Pendulum problem using conservation of energy

[@iamdynamight777]

Thank you so much! Even though I'm writing 10 years later, I can't explain how grateful I am for this video

[@rupal3628]

Thank you very much for the detailed and easily understandable explanation. I am taking an online Mechanics course and I realized its quite difficult to learn Physics through only a textbook or online articles. This definitely saved me a lot of time and frustration.

[@ethanmackie4693]

I’m mesmerised by how well he can write backwards

[@makanacaton3848]

Matt! Thank you so much! My name is Matt too, however, I am stupid:( . Have a great day!

[@honeydew1411]

😭😭😭

[@jvillalobos6988]

amazing!!!! it couldn't be explained simpler than this!

[@tony-iq3hl]

Explained clearly, helped me a lot for my assignment. Thanks!

[@yoprofmatt]

You are welcome. Keep up with the physics! Cheers, Dr. A

[@ronaldcredo2689]

Your presentation is unique! Brilliant explanation!

[@yoprofmatt]

Thank you kindly! Cheers, Dr. A

[@MrCyphox]

rah my man's intro lit af 😂😂

[@dr.mohamedaitnouh4501]

It is important to know that the calculations are made at the bottom h=0 so v is the velocity of the mass along the bottom point when it is launched at rest from an angle \theta. This is not the formula for the velocity or kinetic energy at any point of the trajectory of the pendilum. Great video openened my eyes!

[@ptyptypty3]

and so it goes................... another clear, concise video.. THANK YOU...

[@yoprofmatt]

To my #1 Fanderson, you're welcome. Cheers, Dr. A

[@zogardgazgull5675]

holy physic emperor, thanks for your great work

[@Dorak74]

Have a test on this tomorrow. Helped me alot. Thank you

[@GianlucaRoccaGian]

thanks for helping me out for my lab ::D

[@morrissabi5170]

You are my best ever sir i ever seen,,a reply to me Will Make me watch all your videos

[@valveman12]

Well done on the explanation and proof. It was clear and methodical.

[@theadel8591]

I first read it 'work done on the explanation and proof' physics has done its job

[@engjellabazaj1878]

You've been really helpful with my homework sir. Thank you very much.

[@yoprofmatt]

That's the goal. Cheers, Dr. A

[@innovationsforall]

Fantastic explanation! Thanks for sharing!

[@12717adi]

this is a great explanation, thank you

[@oyindenyafatarakeme]

just one word------> SUPERB!!!

[@junuthapa2007]

Good video. If you could include energy bar chart, this will be good.

[@rodericksibelius8472]

When the swing is raised and released, it will move freely back and forth due to the force of gravity on it. The swing continues moving back and forth without any extra outside help until friction (between the air and the swing and between the chains and the attachment points) slows it down and eventually stops it. WHAT THEN IS THE FORMULA TO KNOW THE TIME WHEN IT PRACTICALLY STOPPED?

[@yoprofmatt]

That's a great question and quite a tough one to answer. But you're absolutely right, friction and air resistance will cause it to stop eventually. Maybe I'll make a video about it. Cheers, Dr. A

[@rodericksibelius8472]

@@yoprofmatt Thanks Doc A. I have been pre-occupied with this idea as the same with the UNDAMPED condition with the RLC circuits both SERIES and PARALLEL, I just can't make the PHYSICS connection of the 3 DAMPING factors involved... Please show all the MATHEMATICAL STEPS going though the mathematics. Than You for Your Response.

[@yosimadsu2189]

@@yoprofmatt I just want to know the difference of both angles in respect of gravity but according to you/physics, it is nothing to do but the friction and air resistance. So it means the friction and air resistance are the important things to eliminate to get the pendulum swing longer? Thanks for the answer 🙏

[@milanacs1916]

Hi sir great video. I just have one question when Rearranging h=L-cos theta why does it become L(1-costheta. Where does the 1- come from?

[@slowvag8v]

could one calculate theta without knowing the initial theta angle? just knowing the value of the initial force in relation to the weight of the object where the angle would be equal to the arc tangent of the relationship between the initial force to the weight

[@noahjmeyer]

this is awesome, thank you! also, did you have to write everything backwards?

[@yoprofmatt]

+Noah Meyer Thanks for the props Noah! I did not have to write everything backwards. See this link: kzbin.info/www/bejne/eYirfqeJg7Crj6M Cheers, Dr. A

[@innovationsforall]

LOL

[@salmanridwan]

thank you professor !

[@tsehayenegash8394]

what is the velocity at the mid way? or what is vx and vy?

[@babavos439]

And what is the angular velocity

[@fizixx]

Did he stop posting vids? He was posting a lot a few months ago, and suddenly stopped.

[@mrb6309]

Where did you get that writing board??????

[@articcircleado]

You are a gift.

[@honeyb3603]

Hi Professor, can you tell me what will be the mass of object at the start of swing, middle on swing and at the end of swing.

[@jaylinn20]

thank you so much, you're amazing

[@mariamoine75]

You are my hero. Thank you so much!

[@Vishnu-dh2xj]

Thank you 😊

[@MrCyphox]

but in proper English, thank you for this video is was really helpful for my midterm

[@yoprofmatt]

+MrCyphox You're welcome. Good luck on the midterm. Cheers, Dr. A

[@Lionfish5656]

How do you figure out the tension in the string @ different angles?

[@carultch]

The purpose of the tension is two-fold, for a pendulum. 1. It needs to oppose the radially-outward component of gravity 2. It needs to cause centripetal acceleration to keep the bob following a circular path To solve part 1, it needs to at least be as large as m*g*cos(theta). To solve part 2, we also need to add onto it, m*v^2/L This means, that in total, the tension will equal: Ft = m*(g*cos(theta) + m*v^2/L) given a simple pendulum with a bob mass of m, a length L, and in a uniform gravitational field of g. The values of theta and v are transient throughout this problem, since v is greatest when theta is zero, and v is zero when theta is at its max. So you would determine v as a function of theta through conservation of energy, in order to get a concise expression for Ft in terms of theta.

[@rickydash6709]

why only translational kinetic energy is considered for a point mass pendulum(1/2 m.v^2) not the rotational part.

[@password6975]

This saved my life, thanks so much!

[@wfvfgfhnj9118]

password 69 ffg

[@AnarchyAce2269]

Thank you!

[@muditt9685]

Unsure if this is a stupid question or not but I was wondering from your (Matt Anderson) perspective are you writing everything inverted onto the board so we see it normally on the video .

[@yoprofmatt]

Nope. I let the software do the inversion. Check it out here: www.learning.glass Cheers, Dr. A

[@miguefigueroa8146]

Thank yo so much for the clear explanation.

[@yoprofmatt]

+Migue Figueroa I get it, "clear" explanation. Very punny. Thanks for watching! Cheers, Dr. A

[@mathadventuress]

Wow this made a lot of sense!

[@yoprofmatt]

Math adventures, Excellent! Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

[@dumbvidz1189]

Will tension give any impulse?

[@tyler-vp1gw]

can you tell me where the force of gravity is the greatest on a pendulum?

[@yoprofmatt]

It's always the same: mg down. Cheers, Dr. A

[@ALLIN-dc5my]

sir can you tell me why you are not taken kinetic energy in second case and I think E here total kinetic and potential please clear my this doubt sir

[@yoprofmatt]

In the second case, the pendulum bob is at its highest point, thus also at rest (v=0) hence no kinetic energy. Cheers, Dr. A

[@saqib3s]

Love this guy

[@Hakundwi1Tshifhulufhelwi]

i did not understand that part of 1-cos@...

[@nikhilvig1954]

would we not conserve the energy like ; mgh + 1/2 mv^2 = mgh + 1/2 mv^2 . where in initialy there was no kinetic energy so initial kinetic energy is zero . so; 1/2 mv^2 = mgh + 1/2 mv^2

[@yoprofmatt]

Your words are right, but the math is wrong. Initial kinetic energy is zero. But final potential energy is also zero. If h is the height at which it started, relative to the bottom of the swing, then you just get mgh=1/2 mv^2. See the 2:20 mark. Cheers, Dr. A

[@DaCom3AK]

Does the force of tension do any work as the pendulum swings?

[@rheedan]

No, for work to be done there must be some displacement (W=F.∆d). As tension acts along the radial axis, and the pendulum's distance from the centre does not change, tension does no work.

[@no1555llo]

THANK YOU

[@marlenekon4392]

Thank you so much. You are a physics g-d.

[@samarendra109]

sir i have just one doubt. why during the energy conservation the rotational kinetic energy is not considered. ie. 1/2Iw^2 . ?

[@carultch]

Because it is treated as a simple a point mass, and its size is insignificant. You can calculate what 1/2*I*omega^2 is, for a point mass on a negligible mass rod of length L, and you will find that 1/2*I*omega^2 also equals 1/2*m*v^2.

[@nick6269]

What would the time derivative of radial acceleration mean? Radial velocity?

[@yoprofmatt]

Believe it or not, it's something called "Jerk": en.wikipedia.org/wiki/Jerk_(physics) Cheers, Dr. A

[@KaranDoshicool]

What's wrong in taking l as l/2 by taking centre of mass of string and ball?

[@carultch]

Because that isn't the center of mass. The mass is assumed to be concentrated in the ball, and the string has negligible mass.

[@JustinEdwards316]

awesome, thanks man! solving for height had me confused

[@yoprofmatt]

+Justin Edwards You're welcome. Keep with the physics! Cheers, Dr. A

[@chemparathi3980]

Is it magic glass or something??? How is it possible

[@supsoup12]

wow, i just realized he has to write everything backwards, thats impressive

[@yoprofmatt]

+Daniel Gurman Not exactly. It's an image flip. See kzbin.info/www/bejne/eYirfqeJg7Crj6M Cheers, Dr. A

[@conzpicuouz]

Is that a clear glass board u are using or a black glass board?

[@yoprofmatt]

Clear glass. Check it out here: www.learning.glass Cheers, Dr. A

[@user-pv3cn7xo1i]

you confused me at 1:21, the energy should equal mgh at Ef. 1/2mv2 is when its at the bottom

[@IzWarped]

The Pendulum starts at an initial height h. Which is also Ei = mgh. Ef is when it's at the bottom so you are correct 1/2 mv^2 is when it's the bottom but were confused about Ei and Ef it seems

[@VintageMinecrafters]

Wishing u could be my physics teacher right now lol

[@ashishsingh0006]

sir which software yu have used to write like this way in yur video

[@yoprofmatt]

Come see us here: www.learning.glass Cheers, Dr. A

[@chandansurendar9107]

thanks a lot

[@yoprofmatt]

Thanks for watching. Cheers, Dr. A

[@EpmMaxim]

thank you!!!

[@yoprofmatt]

EpmMaxim, You're very welcome. Glad you're enjoying the videos. You might also like my new site: www.universityphysics.education Cheers, Dr. A

[@Gabriel_Naomi]

love u andy

[@debjaniroy13701]

Helpful

[@arjunkbiju3636]

Thank You sir...........

[@kunwarpratyush8909]

Does he write reverse?

[@mohamadokab8749]

no they use a mirror to reflect it

[@danielxu8790]

Good physics, and backwards writing, daaaaamn

[@yoprofmatt]

No daaaaamn. Physics trick: kzbin.info/www/bejne/eYirfqeJg7Crj6M Cheers, Dr. A

[@bhaktiwashilkar8590]

why.... E (initial) is equal to E (final)

[@TehIdiotOne]

Conservation of energy. No energy can be created nor destroyed, it can only be converted into other forms of energy.

[@bhaktiwashilkar8590]

ohhh thank u so much!😄

[@alchemy1]

It starts out by claiming/declaring that Ei = Ef. That is why it is never true at all in the actual world. Just on paper only. Any person who is paying attention will just keep on going minding his own business and that includes yourself. And any attempt to to bring other issues such as friction, air resistance is still false. There is so much more and all that is just left out. First can anyone define what is this thing and whoever came up with it...."closed system"? Does it even exist? If it doesn't, it is finished. And if it does exist... that is another can of worms...

[@jasonappleby7040]

Anyone else not get 1m/s. I got 3.14...

[@yoprofmatt]

Tried it again and still got the same answer. Did you use sin instead of cos? Cheers, Dr. A

[@jasonappleby7040]

Calculator was in radians, always worth a check to make sure :) When in degrees we get 1

[@yoprofmatt]

Excellent, thanks for the update. Cheers, Dr. A

[@CDinkle]

@@yoprofmatt When I am in degrees, I get .317, when I am in radians, I get .99.

[@AK-di4rv]

very through!

[@yoprofmatt]

Glad to hear it. Keep up with the physics! Cheers, Dr. A

[@rox3ana]

CHEERS

[@yoprofmatt]

And to you! Cheers, Dr. A

[@drlucky5474]

Do you believe that energy can not be created? I do not think so. If so, why it exist? It means that energy can be created right?. If energy can not be created, then it will not be there. Think about it.....

[@yoprofmatt]

Dr Lucky, Thanks for the comment. In physics we like to talk about theories that relate to experimental predictions and ultimately experimental results. If we cannot test it, it's not physics. Since all the energy in our universe was released by the Big Bang, you might say that the Big Bang "created" energy. But that would imply that you could see what was there before the Big Bang. And unfortunately, we have no way of seeing anything before the Big Bang. Thus that does not qualify as physics. So this is our current understanding: The total energy in the universe is a fixed quantity. We cannot create more, nor destroy it. We can simply move it around. Cheers, Dr. A p.s. Physics, like all science, is constantly evolving. Just because our current understanding is this does not mean our future understanding will be the same.

[@RovanRCTube]

Dr Lucky, I don’t believe energy can be created, it can only be changed in form. At least on earth. But I also believe that our knowledge of physics has us stuck in a box. Although I believe we are 90% correct, there are things we do not know. Perpetual motion is not possible, however there are ways to use physics and combining known forces to convert energy. So extended motion is a factual thing, just look down at the rotating earth. So I believe that some day we will find a way to get over unity without defying physics, and maybe even leaving the laws of thermal dynamics in tact. But it seems that our teachings are holding us back.

[@lukecaetano8405]

I love u

[@yoprofmatt]

Luke Caetano, Thanks, love you too. You might also like my new website: www.universityphysics.education Cheers, Dr. A

[@lukecaetano8405]

@@yoprofmatt I'm an aerospace engineer touching up for statics next semester so will do!

[@zeljkotutunovic6812]

Fake.......

[@yoprofmatt]

Me or the pendulum? Cheers, Dr. A

[@sidneyshaw1608]

lol ok.... *yawnnnn*