Physics 9 Conservation of Energy (9 of 11) Sliding In A Bowl

[@MichelvanBiezen]

TheAllofall, Your friend is correct. Technically the normal force will be mgsin(theta) + (mv^2/R) The second term comes from the centripetal acceleration. In this problem I ignored the second term for the purpose of illustrating the technique

[@JSB_physics-made-easy]

Sir, as you know there is enough increase in speed so effect of it on normal reaction and consequently on friction and work done can not be ignored. Can you kindly include velocity factor and reload next video

[@rachelpiteck]

I've been stuck on a problem similar to this and couldn't figure it out. This broke it down Perfectly! Thank you!!!

[@MichelvanBiezen]

Glad you found our videos. 🙂

[@annasarahkramerfazendeiro8030]

Thank you! So the horizontal speed is the same as the free falling speed if no friction is involved. I was really starting to doubt my judgement in class today ... I guess I did not explain my question properly to the teacher. It is great to be able to arrive here and find an answer to practically every question. Makes me wonder about the percentage of engineers that exist mainly thanks to you!

[@kraftwerk951]

Hi Michel! Thanks for sharing these awesome videos, I'm learning a lot from you. I'm studing at the University of Pavia, Information Technology and Electronic Engineering. I'm integrating theorical parts by using my physic's book but I've found something weird on it. My book says that the work done by the friction is equal to -mg*mu*integral from A to B of the length of the curve ( ds ). Well it started saying Work=Integral from point A to point B of ForceVector*DOT*displacement. Since the friction does work against the path I'm running on, and using the dot product definition we have abs(F)*abs(ds)*cosTheta. Since the friction is pointing ever in the opposite direction of the displacement, we have 180 degrees angle and therefore cos180=-1. So -mg*mu*integral from A to B of the length of the curve ( ds ). And then why are we calculating the path using theta, instead of calculating the effective path of the curve ( so R*pi/2 )? Thanks!

[@MichelvanBiezen]

Strictly speaking, when the direction of the path changes like in this example, the work done must be integrated along the path, so that the angle between the force and the path direction can be expressed as a function of position.

[@swimcoach3702]

If we knew nothing about how mu works until now, the equation would tell us that it must be a number between 0 and 1. Less than 0 gives an impossible result and greater than 1 gives us a super friction that would speed up the velocity greater than gravity.

[@MichelvanBiezen]

Glad you liked it.

[@jayakumarelpula189]

Excellent explanation, professor 👌👍

[@MichelvanBiezen]

Thanks! 😃

[@honestman276]

Thanks. From Bangladesh.

[@MichelvanBiezen]

Welcome to the channel!

[@onielrodriguez9194]

Hi professor Michel van Biezen, I have several questions regarding your methodology. Why do we need to use the angle given and then find a corresponding angle, isn't easier to assume an instantenous ramp in which the angle can vary and then work from there? Also, isn't mg×cos theta usually refer to the normal force when it is on a ramp? Why would it be different here? And also I did the problem in the exact same way you did but used the assumption that the normal force was mg×cos theta and integrated properly, and I still got the same exact answer you did. Was this because it is genuinely another way of doing it or because it just happened to be right by accident as it is in some cases?

[@MichelvanBiezen]

There are often multiple ways to do the same problem. Use the method you are most comfortable with.

[@onielrodriguez9194]

Thanks!

[@enriquesoler3150]

I just want to make this clear, when you were solving for energy lost, why is W=energy loss?

[@MichelvanBiezen]

Energy lost equals the work required to overcome friction.

[@jimbiolsi4276]

Hi Michel, why is the normal force equal to mgsin(theta)? I thought there needs to be centripetal acceleration so normal force would need to be more than mgsin(theta) by mv squared over r. thanks Jim

[@MichelvanBiezen]

The normal force will be mg sin(theta) or mg cos(theta) depending on which angle you are using.

[@shreyanshpandey.3281]

I also have the same doubt . Actually this question was given in my class and I did the same mistake.

[@victorburacu9960]

You are right. Michael makes the same mistake as in another of his videos by not mentioning mv^2/R; not allowed. He could have cancelled this term (assuming for example that R is very large) after writing the equation of motion correctly.

[@mals62]

what would be the conservation equation if the object was being pushed up to a height?

[@MichelvanBiezen]

The equation would be exactly the same. You just have to recognize the initial PE and KE and the final PE and KE and how much energy is lost due to friction

[@ENOCKcheruiyotKurgat]

i appreciate sir well understood

[@MichelvanBiezen]

You are welcome. Glad you found our videos. 🙂

[@ENOCKcheruiyotKurgat]

yes i really do@@MichelvanBiezen

[@mayureshkathe8311]

Dr Biezen, can you please make a video on the concept of rolling friction ?

[@MichelvanBiezen]

Take a look at this playlist starting at video 44: MECHANICAL ENGINEERING 11 FRICTION kzbin.infoplaylists?sort=dd&view=50&shelf_id=12

[@anhlam299]

hey Michel, instead of using Rd(theta) can I just use (1/4)pi.R as the distance that friction acts on the object ?

[@MichelvanBiezen]

No, you must use the method in the video because the friction force continuously changes as the block slides down.

[@nellvincervantes3223]

Sir micheal van biezen, may I suggest that you make a video about finding a force along a section line on structutal frame? If I can send pics here, I would send the image here.

[@elis2506]

guys i'm pretty sure his equations for the horizontal force and the normal force are wrong. Rather than mgsin(theta), normal force would be mgcos(theta) and horizontal force would be mgsin(theta). That's how I've been learning it in physics class and I even looked it up cuz I doubted myself when I saw him write that.

[@MichelvanBiezen]

The equations in the video are correct. Use the definition of the cos and sin cos = adjacent side / hypotenuse sin = opposite side / hypotenuse (It depends on which angle you are considering)

[@elis2506]

@@MichelvanBiezen oh, yeah. I guess the way I learned it was considering another angle. Sorry for doubting you. Also, this video was posted in 2013 and you're still responding! That's amazing and I admire your dedication to students and answering questions. I'll definitely be coming back to your channel when I don't understand my physics class :)

[@andrexjudedagolsapo9410]

can you make an example with that method?

[@MichelvanBiezen]

Sorry, I didn't understand "which method" you were referring to.

[@andrexjudedagolsapo9410]

i mean make a problem and solve with that method😅

[@MichelvanBiezen]

I think you are asking for another example? We have lots of examples where we use the conservation of energy method. This example here is a special example, since the angle continuously changes. That is why we need to use integration.

[@antoniomota7454]

Hi... is this the complete solution? (i.e. with considering v^2/R): ( 2*g*R^3*(1-niu)/(R*(R+niu)) ) ^ (1/2) Thanks. -Antonio

[@MichelvanBiezen]

No this is not the complete solution, we should add the additional normal force produced by the centripetal acceleration.

[@antoniomota7454]

... that was expression for V. When niu = 0 (lossless) it also gives V=(2*g*R)^(1/2), but that doens't mean it's correct, of course :)

[@MichelvanBiezen]

I should redo this problem taking into account the centripetal forces. Thanks for pointing this out.

[@antoniomota7454]

Thank you for the videos, with clear and systematic approach. I love physics, though I'm kind of slow in working the solutions :/ but I never give up on a challenge, which I guess is a quality :)

[@khm5869]

Could you please explain why can't we only consider the initial and final position of the object? In that case, N will be equal to mv^2/R and we can ignore the weight, since there is no weight acting on the perpendicular direction at the initial position.

[@MichelvanBiezen]

Khawla, There is "weight" acting perpendicular to the motion. It is actually the centripetal force which acts towards the center of motion and which causes the force due to friction.

[@ahmedal-ebrashy3691]

Thank you very much!

[@rohithprabakar7055]

can this example be related to simple harmonic motion in any way? If so, can somebody explain it to me.

[@MichelvanBiezen]

Yes, but only if the oscillations are very small such that the angle with the vertical changes only a little. (Same principle as with a pendulum).

[@kindjupiter]

Thanks

[@chifo60]

Hi, Michel Can you tell me why the weight doesn't make any work?

[@MichelvanBiezen]

The potential energy is taken into account on the left side of the equation, which is then converted into kinetic energy and heat. (When we use this energy conservation equation, one type of energy is converted into another type of energy, and thus the work does not need to be calculated).

[@chifo60]

@@MichelvanBiezen But in the video you said that there's no force making work and I don't get why

[@MichelvanBiezen]

The normal force cannot do any work since it is perpendicular to the direction of motion.

[@iamjustapoorboii598]

hey what if niu is greater than one as in the case of cast iron(dry)... then the eqn. won't hold good

[@MichelvanBiezen]

That should not make any difference.

[@kwakuappiah9996]

thanks a lot.

[@valerkand]

Stop on 01:00 ... solving ... 3 min paper work ... solution: V^2 = 2g (h - mu R). Right?

[@valerkand]

Heh, yes h = R is the height equal to the radius, so we have V^2 = 2gR (1 - mu).

[@misbauddin1383]

why it is mg cos for the parallel ?

[@MichelvanBiezen]

+misba uddin Definition of the cos (theta) = adjacent side / hypothenuse

[@misbauddin1383]

thanks

[@TeenGohan9798]

i don't really get why dx=Rdθ help pls

[@MichelvanBiezen]

The best way to visualize that is to make dtheta equal to 2 x pi (360 degrees), what does ds become?

[@TeenGohan9798]

umm i don't know :P

[@MichelvanBiezen]

Note how the circumference of a circle (s) is equal to the radius x angle (R x 2 pi) --> s = R x theta where s = C in that case

[@TeenGohan9798]

oh i seeee now it makes sense , very well explained thank you very much ;)

[@nellvincervantes3223]

dx is the small displacement travelled by the block. Note that friction force (f) and dx are vector quantities and their dot product results to infinitestimal heat loss dW. dW = fdx Remember the circle arc formula, S = r(theta) So in this case, dx = S, and theta = d(theta) for a infinitestimal circle arc length travelled, dx, there is also an infinitestimal angular displacement travelled, d(theta)

[@learnsomethingnew5316]

Here we dont have taken centripetal force why

[@MichelvanBiezen]

We left that out to simplify the problem. It is much more complicated when adding the centripetal force.

[@learnsomethingnew5316]

@@MichelvanBiezen sir I have a lot of confusion in this question a small object of mass m slides along the surface of a bowl what is its angular frequency about the lowest point ?? Thanks for your reply .

[@satyamkumar567]

I think u missed the centripetal force

[@MichelvanBiezen]

Yes, we ignored the centripetal force. Including the centripetal force makes the problem a lot more difficult.

[@kumarsanman3]

HIGHLY COMMENDABLE VIDEOS BUT SOLVE QUES. OF HIGHER LEVEL ALSO

[@MichelvanBiezen]

That is our plan for the future.