A circular ring of mass M and radius R is placed in YZ plane with centre at origin. A particle of mass m is released from rest at a point x = 2R. Find the speed with which it will pass the centre of ring .

Explanation

As shown in figure, first we find potential at A due to the ring, it is given as
`V_(A) = -(GM)/(sqrt(R^(2)+(2R)^(2))) = - (GM)/(sqrt(5R))`
Now potential at origin O due to ring is
`V_(0) = -(GM)/(R )`
When m moves from A to O, work done on it due to gravitational force is
`W = m(V_(A)-V_(0))=m(-(GM)/(sqrt(5)R)+(GM)/(R ))`
`= (GMm)/(R )((sqrt(5)-1)/(sqrt(5)))`
This work done gravitational force on m must be equal to the increase in kinetic energy of the mass m, thus we have
`(1)/(2) mv^(2) = ((sqrt(5)-1)/(sqrt(5)))(GMm)/(R )`
or `v = [(2(sqrt(5)-1)GM)/(sqrt(5)R)]^(1//2)`
This problem can also be solved simply by using energy conservation. Thus initially when m was at rest at point `A_(2)` the total energy of system is only gravitational potential energy given as
`E_(i) = m. V_(A)= - (GMm)/(sqrt(5)R)`
Finally when m pass through O, the total energy of system is
`E_(f) = (1)/(2) mv^(2)+mV_(0)`
`= (1)/(2)mv^(2)-(GMm)/(R )`
As no external work is done on the system in this case, the total energy of system must be conserved, thus according to energy conservation we have
`E_(i) = E_(f)`
`-(GMm)/(sqrt(5)R) = (1)/(2) mv^(2) - (GMm)/(R )`
`v = [(2(sqrt(5)-1)GM)/(sqrt(5)R)]^(1//2)`