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Question
Derive an expression for the difference in tensions at the highest and lowest point for a particle performing the vertical circular motion.
Solution
- Suppose a body of mass ‘m’ performs V.C.M on a circle of radius r as shown in the figure.
- Let,
TL = tension at the lowest point
TH = tension at the highest point
vL = velocity at the lowest point
vH = velocity at the highest point - At lowest point L,
TL = `("mv"_"L"^2)/"r" + "mg"` ….......(1)
At highest point H,
TH = `("mv"_"H"^2)/"r" - "mg"` ….......(2) - Subtracting (1) by (2),
TL - TH = `("mv"_"L"^2)/"r" + "mg" - (("mv"_"H"^2)/"r" - "mg")`
= `"m"/"r"("v"_"L"^2 - "v"_"H"^2) + 2"mg"`
∴ TL - TH = `"m"/"r"("v"_"L"^2 - "v"_"H"^2) + 2"mg"` ….......(3) - By law of conservation of energy,
(P.E + K.E) at L = (P.E + K.E) at H
∴ `0 + 1/2"mv"_"L"^2 = "mg".2"r" + 1/2"mv"_"H"^2`
∴ `1/2"m"("v"_"L"^2 - "v"_"H"^2) = "mg.2r"`
∴ `"v"_"L"^2 - "v"_"H"^2 = 4"gr"` .......(4) - From equation (3) and (4),
TL − TH = `"m"/"r"(4"gr") + 2"mg"` = 4mg + 2mg
∴ TL − TH = 6mg
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