66Could you make the question a bit more clear. I mean how exactly would you vary the angle A? How are these circles kept and so on..
The mass density per unit area of each circle is " σ " , and all the surfaces are frictionless . It is also given that the radius of each circle is " R " . Find the force required to keep the two bottommost circles together as a function of the angle " A " . For what angles is this force maximum or minimum ?
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66
1Sir , the radius of the topmost circle is not given , so I guess the radius is varied to make the angle " A " vary . The bottommost balls are kept on a frictionless surface , and friction is also absent at contact points . The force applied is horizontal .
49r+RsinA=2Rsin(Ï€-2A)
gives, r=R(1-cosAcosA)
2NsinA=σ.πr2.g
N=πσg2sinA.r2
NcosA=Fmin = πσg2sinA.r2.cosA
\small thus,\;F_{min}\;=\frac{\pi \sigma gcosA}{2sinA}.R^2\left(\frac{1-cosA}{cosA}\right)^2 = \frac{\pi \sigma gR^2\left(1-cosA \right)^2}{sin2A}
What is the answer given?
1It's alright , but the factor of two won't be there in the numerator . Thanks .
66Well I was misguided by your statement when you said that each circle have a radius R
1@ vivek ... from the sides (horizontally )
like imgagine drums one over the other