Mechanics> momentum...

The potential energy of a particle in a certain field has the form U=a/r to the power 2- b/r,where a and b are '+'ve constants and r is distance from centre of the field.find r corresponding to equilibrium position of particle????????

Ankita Choudhury , 16 Years ago

Grade 12

1 Answers

AskIITians Expert Hari Shankar IITD

Hi,

At equilibrium, the particle will try to have minimum potential energy.

To find the minimum of any function f with respect to x, we need to put df/dx = 0.

Here, we need to find the minimum of the potential energy function U = a/r² - b/r.

So, we have to differentail U with respect to r,

and then put dU/dr = 0 to find out the position where U is minimum.

U = ar^-2 -br^-1

dU/dr = -2ar^-3 - (-br^-2) = (-1/r²)(2a/r - b)

Putting dU/dr = 0,

(-1/r²)(2a/r - b) = 0

or, 2 a/r - b = 0, because 1/r² = 0 will give r = infinity, which is anyway true

Hence, r = b/2a is the equilibrium distance

Last Activity: 16 Years ago

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Mechanics> momentum...

The potential energy of a particle in a certain field has the form U=a/r to the power 2- b/r,where a and b are '+'ve constants and r is distance from centre of the field.find r corresponding to equilibrium position of particle????????

Ankita Choudhury , 16 Years ago

AskIITians Expert Hari Shankar IITD

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