Obtain expression of energy of a particle at dfferent positions in the vertical circular motion.

Explanation

Let , in diagrm, `A to ` Top position, ` B to` Bottom position, `Cto` Midway position, `V(1) to` Critical velocity or minimum linear velocity at, A, `V_(2) to` Maximum velocity at, B `V_(3) to ` Velocity at midway position, C, `T_(1) to` Tension in the string when object is at A, `T_(2)to` Tenstion in the string when object is at B, mg `to` Weight of object tied at one end of massless inextensible string.

Linear velocity of object at highest point A:
`T_(1)+mg=(mv_(1)^(2))/(r)`
Let `T_(1)=0`
`:. mg=(mv_(1)^(2))/(r)`
`v_(2)=sqrt(rg)`
Linear velocity of object at lowest point B :`" "....(1)`
According to law of conservation of energy,
Total enegy at A= Total energy at B
`Delta PE_(A)+DeltaKE_(A) =DeltaPE_(B)+DeltaKE_(B)`
`mg(2r)+(1)/(2)mv_(1)^(2)=mg(0)+(1)/(2)mv_(2)^(2)`
`2mgr+(1)/(2)m(sqrt(rg))^(2)=0+(1)/(2)mv_(2)^(2)" "` [ from equation (i)]
`4gr+rg=v_(2)^(2)`
`5gr=v_(2)^(2)`
`v_(2)=sqrt(5rg)`
Linear Velocity at midway point C :
Again, accroding to law of conservation of energy
Total energy at A= total energy at C
`Delta PE_(A)+Delta KE_(A)+Delta PE_(C)+DeltaKE_(C)`
`2mgr+(1)/(2)mv_(1)^(2)=mg(r)+(1)/(2)mv_(3)^(2)`
`1mgr+(1)/(2)m(sqrt(rg))^(2)+mg=r+(1)/(2)mv_(3)^(2)`
`4rg+rg=2gr+v_(3)^(2)`
`v_(3)^(2)=3rg`
`v_(3)=sqrt(3rg)`