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प्रश्न
Using energy conservation, along a vertical circular motion controlled by gravity, prove that the difference between the extreme tensions (or normal forces) depends only upon the weight of the objects.
उत्तर
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 acceleration due to gravity is g. At every instant of its motion, the body is acted upon by two forces, namely, its weight `vec"mg"` and the tension `vec"T"` in the string.
Let `"v"_2` be the speed of the 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.
∴ `"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 v1 be the speed and T1 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.
∴ `"T"_1 = "mv"_1^2/"r" - "mg"` .....(3)
and the total energy at A = KE + PE
`= 1/2 "mv"_1^2 + "mg"(2r)` ...(4)
Then, 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) + 2 "mg"` ...(5)
Assuming that the total energy of the body is conserved, the total 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)`
∴ `"v"_2^2 - "v"_1^2 = 4 "gr"` ...(6)
Substituting this in Eq. (5),
`"T"_2 - "T"_1 = "m"/"r" (4 "gr") + 2 "mg"`
`= 4 "mg" + 2 "mg"`
= 6 mg
Therefore, the difference in the tensions in the string at the highest and the lowest points is 6 times the weight of the body.
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