1since dq/dx=n ....q=nx
q^2=n^2 x^2
net force on body..................
k(n^2 x^2)/ x^2 - f
kn^2-f..........
i think instanteneously it will start pure rolling,as initial velocity is 0
(kn^2-f)/m=5f/2m(acm=ralpha)
f=2/7 k n^2
There are two neutral solid spheres of radius r with a seperation d
r< < d
They are supplied charges constantly such that dq/dx=n for 1st and dq/dt=-n for the 2nd sphere.
After what time will the spheres stop pure rolling?
assume masses to be M and coeff of friction to be k
What will be the distance traveled by the spheres during this time?
Edited teh question to make the diferential equation solvable :)
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6 Answers
106As r<<<d
F = (nt)24Πεx2
f = kMg
in limiting case of pure rolling ,limiting friction will act and the rolling will be of forward slipping as v will increase while increase in angular velocity will have a constant magnitude
F-f = Ma .. (i)
and fr = 2Mr2α5 .. (ii)
=> f = 2Ma/5
=> a = 5F/7M and flim = 2F/7
=> kMg = 27(nt)24Πεx2
Conserving energy (neglecting grav potential energy)
2*(12Mv2(7/5)) = (nt)24Πεx2
hmm thinking wat to do now
62asish.. be careful.. you cant conserve energy..
electric field is varying.. it is not constant...
11it will never stop pure rolling as the net force acting on com is constant kn^2-f..............if fmax <2/7 k n^2........it will never do pure rolling........