Rolling without slipping problems | Physics | Khan Academy

In this video David explains how to solve problems where an object rolls without slipping.
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[@djcrazybeast5277]

This is the complete opposite of what I was trying to do I meant “how to roll weed without slipping fingers” but I found this educational video :( I feel like a drop out lol

[@presiankostov9388]

You should moist your fingers. Do it by gently exhaling on your fingers. The warm air you exhale should create a thin layer on your fingers. Another advice don't wash your hands with soap before rolling, because the soap tends to wash off your natural oils off your fingers and it drys your skin as well. Happy rolling!

[@jimmyhk1226]

@@presiankostov9388 Genius!!!!!

[@anusheelsolanki1]

@@presiankostov9388 ahh, I see a well experienced man there😂

[@y33tboy97]

yeah and i have a physics test tomorrow. 🗿

[@NotSoStNick]

@@presiankostov9388respect the technique!

[@pinruihuang8463]

Thanks Khan Academy, I would probably have failed some classes without you guys. This is why I decided to donate and everyone should.

[@taaaaaaay]

Came back here after 3 years of college. Life is good guys! Push hard!

[@aeon5567]

Been 7 yrs....still pure gold❤❤..Thanks Sir

[@rownitasheikh8343]

david sir;your teaching method is really unprecedented!

[@karankakkar3999]

Imagine playing with a 5kg, 4m wide yo-yo

[@masol3726]

Imagine eating 72 watermelons and giving 30 watermelons to your friend.

[@nevis4567]

Give this one a medal *applauds*

[@yehuawang7553]

Another episode of life saver 1 wk before test. When ever I thought i am into college and won't have any help from Khan Academy again, they can always surprise me with the collection of topics taught...thank you!

[@LyricsCubeDude1]

Why do i still bother coming to Physics class when i have this videos teaching me the same stuff but do it better

[@JonFenton-po7iv]

I like my physics grippy bruh

[@soccerboy7112]

I just had a quiz on this and I didn’t know how to put the rotational KE and linear KE together until I saw this, post quiz 😔. Thanks man! So I has different values for the shapes given, in this case I=1/2mr^2.

[@guiolaio1514]

I doesn't depend only on the shape, for example, a hollow sphere has a different value for I than a "full" sphere with the same mass

[@anusheelsolanki1]

5:06 The center of mass was not rotating around the centre of mass, cause it's the center of mass.

[@dmago8]

God of physics🙏😌

[@FraserSouris]

"The Ground is the String"

[@bostangpalaguna228]

so beautiful... 2 different scenarios, 1 totally the same calculation

[@YitzharVered]

Super duper useful, thank you very very much!

[@brunobordon9367]

thx you bro

[@neerajparchand5190]

Please can I know what software you use to explain things in such a creative and perfect way?? A very nice explanation David Sir!!

[@sashalatchaev2050]

Its Sketchbook Pro, and now its free!

[@NiratPatel]

In the first problem you must use h=6m as the centre of mass is 6m from the ground

[@arjunjn5703]

no dude it is 4 m, the CG doesn't hit the ground, 4 m is the distance which CG falls, look the figure carefully.

[@ishita3295]

It was quite helpful

[@jackflash8756]

So with the rolling on the inclined plane without slipping example, the frictional force is creating a torque about the COM which will cause an increase in the rotational Kinetic Energy of the rigid cylinder. Gravity cannot be causing the increase in rotational KE because it acts through the COM. But the frictional force is not doing any work on the body because the point of contact is stationary. Yet the rotational KE is increasing while the friction also acts to reduce the translational acceleration caused by gravity on the COM, such that the increase in rotational KE is matched by a decrease in translational KE of the COM. It almost seems that the frictional force is transforming translational KE into rotational KE without there being any net work done between the inclined plane and cylinder.

[@fouadnara1215]

Link for the playlist please!?

[@adrianho7165]

Then Net external force on the yoyo is not equal to mg? Because F = ma_c.m., while the centre of mass is not accelerating at g ms^-2. If it is accelerating at g, then v_c.m. = (2gh)^0.5

[@henrybristow1928]

Solid vid👌

[@tealahaj8268]

thanks a lot ♥♥

[@andrewgalbraith1858]

In the first example, wouldn't the CM of the yoyo still be 2 m high when the yoyo hits the ground? Edit: What I meant was that the initial equation should be mg(h + r) = 0.5mv^2 + 0.5Iw^2 + mgr, but that's equal to mgh = 0.5mv^2 + 0.5Iw^2 anyway. I was confused about whether h was measured to the yoyo's edge or center.

[@bobmarley9905]

nah cuz height of hoop was measured from bottom-end of yoyo, so CM also traversed same distance/height as it fell to ground... Plus u can define ur potential energy to be zero anywhere (so we make it zero when CM is radius 'R' above ground & mgh when it's at top before falling)

[@TeamJessieAngow]

Thank you this was so helpful!

[@adityajoshi8578]

Thank u so much 😇 😇

[@VineetKrGupta]

Thats some nice stuff !!! Thank you sir it was very helpfull

[@simplydry6506]

God bless you

[@형경이-s1j]

I have a question. In the last two questions, wouldn't the first one rotate quickly under gravity, but the second one rotate obliquely and slowly down? And when rolling without slipping, the part that touches the ground has zero speed, and the top layer has the fastest speed, but when you rotate at the same distance from the axis of rotation, isn't the speed the same?

[@delaneymarie9281]

so helpful!

[@abhishekchunduri8285]

Great explanation!

[@lordmeme8432]

Important Example starts at 13:30

[@astha192]

Dat made me love the concept!!

[@farooquekhan9537]

DAVID SIR ,THANKS FOR THIS HELP,BUT ICANT UNDERSTAND IT WITH FUIIY ENGLISH.PLZ TRY TO CONVERT IN SEMI ENGLISH. YOU ARE VERY GREAT.I HOPE ,YOU WILL DO IT.PLZ .PLZ.PLZ

[@fatimaisra9143]

Farooque Khan I think he only knows English so he can't "convert" it to semi English

[@RajShekhar-jy2zi]

What do you even mean by semi English

[@justdrew320]

Thx

[@bhinwaramjat8767]

Sir, kindly answer my question, why didn't we use g'=gsin(thita)

[@nevin8604]

Shouldn't we have taken g as g sin theta in the last example, as part of the gravitational force woulf have been cancelled by the normal force..?

[@hannakennedy3720]

excellent video

[@nullbeyondo]

14:26 Is the velocity of this center of mass = the final velocity?

[@adityaaggarwal2859]

usually thats true but it depends on the question wildly

[@placementcell6414]

Are all the videos available on youtube

[@徐欣源]

nice video

[@aniarablechannel4668]

By the parallel Axis Theorem, wouldn't the rotational inertia equal 2MR^2?

[@tylorangel2464]

7:20

[@klevisimeri607]

Pro tip: Th bottom point has V=0 (zero velocity) but not zero acceleration ( a ≠ 0) so that point moves up the ⚾ baseball but it doesn't slide.

[@joarhussain]

Thanks!

[@katerinapalacek1771]

what happens if the object is rolling down the plane and there is slipping, what would not hold true

[@হিরোআলম-ণ৮ভ]

Nice

[@nourharb7878]

Excuse me but how would it start rolling in the first place without friction in the last problem?

[@ZioAlboz]

It has to do with torques of the normal force and gravity. Projection of the CM as you can see goes out of the touching point between the cyilinder and the plane. Net torque is different from zero, thus you get an increase in angular momentum and so you get rotation. That's what I think it is. Probably you could explain it in a more rigorous way. But was just giving an idea

[@nourharb7878]

Thank you

[@ZioAlboz]

Actually yes, without Friction, normal force is not even giving any torque, just gravity! You're welcome buddy.

[@davidionce8943]

hey khan academy, if the ground or 4 metres under the yoyo is where potential energy is 0 J, then shouldn't the potential energy of the yoyo before it is dropped be "h+r", (4 metres for the height and 2 metres for the radius of the yoyo? so wouldn't the yoyo have potential energy of mgh where h=6 and not 4?

[@TheErdem29]

At the end, only the outside of the cyclinder is touching to the ground. This means center of mass is still r=2 meters high from the ground. first h(center of mass)=6 and last h(center of mass)= 2. So (delta)h = 4 meters.

[@receng2772]

Does it initially Potantial Energy equals to Mg(H+R) ?

[@Mrwiseguy101690]

No, because the object will only fall a distance of h.

[@seharmughal3846]

Rolling motion of a body is holonomic or non holonomic

[@hure2139]

Why is ωr = V(center of mass)? Doesn’t ωr = V(tangential)? And 2V(center of mass) = V(tangential) So, ω = 2v(cm)/r

[@farooquekhan9537]

i am wating for your reply

[@ahmedhussain4665]

At the end, shouldn't the value of gravity be g*sin(theta)???

[@arifmmm1]

see " polygon model of rolling friction"

[@HP-ie4sf]

why do all of the masses cancel in the end? what is the math behind that?

[@erickcastellanos6814]

hello swaney's class

[@kratosrivia2498]

Erick Castellanos help

[@philipmathews72]

don't we need to take the height wrt to the centre of mass

[@vanajalekkala5098]

The majority of viewers are indians. This just shows the fault in our education system

[@karanahuja0328]

Tru

[@harshraj3255]

isn't tension a non conservative force??how can you apply energy conservation

[@ekkiralac]

Tension is not an external force to the system..

[@apamapam4544]

Why do you cancel all of the masses if you have one on the left side and two on the right???

[@patsenas9481]

Wizardry, black magic

[@banand74]

So... basically v and omega aren't propotional and yet we use that equation to solve probs?

[@owenm5110]

The classic two meter yoyo hahahah

[@kapilduggar94]

It's not weird . It's concept

[@chandanbiswal1636]

Hindi plz

[@adityajoshi8578]

Thank u so much 😇 😇