For Brain Teasers & resources for JEE & NEET preparation, join PG Telegram - t.me/physicsgalaxyworld Best online courses on JEE & NEET, download Physics Galaxy mobile app. Download link : bit.ly/PhysicsGalaxyApp Best Books for JEE Advanced Preparation - Bestseller book of 700+ advance illustrations. Check at - amzn.to/3qyIK8H Best book of chapterwise PYQs of JEE Advanced with detailed comprehensive solutions. Check at : amzn.to/3ZXotZs
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
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
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
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♥
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 ???
@@User-k2v6o as soon as the object leaves the rotating disc, its path is a perfect straight line but before that it has some angular displacement and thus its path is a bit curved when it is in contact with the table but after that it is a straight line
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.
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.
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
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.
@@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.
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 .
I think the path 2 will be correct As the block slides, r increases so does the force due to increase in omega and centrifugal force. Hence, the block will accelerate in the direction of resultant force of the above mentioned forces untill it leaves the disc.... And when it leaves the disc it will stop accelerating and will move with a constant velocity ( not considering air friction for simplicity ) I am still not sure why the path appears to be curved as when block leaves the disc it should not be accelerating but the path says otherwise as the vector for position of block keeps changing Help would be greatly appreciated
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
According to me block will move in the direction of path 3, kyuki jo friction ka kaam hota hai na wo ye hota hai ki jo relative velocity hoti hai na dono surfaces ke bich me use minimise kar sake ya ese bol sakte hai ki friction ki tendency dono surface ko sath chalane ki hoti hai but uski ek limit hoti hai or us limit se jyada wo kaam nhi kar sakta. Ab dekha jae toh alpha ki wajhe se A centripetal and A tangential badh jayega or ek time wo hoga jab wo 'Fs' max se jyada ho jayega toh ab jo centrifugal acc. hoga wo us friction se jyada ho jayega jo ki cetre ki taraf lag rha hai or same hi tangential ke case me hoga . Ab jo centrifugal acc. Hai na wo block ko bhar ki side leke jayega or jo tangential ki taraf jo friction hoga wo bhi lagta rahega tangential acc. Ki side lekin ab tangential friction ab relative motion ko rok nhi payega ab jo centrifugal acc. or 'f' tangential hoga unke net force ki wajah se block 3 path pe jayega 👍
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.
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
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 ✨👍.
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.
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.
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 .
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
Since the disc is moving in clockwise direction and by looking the centrifugal acceleration and tangential acceleration,according to me path 2 is correct. This is an amazing concept sir Understand this concept for the first time😊 and Happy Guru Purnima sir. Please bless me😊😊
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..!!
Sir ye to simple hcv ke questions ka thoda modification tha jesa mereko laga Ham agar disc ke frame mein chale jae to humko block kisi angle pe fisalta hua lagega normally nahi to grnf frame mein sidha khisakta hua lagega 9:18 sir mereko lagta hai ki 3 path mein jaega feel krke bola sir shi he kya 😅?
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. 🙏🙏
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
path 3 block should follow because mw^r now acting inward direction the componet friction opposite but equal magnitude shown in diagram so certainly block can follow path 3 even if considering corollie forces or not
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
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
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.
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.
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
Since the friction force isn't sufficient to keep the circular motion of the object so it should be accelerated in the opposite direction of force of friction, which is option 4
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
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?
I originally marked the answer "False" and thought what you just explained. But now, I'm confused about one possibility- If the disc originally has Omega=0 and then it starts rotating, i.e. we increase its angular speed, at the beginning of the motion, omega =0, but omega is increasing, so alpha = something. Since Omega=0, there is no friction component acting on the block along the radial direction, but since Alpha= something, there will be a tangential acceleration which must be provided by the frictional force. So, in this case, if Alpha is large enough so that the friction equals (mu)mg, the block will start sliding in the direction opposite to the direction of friction. Concludingly, we can say that friction is acting only in the tangential direction, so the block should slide back tangentially. s.n.: All of this is the description of just the instant after the commencement of the rotation of disk. Please Enlighten.
@@Anurag_jaiswaL the question doesn't mention about the whole of motion. it just asks whther the block may slide along the tangent, no matter how long the time may be. it DOES slide tangentially if this case maybe considered. u r aabsoluuuutely correct- friction will start acting radially. but what i'm tryna highlight is JUST the instant.
very good and conceptual video but ye sliding tendency wala samjh nahi aaya? 🤔vo bahar ki taraf centripetal force lagta hai aur ander ki taraf centrifugal. tangential force bhi samajh gaya but vo 2 aur 3rd path karte hue jata hai ye kaise hua?
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 ❤️🙏🏻...
I think path 2 is correct because due to centrifugal force the block will start moving outwards but still will be moving tangential with the disk am i correct
1, bcz if the object overcame the friction, it will slide opposite to the direction of angular vel, also the path will be less curved bcz of radial friction
I think it should be moving in path 2 bcz..the component of friction in horizontal is increasing as it depends on omega which is increasing..and the tangential acceleration is constant throughout ..which implies it will have constant velocity in vertical direction but increasing in horizontal dirn...so it should be following path 2..
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
9:33 I tried to determine this practically using a utensil and a small toothpick box. The result seems to be 2nd situation. Now I can try explaining, due to the movement along the circular path the, sliding tendency is always away from the circular path, as the block moves away from the path, the radius r increases, increasing the accelerations on the body and it slides faster leading to a curved in throw out. (I am not sure this is a complete explanation, someone help me out with what I am faulty at)
Its gonna be 2, I found by imitating the situation. Take your right hand put it on your right cheek such your fingers are pointing parallel to your nose (just keep them in plane of line of symmetry of your face). Twist your torso right hand side while extending them fully forward simultaneously. The path traced by your hand is the path of the object[ approximately of course], that is 2nd ;)
For Brain Teasers & resources for JEE & NEET preparation, join PG Telegram - t.me/physicsgalaxyworld
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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
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
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
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♥
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
@@User-k2v6o as soon as the object leaves the rotating disc, its path is a perfect straight line but before that it has some angular displacement and thus its path is a bit curved when it is in contact with the table but after that it is a straight line
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.
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.
That's right
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
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.
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 .
I think the path 2 will be correct
As the block slides, r increases so does the force due to increase in omega and centrifugal force. Hence, the block will accelerate in the direction of resultant force of the above mentioned forces untill it leaves the disc.... And when it leaves the disc it will stop accelerating and will move with a constant velocity ( not considering air friction for simplicity )
I am still not sure why the path appears to be curved as when block leaves the disc it should not be accelerating but the path says otherwise as the vector for position of block keeps changing
Help would be greatly appreciated
My answer is also same
no bro 4 is possible
3 should be answer
@@royalX555 no bro 2 is correct
just annalyse it with taking concept of centripetal force and v=rw velocity at that instant
yes bro doubt is correct why it will be curved path bcauz after leaving disc there will be no force on the particle
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
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
Anwer may be path 3rd as we have to calculate w.r.t ground.
We use the concept of relative
.... Motion
yes sir ye HC verma sir ne bhi exp. kar ke reason pucha tha apne schoolarship test me this year only
According to me block will move in the direction of path 3, kyuki jo friction ka kaam hota hai na wo ye hota hai ki jo relative velocity hoti hai na dono surfaces ke bich me use minimise kar sake ya ese bol sakte hai ki friction ki tendency dono surface ko sath chalane ki hoti hai but uski ek limit hoti hai or us limit se jyada wo kaam nhi kar sakta. Ab dekha jae toh alpha ki wajhe se A centripetal and A tangential badh jayega or ek time wo hoga jab wo 'Fs' max se jyada ho jayega toh ab jo centrifugal acc. hoga wo us friction se jyada ho jayega jo ki cetre ki taraf lag rha hai or same hi tangential ke case me hoga . Ab jo centrifugal acc. Hai na wo block ko bhar ki side leke jayega or jo tangential ki taraf jo friction hoga wo bhi lagta rahega tangential acc. Ki side lekin ab tangential friction ab relative motion ko rok nhi payega ab jo centrifugal acc. or 'f' tangential hoga unke net force ki wajah se block 3 path pe jayega 👍
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.
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
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 ✨👍.
@@shashvatpandey3028 yyess
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.
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.
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 .
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❤
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
Since the disc is moving in clockwise direction and by looking the centrifugal acceleration and tangential acceleration,according to me path 2 is correct.
This is an amazing concept sir Understand this concept for the first time😊
and Happy Guru Purnima sir. Please bless me😊😊
I think so too
How can you consider centrifugal force if you want to see the trajectory of block relative to ground
@@ADJEE2024 Every force must be considered in my opinion. It's just that the reference frame for its displacement is different.
3 case would be correct at 9:28
It helped me a lot sir.
(JEE-2024 Aspirant)
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..!!
Sir ye to simple hcv ke questions ka thoda modification tha jesa mereko laga
Ham agar disc ke frame mein chale jae to humko block kisi angle pe fisalta hua lagega normally nahi to grnf frame mein sidha khisakta hua lagega
9:18 sir mereko lagta hai ki 3 path mein jaega feel krke bola sir shi he kya 😅?
Thank you sir ... Happy Guru Purnima 🙏 with your help I got into IIT KANPUR
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.
By solving we get x=e^-t and Y=e^-2t considering origin at the moment were v=rw which implies y=x^2 path 2 is correct
First thing that I learnt is that,
Friction do not oppose relative motion, it opposes sliding tendency, thus always in opp drxn of relative sliding
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. 🙏🙏
1. Sir pehle se thoda sa soch liya tha
2. Ans 3 hoga lgta hai (relative motion) after detach f=0
thank you sir great thing to learn 🙂🙂
I think 4th path should be followed by block w.r.t observer on ground
Happy Guru Purnima sir 🙏🏼❤️😌
Thank you sir for clearing the. Concept ❤❤
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
Sir answer is 4. Please confirm.
path 3 block should follow because mw^r now acting inward direction the componet friction opposite but equal magnitude shown in diagram so certainly block can follow path 3 even if considering corollie forces or not
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
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 am very thankfull for your this kind of content. Please continue eye opener series like this vedio and wake up series.
B is right
Top view me disc clock wise hai ghoom rahi block
Slide bhi clockwise hi hoga
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 ??
abhi coaching me friction start bhi nahi hua hai, lekin tab bhi sab samajh aa gaya... wow sir!
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
Sir because definition of fric. tells us that fric. can only act opposite to vel. here fric is non tangential to path so velocity wouldn't be also.
Ashish Arora, got lessons from him in Bansal Kota ages back. Good to see fit & fine
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.
5:24 sir @t (tangential acceleration) ke opposite friction hoga na? Why in the same direction?
Maine socha friction to cake walk but Aaj sir ne aanke khol di ❤❤❤
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.
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
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
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 🙏
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
Since the friction force isn't sufficient to keep the circular motion of the object so it should be accelerated in the opposite direction of force of friction, which is option 4
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
If it start sliding then in ground frame, path of body will be in 2nd direction..
Nice concept ... Love You sir ❤
How can we show that centrifugal force is a pseudo force?
Fs =mv2r ,fs-mv2r=0
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?
I directly imagined the trajectory without thinking of lifting my pen
Happy Guru purnima sir ji, u have been the beacon of knowledge... will be forever in debt . 🙏🙏🙏.
I originally marked the answer "False" and thought what you just explained.
But now, I'm confused about one possibility-
If the disc originally has Omega=0 and then it starts rotating, i.e. we increase its angular speed, at the beginning of the motion, omega =0, but omega is increasing, so alpha = something.
Since Omega=0, there is no friction component acting on the block along the radial direction, but since Alpha= something, there will be a tangential acceleration which must be provided by the frictional force.
So, in this case, if Alpha is large enough so that the friction equals (mu)mg, the block will start sliding in the direction opposite to the direction of friction. Concludingly, we can say that friction is acting only in the tangential direction, so the block should slide back tangentially.
s.n.: All of this is the description of just the instant after the commencement of the rotation of disk.
Please Enlighten.
i think the moment disk gets omega greater than 0 friction will start acting in radial direction and it will effect the motion of block.
@@Anurag_jaiswaL the question doesn't mention about the whole of motion. it just asks whther the block may slide along the tangent, no matter how long the time may be. it DOES slide tangentially if this case maybe considered.
u r aabsoluuuutely correct- friction will start acting radially.
but what i'm tryna highlight is JUST the instant.
Nice elaboration sirji❤
What a mind-blowing concept sir.
very good and conceptual video but ye sliding tendency wala samjh nahi aaya? 🤔vo bahar ki taraf centripetal force lagta hai aur ander ki taraf centrifugal. tangential force bhi samajh gaya but vo 2 aur 3rd path karte hue jata hai ye kaise hua?
I think the answer should be no.2 as we are observing from non inertial frame block will have a pseudo force in the direction of 2nd force
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 ❤️🙏🏻...
I think path 2 is correct because due to centrifugal force the block will start moving outwards but still will be moving tangential with the disk am i correct
Happy Guru Purnima Guru Ji ....love from Bihar 💖🙏🙏🙏
Sir this series is very important for us😊😊😊
1, bcz if the object overcame the friction, it will slide opposite to the direction of angular vel, also the path will be less curved bcz of radial friction
I think it should be moving in path 2 bcz..the component of friction in horizontal is increasing as it depends on omega which is increasing..and the tangential acceleration is constant throughout ..which implies it will have constant velocity in vertical direction but increasing in horizontal dirn...so it should be following path 2..
Thank you sir for such a nice eye opener concept
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
9:33 I tried to determine this practically using a utensil and a small toothpick box. The result seems to be 2nd situation. Now I can try explaining, due to the movement along the circular path the, sliding tendency is always away from the circular path, as the block moves away from the path, the radius r increases, increasing the accelerations on the body and it slides faster leading to a curved in throw out. (I am not sure this is a complete explanation, someone help me out with what I am faulty at)
Yeah I tried too and also getting path 2 but I am not able to explain/understand with fair degree of clarity
Its gonna be 2, I found by imitating the situation. Take your right hand put it on your right cheek such your fingers are pointing parallel to your nose (just keep them in plane of line of symmetry of your face).
Twist your torso right hand side while extending them fully forward simultaneously. The path traced by your hand is the path of the object[ approximately of course], that is 2nd ;)
Happy guru Purnima sir ji
Please Give your blessings sir🙏🙏
I think 2nd path is right because particle will have sime tangential velocity and also the centripetal will balance friction after some time.
video dekhne se pahle mai yahi soach raha tha ki tangentialy hi hoga . thank you sir for clearing this concept
Path of sliding is along 4.
Thank you sir
Sir according to me it may slide in southwest but in some parabolic path
Sir Iam able to understand properly thanks sir ..❤️
pathv4 is the answer because in ground frame it will also move in the tangential direction
Sir can we used pseudo force concept in sliding tendency if we usef pseudo force then the direction of sliding tendency will directly come correct
HW Question - 2 will be the Answer.. 10:17
Yes sir if I had the question then I would have chosen that the body moves out radially but now I get it why I will not go straight
Please bless me sir for jee main 2024 !! 🎉🎉 HAPPY GURU PURNIMA🎉
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 mera pura basics hil gaya
Thank you sir we want more video like this
Love you 🖤
Nice concept, I already knew it though.
Can someone explain how the block would go wrt to ground frame??
Sir I have done with all the physics galaxy volumes ...shall I buy pathfinder or any other book?