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Rigid Bodies Work and Energy Dynamics (Learn to solve any question)
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- čas přidán 6. 08. 2024
- Let's take a look at how we can solve work and energy problems when it comes to rigid bodies. Using animated examples, we go through kinetic energy, work, the principle of work and energy and much more, with step by step solved problems.
🔹Rigid Bodies and Rotation About a Fixed Axis: • Rigid Bodies: Rotation...
🔹Kinematics Work and Energy: • Principle of Work and ...
Principle of Work and Energy(00:00)
Kinetic Energy (00:21)
Work (01:45)
Mass moment of Inertia (03:58)
The 10-kg uniform slender rod is suspended at rest... (04:13)
The 30-kg disk is originally at rest and the spring is unstretched (06:35)
The disk which has a mass of 20 kg is subjected to the couple moment (08:12)
Find more at www.questionsolutions.com
Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2014.
man, you just put 2 weeks of lectures from my professor into 10 min video. I wish my professor can be like this and use the rest just to do practice problems
Professors go into a lot of theory and proofs, which is also very important :) I am really glad these videos helped, and thank you for taking the time to comment. Much appreciated.
man i've commented before and I will comment again, these videos...by far....are one of the best educational (academic level) videos relavent to engineering/physics degrees I've came across. You are a legend, an absolute legend. Ive said that before and I'll say it again. I hope your channel grows biggggggg, and becomes one of those "oh have u watched khan academy's video on __?'. But this time we can say this : "Oh have you watched question solutions's anaswer on this"?
Man this comment made me smile! Thanks for taking the time to write this. Really appreciate it and super glad to hear these videos helped you out. I wish the best for you in everything you do.👍
@@QuestionSolutions You da bestttt
Mechanics final in 5 hours!! Ty so much for all these videos, really has helped so much throughout the semester!
I hope you did amazingly on your final! Thank you for your kind comment. I wish you the best with your future endeavors and keep up the good work!
Thanks for these well made videos. They are particularly relevant for me right now seeing my Dynamics exam is coming up :) I can tell a lot of time and effort gets put into these videos and it surprises me that they don't get enough views.
Thank you very much for this comment, it was really nice to read! I appreciate every word and I wish you the best with your exams!
Also, if possible, please share these videos with your friends/classmates. Maybe it'll help them too and it'll help the channel as well :)
@@QuestionSolutions Of course. These videos are among the better, more useful lot on youtube, and I think it would be of great value to people if they saw these kinds of videos :)
@@ShoncayPlays Much much appreciated! :)
This is literally the "learn to solve any question". Thank you sir💙.
You're very welcome! 💙
The best physics educational channel on youtube, period
Thank you!
youre literally the GOATTTT. Thank you so much!!!
haha, many thanks! 😅
@@marekjanik5439 meeeeeeeeee
@@marekjanik5439 greatest of all time
Thank you so much I cannot express my gratitude!!!!
You are very welcome!
You are amazing, that's it, i don't have enough praises for you brother, god bless you ❤️
Thank you very much! That was a really nice comment :) Best wishes with your studies and keep up the great work! ❤
Excellent! Thank you ever so much!!
You're very welcome!
Your entire series is so helpful to me. I am really really grateful. I'm not exaggerating when I say that I would not be able to pass my exams without you. Thank you!!!
Thank you for taking the time to write such a nice comment. I am really happy to hear that these videos helped you out. I wish you the best in your future school work and all of your other endeavors! Keep up the great work :)
@@QuestionSolutions :D
@@philipbecker3332 :D
Great video. Thanks again
You're very welcome!
Fantastic video! One suggestion would be if you think it would work with the way you do the videos is to show the full problem and write down all the given values upfront before you solve. I like to study for my exams by watching your videos. I pause them before you start the problem and I try to solve it first and then I watch you do it to make sure that I did it right. I really appreciate the content you create. Have a nice day!
I will definitely keep what you said in mind for future problems if it works within the question. I am glad these help, and you are doing it the way I hope every student does. Solve it first, and then watch to see if it makes sense. I wish you the best with your exams :)
The video has helped me . Thank u so much
You're welcome 😊
You are excellent keep up the good work.
Thank you so much! Best of luck with your studies.
This really helped, thank uu so muchh!!
You're very welcome!
Great work. You really helped me with my finals!! Keep going!!😊
I hope you do really well on your finals , best wishes! :)
that was basic and understanable. thx
You're welcome!
why is the work done by weight only multiplied by half the distance of the rod ? 6:20
The weight is applied at the center of the rod, so half way of the rod (at the center of mass).
Do you have a video somewhere for calculating mass moment of inertia? Such as for non-common shapes, combinations, and ones which require integration?
Yes, please see: czcams.com/video/zmzUdFFCFkc/video.html
A two hour lecture was shaved off considerably with this discussion I very much appreciate your work sir
Thank you very much. Best wishes with your studies!
superb totaly
Thank you!
comment for the algorithm. ps ur videos are insanely good
Thank you very much! I really appreciate it.
Hey Man awesome video and help!
how do you get 6.11 in the first question? I try to follow the steps and i get 7.5454 rad/s, and it seems i get the same values for T2 and UF, UW and IO.
You're getting a numerical error for just ω but everything else is matching up? Most likely a calculation error, please try again. So we have 225(3.14)-147.15=ω^2 ---> 559.7=15ω^2 ---> ω^2 = 37.31 ---> ω = 6.108 rad/s. I am not sure where you're doing the error, did you plug in 3.14 for pi? Let me know if it works out.
thanks for those videoes i wish if could make for Machine desgin
I will keep it in mind 👍
why is the spring in the second problem constant wouldnt it exert more force as s increased?
I'm not sure I understand your question. It's just a normal spring that follows hook's law and the stiffness is given to us in the question. That stiffness is used at 7:53.
Question on example 1:
Good day sir. When the rod is in a tilted position does the weight not depend on the angle b/w the horizontal and the bar, thereby not making it a constant force?
The weight for an object is straight down at the center of mass, regardless of the angle.
Thank you very much sir
@@bluejay7331 You're very welcome!
is it possible to solve these excercises with the previous methods, because all th problems look the same and i find it very difficult to know wich stratigy to use
You can solve these problems in multiple ways. Use the method that gives you the correct answer in a short time.
THank you for the videos, but quick question @ time stamp 7:30 how did you get theta= 2sg.
that's like 4*theta
So we know that r = 0.5 m, that's the radius. If we plug it in, we have s_G = θ(0.5). Now we isolate for theta. We get θ = (s_G)/0.5 Now it's just getting rid of the fraction, since we can write (s_G)/0.5 as s_G x 1/(0.5) ==> 2s_G.
@@QuestionSolutions I still dont get it
Please see your professor/TAs during their office hours. It's a lot easier to help with these sort of problems in person. @@danialexquande7813
great video, however, I think it is better to start off not with F*cos(theta)*ds but with the scalar product vector(F)*d vector(s). Does the same thing but is more fundamental
Well not really, if the force is given at an angle, you need to break it into components if there is only single directional movement. While it seems simple to some, other students have a hard time grasping this, which is why cosθ is included. Plus, by the point a student gets to rigid bodies, they would have already covered work when it comes to particles, so they are very familiar with (F)(cos θ)(S).
@@QuestionSolutions true .. for me, I like the arrow-y and point-y quality of vectors. It also ties in nicely with linear algebra
@@ThomasHaberkorn I agree, I like that as well!
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Awesome :)
Can you explain how does the weight try to push the rod counterclockwise in the 1st problem?
Easiest way to understand this is to hold a pencil between your fingers straight down. Then with your other hand, push on the tip so that it is how horizontal and gentle release the grip between your fingers to allow free movement without it falling. You will see that the pencil will turn counter-clockwise and come back to the initial position, which is straight down. I mean another easy way is to realize weight doesn't push things up. It brings it down, so when the rod is at the 90 degree position (horizontal), the weight will try to bring to back to the vertical position. It's not pushing it upwards 😅
@@QuestionSolutions So were analyzing the effect of the weight from the final (horizontal) position doing work to bring the rod back down to a vertical (initial) position, and that work is CCW, which is against the direction we "want." Is this correct?
@@QuestionSolutions yes but only the y-component of weight does work... no?😅
Yes, when the rod is rotated 90 degrees, weight pulls it down, turning it counter-clockwise. I am not sure what you meant by "but only the y-component of weight does work" or maybe you're thinking too deep into this. Just imagine a rod held loosely by a pin and you let go of it. It falls down with the whole body turning about the pin. It can't go clockwise, which would mean it's floating up. I hope that helps.
@@QuestionSolutions yes i understood why it's negative. My objection was on why take into consideration the whole weight along the movement. I just think that along the movement from position 1 to 2 only a component of the weight produces work which is a function of the angle theta. But perhaps I have in mind the work of the moment produced by the weight.... Which is mgl/2 times π/2 rad
LIFESAVER👩🏻💻
Glad to hear! Best of luck with your studies.
I don't understand how you calculated work for weight. For the force we said s=r*theta, but for weight you didn't? Weight is creating resistance the entire 90 degrees, too - why wouldn't we multiply it by s=r*theta, too?
Please use timestamps so I know where you're referring to. Thanks!
He used s=r*theta because the 150N force always acts in the direction of the displacement (the curved path). The work done by the weight will only be in the y direction as weight always acts downwards. Work = Force*distance and the distance in the y is 1.5m
how does the rod gain a displacement of 1.5m when it doesn't move it is pinned at its axis ?
For the future, please use timestamps so I know where you're referring to. Also, displacement here is the distance from the rotational axis, so point O, to the center of mass, which is 1.5m away. Remember that displacement is simply the difference between the final and initial position, so 1.5 m - 0 m = 1.5 m (assuming point O to the be the center of our axes) 👍
In regards to the last example, why does the weight not do any work?
The weight is effecting the pin and the pin doesn't move. So no work is being done. Just like the other 2 reactions at the pin, Ox and Oy. If there was no pin and the wheel was allowed to move in the vertical direction, then work will be done by the weight. 👍
@@QuestionSolutions I thought as much, I just wanted to confirm. Thank you for the assistance, appreciated.
@@arjungovender3248 You're very welcome.
how come last part of the questions for the moment work is not negative work ?
Could you kindly provide a time stamp to where you're referring?
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❤
For question 1 why wasn't 1/2mv² added to 1/2 Iow²
In general, linear velocity doesn't need to be calculated for problems like these, especially when we are considering angular velocity.
How do we know that velocity of rod will be zero when it is rotated by 90 degrees in the first question?
I am not sure what you mean? Are you talking about linear velocity? Because we didn't consider that when we wrote our equation, we only considered angular velocity, which we found to be 6.11 rad/s. Please let me know a time stamp of where you are referring to, then I can help more, thanks! :)
@@QuestionSolutions linear velocity
Kinetic energy has two terms one that of linear velocity and one that of angular velocity so we need to consider both my question is you considered the linear velocity term of KE to be zero when it rotates by 90 degree how you know that it is not given in question
@@QuestionSolutions 4:58
@@adityapandey8096 Ohhh, please look at 1:20. If it's not found about the center of mass, we can simply use T=1/2(I)(ω^2). Take note that point O is not the center of the rod.
at 5.18 I0=1/3ML^2 wont it be like this 1/12ml^2
Please see: cdn1.byjus.com/wp-content/uploads/2020/11/Moment-of-InertiaArtboard-1-copy-16-8.png
what is moment at ic
I don't know where you're referring to. Please use timestamps.
@@QuestionSolutions 1:41
the IC point is refer to which one?
@@TommyLee-pi7fo Please see: czcams.com/video/1aQ9EZGMdDk/video.html
i dont really see why Uw is negative, since its right in the center and going down? why would we assume its anti clockwise force.
I need a timestamp so I know where you're referring to please.
@@QuestionSolutions I have the same confusion. I believe they are referring to 6:26.
@@szeloklau7380 So we are accounting the distance from O to the center of mass, so the vector would go from O to center, which is downwards. :)
@@QuestionSolutions I see, but how does that show that the work on the weight is trying to push the rod counterclockwise?
@@szeloklau7380 So if you look at the very beginning of the problem, I animated the rod moving. When it moves 90 degrees, it's now straight up. Now which way will the weight try to move the bar? Clockwise or counter-clockwise? It's going to pull down, turning it counter-clockwise. I hope that makes sense.
7.58 minute , why is uf negative ?
The spring tries to stop the wheel from going forward, so it does negative work.
@@QuestionSolutions heeee thanks teacher, so nice video
The work for the moment , 2@+4 N.m
I assume you're talking about the 208.18 J? The moment applied is turning the wheel, so the work is positive.
So any moment clockwise is positive work and and anticlockwise is negative work?
@@chenmoney1920 No. So let's say you have a box and you push it left, that's positive work. If you push it right, that's positive work. The friction, whether the box moves left or right, does negative work (because its stopping the block from going, or trying to stop it). Now in this question, if we spin the wheel to the left, or the right, both will create positive work since it does what we want it to. If we considered friction in this problem, that will do negative work.
why multiply by 1.5 and not 3 6:27
Weight vector is applied at the center.
i think there is a mistake in work by spring...force and displacement are at 180 degree so Fs= - K/2(s^2)....
Please provide a timestamp so I know where to look. Many thanks!
At 2:46... Force and 's' are 180° so ...
@@nayankuikel7340 I am not sure by the 180°, since spring work is based on compression/elongation multiplied by the stiffness of the spring. Regardless, the work done by a spring is not always negative, it's easier to keep it in mind as positive, and change it to negative when it's slowing a particle down or in a case where it does opposite to what you want it to do. 👍
Your explanation is excellent but I dont like rigid bodies and especially moment of inertia to ve honest :(. I am a civil engineering student and moment of inertia is after me where I go :/
I think it's not just you, a lot of students seem to dislike dynamics courses at the beginning. I encourage all of them to do as many practice problems as possible. I think that really helps and when you build your confidence, you will like the subject :)
@@QuestionSolutions I hope so
Im sorry but I have a question about Sg at 7:35. I understand why theta = Sg/0.5; however, why does it also equal 2(Sg) or how does it become 2(Sg)?
So I am going to say "sG" is "x" to make it easier. So we have x/0.5, what this is saying is (1/0.5)*x. So you can divide 1 by 0.5, which gives you 2, so you have 2x. In other words, you can move the "x" out of the fraction but we don't do that because it's just 1/0.5, so instead of writing 1, we just put the x there. So again, sG/0.5 = 1/0.5 * sG.
@@QuestionSolutions ahh interesting. I appreciate the response👍
@@geraldsanon3493 You're very welcome!
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Many thanks! It's truly appreciated.
now i can pass with BB
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