Yo-yo in vertical circle example | Centripetal force and gravitation | Physics

Yo-yo in vertical circle example | Centripetal force and gravitation | Physics | Khan Academy

Рет қаралды 175,133

8 жыл бұрын

In this video David gives some problem solving strategies for centripetal force problems and explains many common misconceptions people have about centripetal forces.
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Пікірлер: 64

@sathibabu9365 2 жыл бұрын

The luckiest person on the earth is David sir because he can tackle the physics very easy

@jeffreytebbens3100 6 жыл бұрын

Consider posting an example where the yo-yo is not at the top, but say, 45 degrees from the vertical.

@MrMalucoxx 6 жыл бұрын

Great video. There's just a problem when you consider that the velocity is equal everywhere, which is impossible because gravitation is accelerating the yo-yo. In accelerated systems like this, the centripetal force changes due to velocity and this change is necessary in order to maintain the circle. To calculate the last tension you would use first the law of conservation of energy to get the velocity and then the calculation in the video (it's valid since the expression of centripetal force is valid for all possible circular motions, in that instant)

@AdityaSathya 2 жыл бұрын

This is correct

@clarisaleu-rodriguez5314 7 жыл бұрын

Thank you! You're explanations are so clear and easy to follow.

@asiyafatima544 7 жыл бұрын

You guys are amazing at explanation !!! Khan academy , you people have helped me a lot in simplifying the concepts and making studies much easier ...

@urunclesam1659 4 жыл бұрын

I’ve been confused for hours as to why mg was positive in my practice problems even when it was pointing down and you just relieved me for my test by 100000 percent thank you for existing :)))

@wyattfelipe1458 2 жыл бұрын

You all prolly dont care but does anybody know a tool to log back into an instagram account..? I stupidly forgot my account password. I appreciate any tricks you can give me.

@karsnogen8141 5 жыл бұрын

Thanks, i really need to do my homework now

@trash5694 3 жыл бұрын

Amazing explanation! Thank you so much

@foureyedcyclopse9412 3 жыл бұрын

Thank you. Very informative and helpful!

@JRay2113 7 жыл бұрын

Sooo clear!! thank you

@alevelsdemystified3410 6 жыл бұрын

Great explanation!

@saramehravar 6 жыл бұрын

Thank You! Great Great Great Explanation

@tk006 10 ай бұрын

I'm in high school currently and want to study astrophysics once I leave. My teacher said this topic will be very useful later on when we look at planetary motion so I want to understand it really well. This video condensed the information I needed perfectly.

@nostyx9601 Жыл бұрын

Thank you sooooo much!!

@Learner2500 5 ай бұрын

U guys are the best thank you

@AhmedAli-km5sp 5 жыл бұрын

Thanks!, very helpful.....:)

@tanveerfaridi7481 6 жыл бұрын

thank u very very much sir it is very helpful for me

@azimbagwan2762 7 жыл бұрын

physics is become easy due to you

@joshuamoodley7600 3 жыл бұрын

Yoh! you just saved me from a really bad semester test in a few hours Thank you!

@vivialord5260 3 жыл бұрын

thanks

@aman2426 6 жыл бұрын

Helpful (Y)

@Dewayneyork1 6 жыл бұрын

So in a since, the string is performing as a sort of artificial gravitational pull. Am I correct?

@whatamitalkingabout3 6 жыл бұрын

a video in which the yo yo was not at the top bottom rightmost or leftmost point would be very beneficial. probably wont be this easy on any test

@anushreesabnis5856 6 жыл бұрын

The velocity increases as the yo-yo reaches the downward curve of it's motion, doesn't it?

@wongshanyan1724 2 жыл бұрын

Question , the speed when the string snaps ?

@perveshikamonga4205 7 жыл бұрын

can we calculate time period in this case...if yes, how??

@lilmarshmallow2204 8 жыл бұрын

i didn't get the last part. how did the formula end up like that?

@thasleenathasli6210 5 жыл бұрын

Can it explains by centrifugal force? ?

@goodboymzobe6460 6 жыл бұрын

Is the velocity at the top equal to the velocity at the bottom?

@danielkwon3062 5 жыл бұрын

isnt there normal force on spinning yoyo? like ferris wheel problems?

@marwangoesfit6398 5 жыл бұрын

isn't the centripetal acceleration formula only supposed to work when the object is in uniform circular motion?

@ezraberger6912 6 жыл бұрын

Ropes can't push

@mimamod1867 7 жыл бұрын

I like how everything is in metric system. :D

@nagysadaka7149 2 жыл бұрын

Why we consider that the speed is constant but T is variable during the rotation ?

@olivetom1599 7 жыл бұрын

And so we can say that earth gravity doesn't change it's impact on the string (and so on the finger) whatever the speed of the yoyo is. (and it's not that intuitive to me)

@jameschen3138 Жыл бұрын

I calculated the Centripetal force as mv^2 / r = 8N. Is the centripetal force different from the tension force in this scenario?

@erererpo 7 жыл бұрын

What application did you use here

@devenarya5766 6 жыл бұрын

i also wanna know that same... thnx for asking

@bezpeglar4134 7 жыл бұрын

he sounds like Owen Wilson :P

@adampaolillo1141 4 жыл бұрын

Why is tension not equal in magnitude to mg??

@tyrafero8590 7 жыл бұрын

what about on sideways!!!

@rofiqel6226 6 жыл бұрын

I think when it's sideways you'd need to use trig

@jesusibarra5498 7 жыл бұрын

Wouldn't the velocity be different at the bottom of the circle? So the answer should be different. Im having trouble finding out how to find the difference in velocity.

@daleenhammoud623 7 жыл бұрын

These problems deal with uniform circular motion, where the circular motion occurs at a constant speed; there is only acceleration because the direction of velocity is changing, but the magnitude (speed) is always constant.

@SunilKumar-zs5vh Жыл бұрын

@@daleenhammoud623 magnitude of velocity is changing brother it is non uniform circular motion ..

@PokeShadow77 3 жыл бұрын

This guy is even better than Salman Khan

@jubairtayib8312 6 жыл бұрын

wrong hey.... the velocity should not be the same in the lowest point...mg*2r + .5mu*u = .5mv*v . i think it should be like that ..... that means tension at every point T not only depends on the component of mg along the string but also depends on the velocity at every point as it will be changed because of gravity.