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 and r is the distance from the centre of the field. Find the value of `r_0` corresponding to the equilibrium position of the particle, examine whether this position is stable. |
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Answer» `U(r)=a/r^2-b/r` Force `=-(dU)/(dr)=-((-2a)/(r^3)+b/r^2)=-((br-2a))/(r^3)` At equilibrium, `F=-(dU)/(dr)=0` Hence, `br-2a=0`, at equilibrium. `r=r_0=(2a)/(b)` corresponds to equilibrium. At stable equilibrium, the potential energy is minimum and at unstable equilibrium, it is maximum. For minimum potential energy, `du//dr=0` and `d^2U//dr^2gt0` at `r=r_0`. Let us investigate the second derivative. `(d^2U)/(dr^2)=(d)/(dr)((dU)/(dr))=(d)/(dr)((-2a)/(r^3)+(b)/(r^2))=(6a)/(r^4)-(2b)/(r^3)` At `r=r_0=(2a)/(b)implies(d^2U)/(dr^2)=(6a-2br_0)/(r_0^4)=(2a)/(r_0^4)gt0` Hence, the potential energy function `U(r)` has a minimum value at `r_0=2a//b`. The system has a stable equilibrium at minimum potential energy state. |
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