85% Students don’t know Basics of Friction | Eye opener JEE PHYSICS
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- Опубликовано: 1 июл 2023
- Friction and Newton’s laws of motion is a fundamental topic of class 11 JEE Physics but 85% students don’t know the basics of friction in JEE Physics. This video is an explanation of some basic concepts of friction by Ashish Arora sir to build a strong OTBT thinking on topic of friction.
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I am sitting in Jossa 2023 and got NIT Allahabad also but still feels soo good of seeing our old good concepts ...
FOR ONE WHO ARE IN PREPARATORY PHASE
trust me it is one of the first whose gonna shape you and your career in a whole new direction...
Good luck buddy's
Hey...I also got mnnit allahabad cse😊😅..Btw what's ur branch..
@@anushkaverma3706how much percentile did you got?
Mechanical
@@anushkaverma3706what's your percentile and rank ...
I got NIT RAIPUR the best for core branches after NIT JSR
Physics skills ...............♾
Meanwhile abj sir
Right 👍
Infinity symbol -> ∞
@@kunaldas9543abj sir is good but cant be compared to ashish arora sir☕☕☕
@@shashishekharmishraiitkanp6928 yes ofcourse god is can't compared with king (human ).
Hint
Here asish sir is king
......xd
Here is my analysis of last part: it moves through path 2.
(Complete analysis is done assuming block is not at the edge.)
As the block start moving, it does so radially outward wrt disk. Now as the disk rotates in clockwise sense, at every moment that the block slides along the line, there is some angular displacement in the clockwise sense. This implies, it cannot travel through path 1 or 4 as Sir mentioned (as in both cases, there is no angular displacement).
Now, it moves through either path 2 or 3. But it must be noticed, that the clockwise rotation is causing the curvature. While staying in contact with the disk, the object tries to have a straight line path, but there is another velocity (that of the disk) that moves it "up" (basically, rotates it clockwise). Hence it gets a clockwise curvature. (Outside the disk, its path is a straight line tangential to the point leaving the disk.
Hence my conclusions:
1. If the object were at the edge, it would follow path 1, but tangentially outward.
2. If the rotation were anticlockwise, it would follow path 3.
3. In current situation it would follow path 2.
If anyone has any question/dispute behind this analysis, kindly reply. It would be really helpful 😃
Bro I have one doubt
When disc rotate in anticlockwise direction of tangetial fric get reverse
As a result block now can take 1 or 4 and due curvature it follow 4 ???
Straight line wala hi kyu
isme same lekin curve wala kyu nhi banega
actually i thought i would go vertically outward but not tangentially...but i did not think of vector sum of both sliding..
my answer for the path is 2 as the block will be having some velocity at the point of leaving and path number 2 is showing exactly that path....
Thank you so much ashish sir..for evolving my mind to this level♥
Nice concept sir..
Thankyou for clearing the concept ❤❤
Even aftet paying 30 to 60 thousand for my tuitions they cant even explain me in this depth and fundamental level of thinking. really we all students are so lucky to have a teacher like Ashish sir.
Happy guru pournima 😊 to Ashish sir and all PG team.
I think it should move along 2 paths as velocity of object = vel. w.r.t disk +vel of disk
There resultant will be along 2 path and since tangential velocity of disk is changing motion of object will also change in that direction
Happy Guru Purnima Sirji..!! Inspite of being so fundamental why we weren't able to think it..? Thank u for providing such thought provoking content..!!
Thank you sir for clearing the. Concept ❤❤
Yes sir 🙏I have solved even advanced level questions on circular motion and pyqs of advanced and nsep but did not think in this deeply manner. I request you to start explaining video format for some good conceptual tbt questions. This would be very helpful for all PHYSICS enthusiasts
Yesss
thank you sir great thing to learn 🙂🙂
The result direction of sliding with respect to ground (at an instant) of the block will be in the direction resultant of ( force of friction and centrifugal force ) at that instant
That is most probably it will follow path 2 .
Nice concept ... Love You sir ❤
What a mind-blowing concept sir.
Nice elaboration sirji❤
Thank you sir for such a nice eye opener concept
Sir as per my understanding, block will move in path 3 becoz to leave the disk block will have to cover some distance radially for that it will take some time, it can't leave it instantly so in that period of time eventhough block is in contact with disk means a friction is still acting on it thus the line along which block will move wrt disk will shift from South West drxn to North West drxn that's why it will follow path 3.
Sir path 2 as Velocity of A w.r.t B = Velocity of A w.r.t ground - Velocity of B w.r.t ground same for acceleration , so along that dotted line is the velocity of cube and and velocity of disc w.r.t so we add them and there is no acceleration of cube w.r.t disc but the acceleration of disc is along "at" and hence path followed is second
PLZ NOT STOP THIS SERIES VERY VERY HELPFUL ❤❤❤❤❤❤❤❤❤❤❤❤
Sir, I am very thankfull for your this kind of content. Please continue eye opener series like this vedio and wake up series.
Sir answer is 4. Please confirm.
This literally was a part of my doubt last year . Most teachers whom I asked simply ignored it and said to memorize it .
Thanks to Ashish Arora sir for so much conceptual clarity ✨👍.
don't go to fake coachings
@@shashvatpandey3028 yyess
9:43 sir the path traced by this object would most probably be the path 1 curve as the friction is acting opposite 2and 3 rd would certainly be not possible the path 4th is curved in outward direction which is possible only if the object was revolving in clockwise direction
Sir Iam able to understand properly thanks sir ..❤️
Sir according to me, it will go to path 3 in ground frame.
Path 1 and 4 isn't possible as the block has sliding tendency but the direction of its motion will still be decided by the friction of the disc.
For 2 & 3, at the moment when we increase the angular velocity of the disc, the new velocity of the block will initially have a tangential component only and the component of friction providing centripetal acceleration doesn't change. So the block goes through path 3.
Lekin ground ke respect mein toh disc angular velocity and centripetal ki direction mein move krega toh friction backward act krega na?
@@karanjoshi6415Yes, disc toh apni direction me hi move krra hai, block bhi usi se direction me move krega friction ki vajah se but path circular nahi hoga block ka.
Sir, according to me it will go to path 3rd, As it has initial tangential component and there is constant friction component in it's opposite direction.
I think the answer is path 3. I compared the friction with gravity and the tangential velocity with velocity of an oblique projectile. The path 3 which is a parabola, should be trajectory w.r.t ground.
abhi coaching me friction start bhi nahi hua hai, lekin tab bhi sab samajh aa gaya... wow sir!
Happy Guru Purnima sir 🙏🏼❤️😌
Happy Guru Purnima to such a fantastic teacher
My revision notes of the guided checklist for chapters like electrostatics, current electricity gravitation ,etc comes in one sheet both sides . Sir please tell me are this much short notes ok or they have to be more short ??
The answer should be 1 because acc to me if the path of block will be in right direction [and downward as concluded] and the disc is moving opp to it hence the direction of block wrt disc must be as stated and yes sir before you explained i though it may move in tangential direction. Thank you sir
Nice concept, I already knew it though.
I don't if you noticed but this concept would be very easy if you just applied pesudo force in rotating frame in the dirn opposite to tengential acceleration
Thank you sir
Thank you sir ... Happy Guru Purnima 🙏 with your help I got into IIT KANPUR
Happy Guru purnima sir ji, u have been the beacon of knowledge... will be forever in debt . 🙏🙏🙏.
Sir it will follow path 2, as V(bd) is lets say -xi -yj and V(dg) is the tangential velocity of disc which will be something +wj, V(bg) will be V(bd)+V(dg), Thus it will outwards and also logically follow path 2
It helped me a lot sir.
(JEE-2024 Aspirant)
Wahhh ❤, thanks sir
Already knew all this....just came up to check if there might be anything i am missing out to learn more about friction....thank u very much sir❤
Happy guru Purnima sir
Seeking your blessings 🙏
You are one of the best mentor in India
And I am very grateful that I found this channel and able to follow your advice .
From botton of my heart I thank you for your effort for us .
Inspiration ❤
As acclrn req in disc frame radial and tangential is proportional to r . So with disc it's slipping direct will remain constant .and if we now comes to ground frame we see path 2 if mass is heavy and disc is spinning fast.
Anwer may be path 3rd as we have to calculate w.r.t ground.
We use the concept of relative
.... Motion
Nice Explanation😊
Sir this series is very important for us😊😊😊
I think it can be path 2 and 1. It can be 2 for simple blick kept and 1 if it is a sphere kept
Path 3 will be correct for observation w.r.t. ground. 1 and 4 are obviously wrong because the disc is rotating clockwise and hence path cannot be in opposite sense. Path 2 implies that "friction" is increasing and not "angular velocity" opposite of which is path 3 which should be the correct answer.
Sir, my analysis says that the block moves in path 2 wrt ground because as sliding starts the circular path in which block moves starts to get larger and wrt ground block has some max. acceleration although the point just beneath the block will have more velocity and acceleration so wrt disc block will seem to go behind of that point but wrt ground the block has a acceleration in tangential direction so it will tend to move forward with increasing circular radius and thus it will move in path 2.
My explination for the problem asked by sir(Though i have the feel of it in my head, but may not be able to completly explain it with the clarity sir may explain. I would try my best though):-
The initial conditions say that the disc is moving with constant angular velocity. Now, when we accelerate it, the block will tend to oppose the acceleration, and its velocity(instantaneous) will be less than the disc, thats why we see it slide backward W.R.T. the disc. But, it will not start moving in the backward direction(W.R.T ground). So, option 1, 4 are wrong. As to compare between 2 and 3, we can think of it as that the box though opposing the acceleration, still accelerates. The velocity of the box must increase as it moves more and more towards the edge of the disc. So the path must curve upwards. So the correct path must be path 2.
Also sir, your TBTs and tips really helps. I would owe you just too much throughout this preperation. If ever possible, I will surely meet you once. I live in Jaipur only. Thanks for all your support, and i seek you blessings on this day of Guru Poornima. 🙏🙏
It was much needed
I think the answer is path 2 as in frame of reference of ground it will have tendency to move upwards and to avoid this a friction will act in vertically downwards direction. Now in horizontal direction it will have tendency to move inwards due to centripetal acceleration, again to avoid this a friction will act in outwards direction. hence it should follow path 2. Please confirm whether my thinking is right or not sir.
Namste Sir ji
Happy Guru Purnima Sir ji
Wish you good health
Love you PG team ❤
yes sir ye HC verma sir ne bhi exp. kar ke reason pucha tha apne schoolarship test me this year only
Is the path 1 right because in relative motion we give the component of velocity of the table to the block in the opposite direction thus after that we will find the component of blocks velocity and the velocity given opposite to the block which comes as 1 option
Happy Guru Purnima guru ji 🙏🙏🙏❤️❤️
Thank you sir🙏
Path 2 is correct
Happy guru purnima sir
Happy Guru Purnima sir
Ashish Arora, got lessons from him in Bansal Kota ages back. Good to see fit & fine
Happy guru Purnima sir ji
Please Give your blessings sir🙏🙏
Happy guru Purnima sir 😊
I think path will be 2 , as when block is in disc it has both centripetal force and tangential force and when it will leave its centripetal acc and tangential acc will be zero but it have some velocity tangential and centripetal by this because of centripetal velocity it take curve and because of tangential velocity it move forwards.
This is my thoughts. If wrong please make me help in correction.
Sir mera pura basics hil gaya
Thank you sir we want more video like this
Love you 🖤
Sir as per my analysis 4th option must be correct because here we are analyzing 2 motions at a time,one is of block along line of friction(let's class that ab),we are assuming that pseudo force is just greater than friction,therefore along ab it ,must have some acceleration, now as it moves away from center of disc,the r of a point,on which it has now reached along ab will increase,since omega is same the velocity of that point must be greater than the previous point of the block,therefore it must move a greater distance along direction perpendicular to ab,hence in the subsequent motion the path would be similar to option 4
My assumptions are:-
Omega is constant after motion along ab starts
Sir I want to add on in last case
that when all qualities are given then i.e say r=5m
Then in which quadrant block will leave the disc
Happy guru purnima Sir❤
1. Sir pehle se thoda sa soch liya tha
2. Ans 3 hoga lgta hai (relative motion) after detach f=0
Sir I have done with all the physics galaxy volumes ...shall I buy pathfinder or any other book?
Happy Gurupornima Sir ✨️
Maine socha friction to cake walk but Aaj sir ne aanke khol di ❤❤❤
He opened the eyes for early rise , of mine who never rise early just for study 😅.
Thank you sir for your astonishing guide.
Happy guru Purnima 🙏
Happy Guru Purnima Guru Ji ....love from Bihar 💖🙏🙏🙏
Happy Guru Purnima, dear Sir.
Was unable to think about this situation. Reason- having weak basics in circular motion .Will go for case 2
I think .....it will follow the 4th path as for relative motion on stopping the box and taking its dirn opposite to that of the net resulatnt tangential accn and cetripetal accn we get the 4th path .....sir please correct me if i am wrong
Very true sir.
If it start sliding then in ground frame, path of body will be in 2nd direction..
at last, we have vel. of coin rel. to disc in ↙ direction, now if apply {vc/g=Vc/d+vd} and we know vd is ⬆ and add vectorially like (↙⬆) then vc would something lie like ↖,i.e option 2 & 3 came into consideration but as net fric. is more shifted to right side, so resultant is more close to mw^2r side, so it should follow 2nd path.
Also I am not able to visulaize the ques at first, thank u sir for this brain teaser
@Physics Galaxy Sir pls check my solution & pin it, if it is correct..it will also help others to imagine, if it is incorrect, pls corrct me
@physics Galaxy i think 3rd is correct did a experiment with plate and rubber
Ans:1
I think 4th path should be followed by block w.r.t observer on ground
I think 2nd path is right because particle will have sime tangential velocity and also the centripetal will balance friction after some time.
I think ki option no 2 should be correct because in ground frame as the body start moving then firstly it will try to move in tangential direction and friction will oppose it but as the body move forward bofy will opse thrle dics in opposite direction of motion so body experience a foce in direction of circular disc so it will follow path 2
The Friction can be tangentially if the initial velocity of block wrt ground is 0, right? Because at that instant tendency is tangential and later after some time centripetal component will also come?
B is right
Top view me disc clock wise hai ghoom rahi block
Slide bhi clockwise hi hoga
3 case would be correct at 9:28
video dekhne se pahle mai yahi soach raha tha ki tangentialy hi hoga . thank you sir for clearing this concept
4 might be the correct ans if rotation is clockwise and wrt observer on ground.......tried the same experiment at home
pathv4 is the answer because in ground frame it will also move in the tangential direction
Nice
Yes sir i was able to think that block won't slide tangentially and i think that wrt ground the block will appear to slide on path 2.
Sir path 3 because v disk vector will have greater magnitude than than v wrt disk and v wrt ground is vector sum of v wrt disk and velocity of disk
Path of sliding is along 4.
Guru Purnima ki bohot bohot hardik shubhkamnaye guruji ❤🙏🏻
Aap ke Gyan owr shiksha hum logo ko yse hi dete rahe❤🙏🏻
Hum apse, fundamental of physics shikhte hai..🙏🏻
Apko naman guruji ❤️🙏🏻...
Ashish sir is GOAT of physics ❤
Sir please make video on system of conducting shells and plates
Apposite to resultant vector
Sir at the time when tangential acc is developed ,, due to moment of inertia when disc rotates the mass moves backwards