Let us consider that a particle of mass #m# is executing a vertical circular motion,centering the point #O# as shown in figure and having been attached to an in-extensible string of length #R#.
Let #P# represents an arbitrary position of the particle , where the string makes an angle #theta # with the vertical. If #v# be the velocity of the particle at this position then considering forces on the particle we can write
For position P
#(mv^2)/R=T-mgcostheta........[1]#
For position L (lowest bottom position where #theta=0#)
#(mv_"bot"^2)/R=T_"bot"-mgcos0^@#,
where #v_"bot"# is the velocity of the particle at L
#=>(mv_"bot"^2)/R=T_"bot"-mg...........[2]#
For position H (highest top position where #theta=180^@#)
#(mv_"top"^2)/R=T_"top"-mgcos180^@#,
where #v_"top"# is the velocity of the particle at H
#=>(mv_"top"^2)/R =T_"top"+mg...........[3]#
Considering conservation of energy at #L and H# we can write
#KE" at H"#
# = KE" at L" -"gain of PE due to lift of height 2R (L to H) "#
#1/2mv_"top"^2=1/2mv_"bot"^2-2mgR#
#=>v_"top"^2=v_"bot"^2-4gR.......[4]#
The particle will complete the circle , if the string doesn't slack at the highest point when #theta =180^@# There must be centripetal force to make this happen. The minimum centripetal force required can be had by putting #T_"top"~~0 # in equation [3].
#(m(v_"top"^2)_"min")/R =0+mg#
#=>(v_"top")_min=sqrt(gR).......[5]#
Inserting this value in equation [4] we can get the minimum required velocity of the particle at position #L#
#=>v_"top"^2=v_"bot"^2-4gR#
#=>(v_"top"^2)_ min=(v_"bot"^2)_"min"-4gR#
#=>(v_"bot"^2)_"min"=(sqrt(gR))^2+4gR#
#=>(v_"bot")_"min"=sqrt(5gR)#
