much for taking the time to write that. I appreciate it, and I am really glad this helped you. To find the angle between the x-axis and the tangential acceleration, please see this example: ua-cam.com/video/e3A-yeWFhWo/v-deo.html . It covers how to find that. :) Best of luck with your studies!
Acceleration has 2 components, normal and tangential. Tangential acceleration is the stuff you usually talk about. Normal acceleration is acceleration when an object travels through a curve.
The 200m/s if initial velocity and it was increased by certain amount of a=0.8,giving final velocity of 220m/s and its what you use as your V?? Check the 3rd example,please!!!
I don't think you understood the question. At point A, the jet has a velocity of 200 m/s, and we need to find our unknowns at point A. So we use 200 km/s. At that instant, there is an acceleration but at that instant, the speed is only 200 m/s. Imagine you're in a car with an acceleration of 20 m/s^2. Your speedometer shows a speed of 40 km/s. If we are looking for values at that instant, we don't figure out what the speed is going to be at a later stage, we use the value shown on our speedometer. Same thing.
You should be able to find all the derivations and proofs of these formulas in your textbook, or any other dynamics book. I don't really want to prove or derive these equations since there are tons of videos done about how these equations come to be. If you would like, I highly suggest Khan academy videos, they show proofs and derivations. I just want to show how to solve problems you face in these chapters, or rather, how to use these equations and when to apply them. Thanks!
You're very welcome. Hmm, I am not entirely sure I understand the "spring connected" part. Is the object on a curved path? If so, you can use principle of work and energy to figure out acceleration and then it's all the same. Transverse acceleration would be the acceleration you get from your kinematics equations, while normal acceleration can be found using the velocity and radius of curvature. I have not really seen a spring connected problem in this chapter, so I assume you're talking about multiple things happening at once. Regardless of the situation, remember, if it's on a curve, there will be normal acceleration, and the way to find it is with velocity and radius of curvature. If you keep that in mind, the rest is usually trying to find what the velocity is, and what the radius is. I am not sure if this answer helped, but if there is a more specific question, let me know, and I'll do my best. Best of luck with your studies.
Hmm, from what I remember, it is almost always given. You can always find it using the equations from previous chapters, like a=dv/dt, or vdv=ads, etc.
“Thanks for watching”? No, we actually should thank YOU for teaching us!
😅 You're very welcome!
@@QuestionSolutions realtalk whoever you are sir! thank you for wasting your time on us
@@flee3695 It's not a waste 😅 Glad it helped!
You just taught me in 6 minutes, what my teacher could not in 3 months. Thank you so much!
That's awesome :) You're very welcome, best of luck with your studies.
Teachers always have a syllabus to cover but thank God that UA-cam teachers don't.
at least u had 3months. After our first 4 lectures, we had our first midterm
Short video, straight to the point, detailed and profound! I loved it!!!
Thanks so much! 👍
Honestly a life saver. Been more helpful than uni lecturers this year. Thank you!
Really glad to hear. I wish you the best with your studies!
This really short and straight to the point, i have been confused for 3 months until today when the confusion is cleared in just 6 minutes. thank you for this helpful content
I am really glad to hear that :) Best wishes with your studies!
This is my favorite tutorial
Channel so far
Very clear audio and animation.
And very responsive to questions
Thank you very much :)
I'm going to recommend this channel to everyone in dynamics. Absolutely phenomenal teaching. Everything is so simplified
Thank you very much, I really appreciate the shares. It helps the channel out and hopefully, it will help other students too :)
bro ive been following you since I was struggling, you have been helpful thus far, I have recommended you to many, I appreciate your work.
Thank you very much and I really really appreciate the shares! It not only helps the channel a lot, but it can also help other students :)
these videos make it seem so much simpler, thank you!!
You're very welcome! :)
My day saver.... you saved my time..thank lot for amazing teaching life saver..
You're very welcome! Best wishes with your studies.
Your questions are pretty good and beneficial!! Your explanation is phenomenal!
Thank you very much! I am really happy to hear that. :)
Wonderful technique. You explain vast amount of things is few minutes. I need Centroid lectures because my semester around the corner. My sem was on 27 th feb. Please upload the lectures of centroid as quick as possible. 🙏🏻
I am really glad it was helpful. Unfortunately, I am unsure when I can upload videos about centroids as I am currently working on thermodynamic videos. Best wishes with your studies!
You truly deserve more subscribers! You are a God send! Thank you so much
Thank you very much! Glad to hear these are helpful :)
Thank you so much I've watched multiple videos on this and you've made it extremely simple and clear. thank you
You're very welcome! Glad to hear it helped.
king of engineering mechanics......best channel for this course
I am really glad to hear these videos were helpful. Keep up the great work!
I have never learned more from a 5 min video in my entire life..
😅 I am glad it was helpful! Many thanks.
you're a lifesaver brother.
You're very welcome!
You taught me in very understandable way ...and i got whole concept ij just 6 mints.👍👍👍👍👍👍👍👍
I am really happy to hear that :) Keep up the good work!
This is amazing. I’m taking dynamics over the summer during a shortened semester and hardly have time to study. This explained N-T component problems better than my professor could and it was only 5 minutes💀. Fkn legendary
Thank you very much! Glad to hear these videos are helpful to you. Keep up the awesome work and I wish you the best on your shortened semester.
I tried so hard to understand this topic yet i didn't understand.... Your video was sooo simple and straight to point..!!! Thank you!!!!
I am really glad this video helped you out. Keep up the great work and best wishes with your studies.
Soooo muchhhh happy to find this channel 😃
I hv exams after 4 days and the syllabus is too much
These 7-8 min vdos are helping me a lot
Thank you so much 😊
You're very welcome! I am happy to hear that these videos are helpful. I wish you the best with your exams and keep up the good work :)
Average high schooler taking engineering course and having problems doing this work that takes me 3 hours. Thank u for making video regarding this topic
You're very welcome! I wish you the best with your studies, keep up the great work.
4:38 1:32 Thanks for helping out I want to ask how I find value of dx
Sorry, but what do you mean by dx? Are you looking for time derivatives?
@@QuestionSolutionshow find the secound derivative dy/dx=0.02x
So you take the second derivative with respect to x. In this case, the second derivative is just 0.02.@@rawans1718
i thought this topic is beyond my capacity of understanding but you proved i was wrong. You are golden, this video is golden
Nothing is beyond your capacity of understanding. I am glad it was helpful. Try to do as many questions as you can and you will have an excellent foundation for future topics :)
@@QuestionSolutions
Yess i will and in fact im doing it now, thank you for your encouraging words. Its such a light in this dark academic years of mine
Don't be too hard on yourself. Dynamics can be a bit difficult but the more questions you do, the more confident you'll be. Just because it's tough now does not mean it will be throughout your whole course. Be confident and give it all you got. YOU GOT THIS!@@cosmosatrosanguineus3171
For real, these words are what i really needed to hear. This year had been too dark and im too im overwhelmed to grasp any of these concepts. Perhaps ive been way too hard on myself and i didnt noticed that. Thanks for shedding light, and restoring my hope. ❤
Are you a first year university student? Even if you aren't, it's actually fairly common to feel overwhelmed especially because university courses move so quickly. Create a nice time table with time allocated for each course and take it one step at a time. Seriously, doing things one step at a time is really good for students so they don't get overwhelmed. Also, when you need help, PLEASE use office hours, talk to your professors and TAs. They are usually sitting in an empty office waiting for students to come and ask questions (few comes rarely). Best wishes, and remember, you really can do this!@@cosmosatrosanguineus3171
Thank you so much for making these videos! They helped a lot!
You're very welcome! I am glad to hear they help. Keep up the great work and best wishes with your studies.
years behind in the comments but this is truly saving me!! thank you!
Glad to hear :) I wish you the best with your studies.
Thank you Mr. Savier
You are very welcome.
Thanks you for good and simple describing it’s practical!
You are very welcome. Good luck with your studies! :)
Best video ever cleared my concept thank you for making such a useful video ❤
You are very welcome! ❤
OMG iwas so scared and unprepared for my dynamics exams but with this video i understood how to answer related questions thank you so much
Glad to hear these helped :) Best of luck with your exams!
This video is very helpful,thank you.
You're very welcome. Best of luck with your studies.
never attanted to class and i have a test in a couple of hrs, thanks a lot
All the best with your test!
So practical explanation!! Thank you so much!!
You are very welcome!
This really helped a lot, thank you
Glad to hear! Keep up the great work.
Man seriously that was such an amazing explanation!
Glad to hear! Many thanks for the kind comment.
@@QuestionSolutions you deserve it!
This is actually so helpful like holy
Glad to hear :) Keep up the great work!
nice..... Love from Srilanka
❤❤
You are a life saver
Thank you :)
thankyou , it was very informative
Glad it was helpful!
If we have teachers like you in our college we can learn more things but we don't have
I am glad these videos are helpful to you! Best wishes with your studies.
Thank you sir for your clear explanation.
You are most welcome. Best of luck with your studies.
Thankyou sooooo much man you a are a life saver ❤
You're very welcome. Keep up the great work and best wishes with your studies. ❤
You have made mea learn in almost 6 minutes while my professor taught me in more than 5 hours
I am very happy to hear that and wish you the best with your studies!
not me watching this few hours before our exam. thanks a lot!
You got this! Best of luck and you're very welcome.
Ha! Same.
In the final plane question, if you convert the X value in the P equation into meters instead of kilometres it gives you an answer of something like 8x10^10m which is obviously way larger than your answer (even after you convert your answer into meters) , so in the calculation even though speed and everything else is in m/s and not km/s for some reason the only correct way to get P is by using km, is there a specific reason for this? previously we could easily work out p by using meters? why is it that this has changed for this specific question? i was always under the impression that you are meant to convert your units within the actual calculation and not once you already have a final answer... could you please assist xd
Great question! There is a small misunderstanding on your part :) The whole P equation is in kilometres, not meters. I think maybe you thought the 0.8 on the bottom is m/s, but that is not it. It is the second derivative of our y=0.4x^2 equation. So to understand, notice that our original equation for the path of the jet is actually an equation given to us in kilometres, not meters. What I mean is, if I substitute 1 into x, we are saying that's 1 km, not 1 m because if you look at the diagram, we see that our distances are given in km. I'll go a bit deeper by giving an example. Let's say I substitute x=2 into the equation. We get y = 1.6 km. That means at x=2 km, y=1.6 km, not 1.6 m. So when we take the derivative of this equation, every equation we get is also in kilometres. This is why you cannot substitute a meter value into the p equation. After everything is done, you can convert your kilometre value into meters. I think it would have been good if the units for the equation are shown, which is usually the case, but since its not, it might be a bit of a pitfall. Regardless, excellent question and I hope this answer helped.
@@QuestionSolutions thank you! this was quite a nuanced point and I had the same issue. Glad there was a satisfying answer
If P equation is: y=.4x^2, would y be in km^2? You are squaring x in km to get km^2?@@QuestionSolutions
I couldn't find videos of this topic in spanish, I'm so happy I found your channel, it's exactly what I was looking for!!! Do you have videos of this topic combined with springs??, sorry for bad English
Yes, there are multiple videos covering a lot of the dynamics chapters. Please look through the playlist :)
Hello sir. Good day! May you keep uploading good videos... thank you so much
You are very welcome. I will try my best :) Good luck with your studies!
Thank you so much
You're most welcome!
Thanks sir
Love from india
You're very welcome. Best of luck with your studies.
@@QuestionSolutions thanks sir for you good wishes
3:57 why we dont use v=4 m/s into the an equation?
We do at 3:45, the lower bound of our integral uses 4 m/s.
Thanks sir! Can you suggest any references book for dynamics
Mechanics for engineers - dynamics by R. C. Hibbeler and K. B. Yap. :)
YES IT DID HEPLED!!! THANX ALOT!!!
AWESOMEEEEE! YOU ARE VERY WELCOMEEEEE!!! 👍
4:48 if i plug in 5000 m instead of 5km, i get a different radius of curvature. why can't i plug it in there so i don't have to convert it back to meters later
The equation is written with respect to km.
You are the best
Thank you very much!
3:43 how did you get the 8 in the equation?
So when you solve the definite integral on the left side, you get v^2/2 - 4^2/2 ==> v^2/2 - 8
Great video . Could you help me with a similar problem. I need to calculate the angle between the tangential component of the acceleration and the vertical line (y-axis) . Is there a way to relate the y(x) equation such that i can calculate the angle at any given point on the curve?
Hey this is the question snipboard.io/aifgKc.jpg
I am not too sure I understand your question, sorry about that. But at 2:13, I show a purple arrow and green arrow. Using those, can you tell me what angle you're trying to find? From what I understand, you want the angle between the green arrow and the y-axis? Are you looking for where it intercepts and the angle there?
Thank you for this video! But how can you solve a problem requiring the normal acceleration, wherein only the constant speed and diameter of the circle are given?
I show the equation at 0:20. It's velocity divided by the radius. Since you have the diameter, you just divide the diameter by 2 to get the radius. Btw, speed is the velocity (without a direction). 👍
@@QuestionSolutions I was just hesitant about my solution, but that is what I also did. Thank you so much!
@@romaalexieespeleta9956 You're very welcome! Keep up the good work :)
how did the radius of curvature in the last example became 87.62? when i plug in the same equation in my calculator it shows 113.05 is this some kind of error? please clarify thanks! 4:50
i got it i was wrong with where i put my ^2 thanksss this has been super helpful for practicing questions
Awesome! I am really happy it worked out for you. Keep up the good work.
The same thing happened with me @@yellowdood8445
I hope you will see this in time, i have a question about the last excercise (with the jet). When you search the radius of curvature, I used as my x value 5000 m instead of using 5 km. This gives a whole other outcome that is impossibly great. Why doesn't it work if you convert the distance to meters. I love your videos btw they help me so much
The equation given is in km.
I am really glad these videos helped you out :)
Good work
Thank you so much 😀
Thanks! This helps a lot! However how do I solve a problem like the one at 2:50 when the acceleration is a function of time instead of a function of distance, and that I need to know the velocity at a specific distance without knowing the time it takes?
Glad to hear it helped! It's hard to answer your question without knowing the givens in your questions, but there should be enough values for you to use your kinematics equations. The overall process remains the same.
@@QuestionSolutions Glad to know, thank you!
@@Lexyvil You're very welcome.
Thank you Air
You're welcome! 👍
Thank you sooooooooo much!!!!!!!!!
You are veryyyy welcomeeeee :)
thankyou sensei
You're very welcome!
If we take 5000m instead of 5km in finding radius of curvature than answe comes different , can you please explain
The equation is given in km, not meters.
very much helpful sir
can you please upload more videos in curvilinear motion advance
Glad it was helpful. There really isn't any more to upload on curvilinear motion, I think there are some solved problems already uploaded as well. I will keep what you said in mind for the future 👍
love it man....
Thank you very much!
bro is there any chance you graduated somewhere in canada? Because all these examples and the format of the videos matches some of my courses for statics, thermo and dynamics. Which makes it even more helpful for me but I am just curious.
Yes, and glad to hear they are helpful. :)
Thank yoooou 😊😊🌸🌸💮
You're very welcome! :)
Thank you sir
But i think there is a mistake In the calculation of an i think you can't divide m/s
With radius in km but all in all I sincerely thank you sire I've watch your video in dynamic from A-Z and they really helped me
You're welcome. Can you give me a timestamp to the location where the mistake is? If there is an error, I'd like to make a pinned post about it. Thanks!
THANK YOU
You're very welcome!
This is freaking awesome stuff. I am studying this for my back log exam and this video seriously had me questioning what the fuck was wrong with my teacher.
I am really glad these videos were helpful. Keep up the great work and best wishes with your studies.
On question 2, what equation did you use to relate acceleration to tangential acceleration? When you did a=at=0.05(10).
It's given in the question, so when questions say a car accelerated at 2 m/s^2, that's tangential acceleration. Normal acceleration only occurs when an object is travelling along a curve. So there is nothing to relate accelerating to tangential acceleration since that's what's given in the question.
@@QuestionSolutions very good video, but i still don't understand why would you multiply it by 10, the acceleration should be constant?
hello, I am having some confusion with P. Do you know why I might be getting 1202.93?? I'm sure I might be miscalculating something, but I have checked me work so many ways. Let me know if you can figure out my problem. Thank you!! Amazinggg videos!
Please give me a timestamp so I know where you're referring to. That way, I can help you better. Thanks!
for the as ds = v dv used in question 2, why do you subtract 8? where dyou get that?
It's a definite integral, so don't forget to subtract after you plug in the values. (4^2)/2 = 8.
does the value of x that is 5km doesn't need to be converted to meters ?
No, since the radius comes out in km, then you can convert to meters.
i love you thank you so much
You're very welcome! ❤
You ever thought about coming to ASU and teaching? We could’ve really used you last semester!
😅 Hopefully, anyone anywhere can watch these and gain a good understanding of the concepts, even if I can't be there physically. Many thanks for your comment!
Thanks.
You're very welcome!
I was about to cry till I found you
Don't worry, you got this! One step at a time, one question at a time :)
Can I ask you about the name of the program that you presentat with it
Your student abdo
The animations are done using after effects and drawings are done on illustrator.
Thnkyou so much!!!!!!!
You're very welcome!
Thanks sir
You're very welcome!
The first example I already knew the tangential acceleration was going to be 3m/s^2 because the sled is sliding down the hill therefore it is speeding up. That means it is accelerating. And it is tangential to the curve. Though you said vA dot is 3 m/s^2, we can assume that speed at point A with a dot on top is the same as acceleration. I even know that the radius of Curvature was going to be in use when I saw the y = 0.01x^2. So I didn’t need much help with that one.
Awesome! 😁
@@QuestionSolutions second one I know what had to do and what equations to use. So I did it myself than when I played the video to see how you did it. I realized that I made little booboo. 😅 I just forgot to make v one half v squared. But again I had the right idea. I just needed to slow down a bit. 😅
@@darrylcarter3691 Great! Take it one step at a time, try it out first, and then check and that's the way to go :) Well done!
Thamk you very much
Thanks for your video and keep safe alwayss
You're very welcome!
Hello! I used the formula that you taught in this video. But I'm not sure if I got the correct answer, here is the problem: A particle having a constant speed of 15 ft/s moves around a circle of 10 ft. in diameter. What is the normal
acceleration? And my answer is 45 ft/s², may I know if it's correct? Thank you!
Yes, your answer is correct. Normal acceleration is found by dividing velocity squared by the radius. So 15²/5=45 ft/s². 👍
@@QuestionSolutions Thank you so much! You really have a good way of teaching.
@@peachmochi2223 Thank you for the kind words. :)
AWESOME VIDEO! Quick question, why is the tangential acceleration equal to the value of the sentence "is increasing at a rate"? Is the tangential acceleration related to the magnitude and the normal acceleration related to the direction?
I guess the easiest answer I can give is this. In almost all the cases, the acceleration you're given in a question is tangential. Normal acceleration is zero unless the object is traveling along a curve. So imagine there is a question and it said a car is accelerating at 2 m/s^2, or the velocity is increasing at a rate of 2 m/s^2. That's tangential acceleration. If the car was on a curve, then you will end up finding normal acceleration. Normal acceleration is related with the radius of curvature and velocity. It will ALWAYS point towards the center of the curve. Tangential acceleration is acceleration along a tangent at an instantaneous point. I hope that helps. Best of luck with your studies :)
@@QuestionSolutions Hell yeah man! Thank you so much! Your videos really helped my out. Keep going please!
@@ReneZalazar01 You're very welcome and I am really glad to hear it helped :)
Thank you so much. Your explanations are crystal clear. Would you please teach us how to find the angle between posetive x-axis and at (tangential acceleration? Thank you in advance. PS. Please upload more videos for dynamic.
Hi, thank you so much for taking the time to write that. I appreciate it, and I am really glad this helped you. To find the angle between the x-axis and the tangential acceleration, please see this example: ua-cam.com/video/e3A-yeWFhWo/v-deo.html . It covers how to find that. :) Best of luck with your studies!
Sir how we're understanding that this velocity and acceleration is normal or tangent?
Acceleration has 2 components, normal and tangential. Tangential acceleration is the stuff you usually talk about. Normal acceleration is acceleration when an object travels through a curve.
In the last example when calculating the radius, you use 5 (km) instead of 5000(m). Why shouldnt you use 5000m?
I was also wondering
The equation is given in km.
The 200m/s if initial velocity and it was increased by certain amount of a=0.8,giving final velocity of 220m/s and its what you use as your V?? Check the 3rd example,please!!!
I don't think you understood the question. At point A, the jet has a velocity of 200 m/s, and we need to find our unknowns at point A. So we use 200 km/s. At that instant, there is an acceleration but at that instant, the speed is only 200 m/s. Imagine you're in a car with an acceleration of 20 m/s^2. Your speedometer shows a speed of 40 km/s. If we are looking for values at that instant, we don't figure out what the speed is going to be at a later stage, we use the value shown on our speedometer. Same thing.
can I see the derivations of these formulas?
You should be able to find all the derivations and proofs of these formulas in your textbook, or any other dynamics book. I don't really want to prove or derive these equations since there are tons of videos done about how these equations come to be. If you would like, I highly suggest Khan academy videos, they show proofs and derivations. I just want to show how to solve problems you face in these chapters, or rather, how to use these equations and when to apply them. Thanks!
Thank you for explaining it on a easier manner. BTW, I have a question what if there's a spring connected on the object what should I do ? Thanks
You're very welcome. Hmm, I am not entirely sure I understand the "spring connected" part. Is the object on a curved path? If so, you can use principle of work and energy to figure out acceleration and then it's all the same. Transverse acceleration would be the acceleration you get from your kinematics equations, while normal acceleration can be found using the velocity and radius of curvature. I have not really seen a spring connected problem in this chapter, so I assume you're talking about multiple things happening at once. Regardless of the situation, remember, if it's on a curve, there will be normal acceleration, and the way to find it is with velocity and radius of curvature. If you keep that in mind, the rest is usually trying to find what the velocity is, and what the radius is. I am not sure if this answer helped, but if there is a more specific question, let me know, and I'll do my best. Best of luck with your studies.
you are amazing
Really appreciate your comment. Many thanks!
4:47
There is a mistake 5000^3
The equation is given with respect to km, if you covert to m, you will get an incorrect answer.
I love you ❤️
❤️
jesus, wish i found you sooner
Better late than never :)
Omaygaddd thanks :")
You're welcome! Best of luck with your studies.
how to get the acc. tangential?
I will do my best to help but I am unsure of what you are asking? Sorry.
@@QuestionSolutions the tangential acceleration
@@QuestionSolutions or is it always given?
Hmm, from what I remember, it is almost always given. You can always find it using the equations from previous chapters, like a=dv/dt, or vdv=ads, etc.
thank you so much, I highly appreciate the time you lend just to reply to my comment sir 😊
Where r you from sir please love from india AP
Canada 👍