According to that answer it's 2R
But I'm sure the answer is 4R bro
Btw how is the point of contact accelerated? Isn't it at rest?
If this is true then my doubt will be cleared
@Viraam_Rao @Ayan_Agrawal @arush_kumar_singh
If it wasn't in acceleration then it wouldn't lift up the next moment.Thats my guess.
Yes but I'm not able to prove that mathematically
it will be at rest
as the state of pure rolling says the point of contact is at rest
Then can you comment on why 2r is wrong?
It rotates about ICoR its instantaneous radius of curvature is 2 r but trajectory it traces may have different radius of curvature
i had the same dout for over a year
it havent cleared yet
Isn't V2/R instantaneous radius of curvature too?
got an idea just now
we should use the formula for radius of curvature
that is R=(1+(dy/dx)^2)^3/2 /d2y/dx2
as its accl is varying over it cant be caluclated using v2/r
What is dy/dx in this situation
it is the formula for radius of curvature for any cure at a point
some higher order maths
You’ll have to find the equation of trajectory of point A and then find dy/dx.
It’s simply gonna be too long but will for sure give the right answer. Always better to use ur v^2/R method
But again V2/r and icor are giving different answers why?
Only reason why Icor is wrong can be ic POC accelerated but can't prove that
You will get some idea from here
It's at rest at that instant.An object can be at rest and still have acceleration at that instant.
There we knew the trajectory equation
Here we dunno that
Btw how to derive the cycloid trajectory over here?
@Ayan_Agrawal @Viraam_Rao @arush_kumar_singh