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Derive an expression for the difference in tensions at the highest and lowest points for a particle performing vertical circular motion.

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Consider a small body ( or particle) of mass m tied to a string and revolved in a vertical circle of radius r at a place where the accerleration due to graviy is g. At every instant of its motion, the body is acted upon by two forces, namely, its weight `vec(mg)` and the tension `vecT` in the string.
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let `v_(2)` be the speed of body and `T_(2)` be the tension in the string at the lowest point B . We take the reference level for zero potential energy to be the bottom of the circle. THen, the body has only kinetic energy `(1)/(2)mv_(2)^(2)` at the lowest point.
`therefore T_(2) = (mv_(2)^(2))/(r) + mg " "...(1)`
and the total energy at the bottom ` = KE + PE = (1)/(2) mv_(2)^(2) + 0`
` = (1)/(2) mv_(2)^(2)" "...(2)`
Let `v_(1)` be the speed and `T_(1)` the tension in the string at the highest point A. As the body goes from B to A, it rises through a height h = 2r.
`therefore T_(1)=(mv_(1)^(2))/(r) -mg" "...(3)`
and the total energy at A = KE + PE
` = (1)/(2) mv_(1)^(2) + mg(2r) " "...(4)`
Thus, from Eqs. (1) and (3),
`T_(2) -T_(1)=(mv_(2)^(2))/(r)+mg-((mv_(1)^(2))/(r)-mg)`
` = (m)/(r)(v_(2)^(2)-v_(1)^(2))+2mg" "...(5)`
Assuming that the total energy of the body is conserved, the totl energy at the bottom = total energy at the top
Then, from Eqs. (2) and (4),
` (1)/(2)mv_(2)^(2)=(1)/(2)mv_(1)^(2)+mg(2r)`
`thereforev_(2)^(2)-v_(1)^(2)=4gr" "...(6)`
Substituting this in Eq. (5),
`T_(2)-T_(1)=(m)/(r)(4gr)+2mg`
` =4mg+2mg`
= 6 mg
Therefore, the difference in the tensions in the string at the highest and the lowest points in 6 times the weight of the body.

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