Rigid Bodies Conservation of Energy Dynamics (Learn to solve any question)

[@rehanrashid9296]

I hope you realize that your effort (especially with the very helpful animations and schematics) don't go unnoticed. You are doing an amazing job in helping to develop engineers and scientists, thank you

[@QuestionSolutions]

Thank you very much for your really nice comment. I wish you the absolute best in all your endeavors!

[@randomthings4344]

I would suggest to add some hashtags in your description like engineering dynamics,mechanical engineering,RC Hibbler Mechanics so that it shows up more often in search bars. Because tooooo many people are unaware of this legendary channel. You have helped me pass many exams just by your videos even if I watched them a single night before the exam.

[@QuestionSolutions]

Thank you very much for your suggestion. I will look into it and add the relevant ones needed. I didn't know those help that much, I figured most people find these videos through a search. Also, really appreciate you taking the time to give that tip and for the comment. I am very happy to hear these videos helped you out, and I wish you the absolute best in your future studies and endeavors.

[@alsix7334]

I have a final exam Wednesday and this has been an amazing help! Keep up the amazing videos!

[@QuestionSolutions]

That's great! I wish you the best of luck with your exams.

[@sanali4425]

Man, you are truly a lifesaver!! My exam is tonight and you have no idea how ur videos are helping. THANK YOU VERY MUCH!!

[@QuestionSolutions]

I hope your exam went well. Thank you and best wishes with everything!

[@mpfuxetochauke1390]

Your videos are massive. I'm gonna use your methods on my Monday exam . God bless you Sir

[@QuestionSolutions]

I wish you the absolute best with your exam! Do your best :)

[@Olivia-hu8eg]

HOW ARE YOU LITERALLY UPLOADING YOUR VIDEOS AT THE SAME PACE OF MY PROFESSOR?!!!!

[@QuestionSolutions]

I've been trying to finish off dynamics for a while now, and finally got to it with the video on conservation of momentum. I hope they all help you!

[@tinkywinky4554]

Hello. I have a question. We said before video, if our particle moves toward downward, then gravitional potential energy is going to be positive, because particle and acceleration of gravitational on the same direction. In this video at 4:25, block goes downward and its same direction with gravitional acceleration but you said its negative. Can I ask you why?

[@QuestionSolutions]

Pulley problems work a bit differently because we use a datum. So here, our datum is our reference point, from that point, the block moves down, below the datum, so -0.6 m, giving us a negative value.

[@real_gifted]

My observations (hack): 1. The main conservative forces are the weight and spring forces. 2. Vg - Weight (of the body in focus) Ve - Force of the spring (on the body) 3. Using basic geometry to determine the change in displacement (angular/linear) for both energies. 4. The universal sign conventions (especially on Vg) 5. Looking out for keywords such as "REST", "Revolution", "Descend", etc (crucial for T1 & T2). 6. DO NOT FORGET TO ANALYSE "T" FOR THE BODY. THANK YOU SO MUCH FOR THE WONDERFUL TEACHINGS! God bless you sir.

[@QuestionSolutions]

Thank you for taking the time to provide your tips and tricks. :)

[@camerongillespie870]

Thank you for another great, in depth, video. Your hard work is commendable and I am enjoying watching your subscriber count increase.

[@QuestionSolutions]

Thank you so much for your wonderful comment. I really appreciate it. :)

[@Olivia-hu8eg]

EVERYTIME HE TEACHES A NEW LESSON I CHECK FOR YOUR VIDEOS AND I FIND THAT YOUVE UPLOADED THEM A COUPLE OF HOURS AGO!!!! 😆

[@QuestionSolutions]

😅 Glad to hear it's there for you. I hope they are helpful.

[@Olivia-hu8eg]

@@QuestionSolutions they really are! thankyou🥰

[@QuestionSolutions]

@@Olivia-hu8eg You're very welcome!

[@albazmovie6267]

thanks, man you helped me pass the course you're the best

[@QuestionSolutions]

I am super happy to hear that :) Keep up the awesome work.

[@TranGiaMinhA]

Thank you for your wonderful videos. They helped me a lots for upcoming exam. Your lecture and the way you delivered are really helpful & suitable for me. Btw, i hope that you would make some videos about fluid mechanics :))

[@QuestionSolutions]

Thank you very much for your comment, really nice of you :) I am really happy to hear they've been helpful. I wish you the best on your upcoming exam. And yes, fluid mechanics is definitely on my to-do list.

[@roxon6929]

Thank you.... Superb video... Love from India

[@QuestionSolutions]

Thanks so much, best of luck with your studies!

[@TheBlindFlame2]

Liking every video before i watch because i know its going to be good

[@QuestionSolutions]

Too nice :) Thank you so much!

[@jakobolson9958]

you are amazing, deserve lots of more views!

[@QuestionSolutions]

Thank you very much! Really appreciate it.

[@r2k314]

Hi. Thanks again for the wonderful videos. There is something I don't understand and can't find an explicit explanation. At 6:42 your equation for kinetic energy includes both the rotational and "linear" kinetic energy of the pendulum, why isn't this double counting, ie., my intuition (such as it is) thinks it should just be the rotational kinetic energy???

[@QuestionSolutions]

You're welcome. Interesting question, and I think a way to understand this is to think about acceleration. Do you remember when acceleration was talked about, there were two components to it? Normal and tangential? So if someone were to ask for the magnitude of acceleration of a car, you have to find both components. Saying just the tangential or just the normal acceleration is not correct, since both components make up the true acceleration. Thinking like that might help you understand the double counting part you're referring to. Both rotational and linear kinetic energy are part of this object. Just like how when a car is going on a straight path, the normal acceleration is 0, in some situation, one part of the rotational or linear kinetic energy can be zero. But in other situations, both of these components/parts give us the full picture of energy in an object.

[@r2k314]

@@QuestionSolutions Ok. I see your analogy. Thank you so much.

[@QuestionSolutions]

@@r2k314 You're very welcome!

[@mab9316]

Beautifull presentations. Thank you.

[@QuestionSolutions]

You're very welcome, thanks for the nice comment 👍

[@yuutokun3566]

When do I use the Io=mk² and Io=0.5mr²?

[@QuestionSolutions]

So if you're given the radius of gyration (k), then you just plug it in and solve. If you are given a uniform disk, you use 1/2mr^2. The first example uses the radius of gyration and the equation for a uniform disk.

[@revaldonkwinika1800]

Excellent. Could you please also do impulse and momentum of rigid bodies

[@QuestionSolutions]

I am actually working on it right now. :)

[@QuestionSolutions]

It is now up! 👍

[@revaldonkwinika1800]

@@QuestionSolutions Thank you so much. You are terrific😊

[@muhdsyafiq2978]

how do we determine the placement of datum line? love your videos btw 👍

[@QuestionSolutions]

It's up to you. Usually, a solid place like where pulleys are attached to a wall, or somewhere easy where we can easily calculate the change in positions.

[@irfnkml..]

Hi sorry for disturbing you. I want ask you , why kinetic energy T2 at 6:42 not use T=1/2 Io w^2. ?

[@irfnkml..]

And i also confuse when we must use T= 1/2 mv2G + 1/2IG w^2 . Thankyou

[@QuestionSolutions]

So the question is giving us all the values with respect to the center of mass, G. That means when we write our equations, it's easier to consider point G, rather than O. @@irfnkml..

[@epik_robie1709]

what's the difference between the regular radius of a disk and the radius of gyration?

[@QuestionSolutions]

Simple answer, you only need to worry about radius of gyration when you find the mass moment of inertia, and it's almost always given. This is not the same as regular radius. It's to do with the distributed area about the centroid axis. I don't have a video on it because it's usually not covered.

[@saturn_Il]

At 6:02, you skipped a step. S = the radical thing only, then you get delta S = S - 1.5. Just might confuse other people since you write S = radical something - 1.5, instead of delta S.

[@QuestionSolutions]

Here, "s" is simply the length the spring stretched. You can use delta s, or just s, or whatever makes it easier :)

[@wakeawake2950]

Nice video 👍😊

[@QuestionSolutions]

Thank you! :)

[@usamaalhasni9300]

7:30 why you dont use the parallel axis theorem , T=0.5Iow^2 ??????

[@QuestionSolutions]

If that works for you, do it. Usually, you will find that there is more than 1 way to solve these problems and get the answer.

[@Parthasorenchess]

7:21 Do we have to find centre of mass?

[@sohaibahmad1207]

Yes because Gravitational potential energy is measured with change in height of centre of mass. But you don't need to worry bcoz it will already be provided for difficult shapes.

[@QuestionSolutions]

If the object is a complex one, you will most certainly be given the location. 👍

[@vedbrahmbhattiit-j4497]

Hey awesome content bro 😁👍🏻

[@QuestionSolutions]

Thank you so much 😁

[@markonikoletic4088]

Great video, but wouldn't the answer for question one be in units of m/s instead of m/s^2? It is a velocity, not an acceleration.

[@QuestionSolutions]

Thanks, and yup, that's a typo. Velocity is always m/s, and acceleration is m/s^2.

[@radiatedbug]

Hello once again, I've got a simple question. For the second question, why didnt we consider the fact that the spring force is resisting the change in motion (pulling on lever), and therfore consider it negative?

[@QuestionSolutions]

So are we talking about once the spring stretched? If so, the force of the spring is trying to bring the bar back to where it was, so the force would be upwards. If we are talking about the force of the spring when the bar is at rest, it's not really doing anything, meaning it's not compressed or stretched, so there is no protentional energy 👍

[@radiatedbug]

@@QuestionSolutions Helloooooo once again, and I meant that from my understanding is that anyyyy force that tries and stop a motion from happening (so in this case the spring pullsss on the rod to revert it back to original position; so forces opposing motion like if we elevate a ball from ground floor to the 10th floor, the work down when an object goes up is negative thus its going to be considered as negative work). So here the spring is pulling the object back so shouldnt it also be negetive work? Also side note: If you dont mind may i contac tyou via email or smthn, I've got a question and i think it would be ore comfortable for you if it is done via email.

[@QuestionSolutions]

@@radiatedbug Springs are a bit tricky because almost always, they will do positive work, unless you throw something at it, and you still want the object to go through the spring, so when it slows it down, then we say negative work. But if you throw something at it, and you actually want it to stop, then it'll still be positive work. Very rare to get questions with springs that involve negative work. Even in this example, the spring is trying to pull the rod back to the original state, so positive work, where as gravity is pulling down, away from the starting point, so we get a negative value. And yes, email is contact @ questionsolutions.com 👍

[@radiatedbug]

@@QuestionSolutions I understood that clearly, very well explained thank you. And I will send you an email within 24 hours hopefully , tahnk you so much once again

[@QuestionSolutions]

@@radiatedbug You're very welcome!

[@arjungovender3248]

Hello, will there be videos posted for power in the work and energy of rigid bodies?

[@QuestionSolutions]

Probably not for a while, currently, I am not focusing on dynamics videos. Though I am sure there are many videos on youtube about the topics :)

[@amjadsamra105]

last problem, how is the distance from O to G 0.35m?

[@QuestionSolutions]

Look at the diagram at 6:25, the distances are given.

[@assaidnael8843]

for the last question you subtract 0,15m. Why ? and how do know it is 0,15m? Thanks

[@QuestionSolutions]

So if you look at 6:25, all the dimensions are given. If the whole pendulum is 0.6 m (top of the diagram) in length, and from the base to the middle of the circle is 0.45 m, then the spring when it's un-stretched is 0.6 - 0.45 = 0.15 m.

[@husseinmohamed1777]

Thank you

[@QuestionSolutions]

You're very welcome

[@e-nes4042]

4.34 velocity m/s^2 ? is it accelaration?

[@QuestionSolutions]

No, it's a typo. Velocity is m/s.

[@e-nes4042]

@@QuestionSolutions thanks 👍🏻

[@MdJunayed]

I am facing difficult questions in my class tests from this topic. Can I send those to you to solve ?

[@QuestionSolutions]

I don't really solve questions submitted since then I have to solve everyone's questions and it opens up a flood gate of problems. Sorry :(

[@MdJunayed]

Oh its ok 😌 keep up the good work 😁

[@QuestionSolutions]

@@MdJunayed Thank you and again, I am sorry! I wish you the best with your studies. :)

[@Makinamühendisi37]

4:34 isn’t it m/s ? I know little but some teachers get mad so fast 😢

[@QuestionSolutions]

Yes, there is a typo there. Good catch!

[@Makinamühendisi37]

@ but doesn’t matter. You are the best. If you wouldn’t teach dynamic, I couldn’t do that. Thank you so much

[@QuestionSolutions]

@@Makinamühendisi37 Thank you, and you're very welcome! :)

[@joshuacharlery5826]

First!

[@QuestionSolutions]

This is the "first" first comment on this channel. 😅 Many thanks! :)

[@Ishrat_Hussain]

how s=0.6 m : at time 8: 16 , I think it's wrong

[@QuestionSolutions]

No , it's correct.

[@shutnawi6635]

thank you

[@QuestionSolutions]

You're welcome!