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

Learn how to solve rigid body conservation of energy problems step by step with animated examples. We cover potential energy, kinetic energy and more.
🔹Rigid Bodies Work and Energy: • Rigid Bodies Work and ...
🔹Rigid Bodies Rotation About a Fixed Axis: • Rigid Bodies: Rotation...
🔹Kinetics Conservation of Energy: • Conservation of Energy...
Intro (00:00)
The spool has a mass of 20 kg and a radius of gyration (02:23)
The slender 6-kg bar AB is horizontal and at rest (04:35)
The 30 kg pendulum has its mass center at G (06:16)
Find more at www.questionsolutions.com
Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2014.
#ConservationofEnergy #engineeringdynamics #mechanicalengineering #rigidbodies
#hibbeler #Dynamics

[@rehanrashid9296]

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[@QuestionSolutions]

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[@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]

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[@QuestionSolutions]

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

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[@QuestionSolutions]

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[@QuestionSolutions]

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[@QuestionSolutions]

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

[@Olivia-hu8eg]

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[@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!

[@sanali4425]

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[@QuestionSolutions]

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

[@albazmovie6267]

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[@QuestionSolutions]

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

[@Olivia-hu8eg]

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[@QuestionSolutions]

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

[@Olivia-hu8eg]

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[@QuestionSolutions]

@@Olivia-hu8eg You're very welcome!

[@jakobolson9958]

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[@QuestionSolutions]

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[@mab9316]

Beautifull presentations. Thank you.

[@QuestionSolutions]

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

[@roxon6929]

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[@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!

[@vedbrahmbhattiit-j4497]

Hey awesome content bro 😁👍🏻

[@QuestionSolutions]

Thank you so much 😁

[@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 :)

[@shutnawi6635]

thank you

[@QuestionSolutions]

You're welcome!

[@wakeawake2950]

Nice video 👍😊

[@QuestionSolutions]

Thank you! :)

[@husseinmohamed1777]

Thank you

[@QuestionSolutions]

You're very welcome

[@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😊

[@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.

[@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..

[@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!

[@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.

[@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.

[@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 👍🏻

[@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. 👍

[@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.

[@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.

[@Diggydogdraddy]

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 👍

[@Diggydogdraddy]

@@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]

@@Diggydogdraddy 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 👍

[@Diggydogdraddy]

@@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]

@@Diggydogdraddy You're very welcome!

[@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. :)

[@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.

[@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 :)

[@joshuacharlery5826]

First!

[@QuestionSolutions]

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