actually the framing of question is wrong, it should say "rod of negligible mass", then this would be correct. But if the rod is "massless" then the net force and torque on the rod is always zeo. Therefore both of hanging mass will exert 0 normal reaction and so they will move vertically with acceleration of g. Therefore mass on the corner will fall off from the rod instantly and the remaining mass will move with acceleration g and rod will in such a way that net force and torque on it is 0. A similar question is in pathfinder, according to which w of rod at time t is 4xgt/(g^2*t^4+4x^2), and angular acc which we can find on differentiating this, on put t=0 in that equation, we wil get at t=0, alpha = g/x, so to maximize its acceleration at time t=0, position of the mass x should be 0.
Yep. If the stick is massless then it can't exert any force tangentially or its ang. Accl approaches infinity. So minimising I of the system ( stick + ball), we get x=0
If ball remains stuck on stick, net torque is going to be 0 ,but this can't practically happen since Torque will exist .isn't that both balls will fall off?
@@Aayan_200 Yes in real life, a rod cannot be massless so even if small, a torque will exist. But in an imaginary scenario, for a massless rod, it will have such a movement which keeps its net torque zero.
Bs ek doubt tha sahi to mene kiya But agr hm alpha ki jagah simple downward acc. Ko maximize kre to x=l aarha hai Bs ye jaana hai alpha ko hi maximise kyu kre simple downward acc. Ko kyu nhi Silly doubt ho maaf krdena🙃
@@Farhan74575 Bhai yaha tu alpha isliye balance kr rha kyuki wo har point ke liye constant hoga Acceleration in the downward direction will also depend on r Tera ek point pe alg acc. ayega aur dusre pe alg usko maximize krne ka matlab nahi
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sahi nikaal liya tha first attempt me differentiate karke
took only 2 minutes
same
Good work.
Same
actually the framing of question is wrong, it should say "rod of negligible mass", then this would be correct.
But if the rod is "massless" then the net force and torque on the rod is always zeo.
Therefore both of hanging mass will exert 0 normal reaction and so they will move vertically with acceleration of g. Therefore mass on the corner will fall off from the rod instantly and the remaining mass will move with acceleration g and rod will in such a way that net force and torque on it is 0. A similar question is in pathfinder, according to which w of rod at time t is 4xgt/(g^2*t^4+4x^2), and angular acc which we can find on differentiating this, on put t=0 in that equation, we wil get at t=0, alpha = g/x, so to maximize its acceleration at time t=0, position of the mass x should be 0.
Yep. If the stick is massless then it can't exert any force tangentially or its ang. Accl approaches infinity.
So minimising I of the system ( stick + ball), we get x=0
If ball remains stuck on stick, net torque is going to be 0 ,but this can't practically happen since Torque will exist .isn't that both balls will fall off?
@@Aayan_200 Yes in real life, a rod cannot be massless so even if small, a torque will exist. But in an imaginary scenario, for a massless rod, it will have such a movement which keeps its net torque zero.
Arrived at the correct answer within 5 minutes, using the same approach! 👍
Thanks learnt something new
Bhaiya concept crunch of conics when ?
1.5 mins only . Very basic question if youve studied mechanics with iitain teacher
Bhaiya pls background black rakha karo plzz white me aakhe khichti che
Bs ek doubt tha sahi to mene kiya
But agr hm alpha ki jagah simple downward acc. Ko maximize kre to x=l aarha hai
Bs ye jaana hai alpha ko hi maximise kyu kre simple downward acc. Ko kyu nhi
Silly doubt ho maaf krdena🙃
Rod is pivot from one end, to fr acceleration ko express hi, r * alpha se kroge
@@jeesimplified-subject yahan pr r kya hoga?
Com ki distance from pivot?
If yes then wo bhi to x pr depend kregi
@@jeesimplified-subjectSir net Torque on a massless rod should be 0 . How is the Torque due to the masses being balanced here about pivot.
@@Farhan74575 Bhai yaha tu alpha isliye balance kr rha kyuki wo har point ke liye constant hoga
Acceleration in the downward direction will also depend on r
Tera ek point pe alg acc. ayega aur dusre pe alg usko maximize krne ka matlab nahi
@@Akuma-kami-koroshisame mene bhi soch liya tha
thanks bro for your explanation
Bhaiyaa i need personal guidance plss tell me how can i avail only personal mentorship
Similar que appeared in JEE mains I don't remember the Year but I was a rod
i did do a silly mistake... last sahi waale step mein 2x^2 aur x^2 ko cut karke x = L/2 kar diya... chomu waala kaam
At dist. L from hinged pt. Near the previous mass.😢 shit galat ho gaya
Sir can u send those quiz for every chapter
Thank u bhaiyya ❤❤
2L/3
took like 2 mins to get the ans
im slow in diff so it wouldve taken only 1.5min
good question
Same bro. It also took 2 minutes to get the answer.