InterviewSolution
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InterviewSolution
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The potential energy of a particle in a certain field has the form `U=a//r^2-b//r`, where a and b are positive constants, r is the distance from the centre of the field. Find: (a) the value of `r_0` corresponding to the equilibrium position of the particle, examine where this position is steady, (b) the maximum magnitude of the attraction force, draw the plots `U(r)` and `F_r(r)` (the projections of the force on the radius vector r). |
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Answer» From the equation `F_r=-(dU)/(dr)` we get `F_r=[-(2a)/(r^3)+(b)/(r^2)]` (a) we have at `r=r_0`, the particle is in equilibrium position. i.e. `F_r=0` so, `r_0=(2a)/(b)` To check, whether the position is steady (the position of stable equilibrium), we have to satisfy `(d^2U)/(dr^2)gt0` We have `(d^2U)/(dr^2)=[(6a)/(r^4)-(2b)/(r^3)]` Putting the value or `r=r_0=(2a)/(b)`, we get `(d^2U)/(dr^2)=(b^4)/(8a^3)`, (as a and b are positive constant) So, `(d^2U)/(dr^2)=(b^2)/(8a^3)gt0`, which indicates that the potential energy of the system is minimum, hence this position is steady: (b) We have `F_r=-(dU)/(dr)=[-(2a)/(r^3)+(b)/(r^2)]` For F, to be maximum, `(dF_r)/(dr)=0` So, `r=(3a)/(b)` and then `F_(r(max))=(-b^3)/(27a^2)`, As `F_r` is negative, the force is attractive. |
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