InterviewSolution
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InterviewSolution
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A disc of radius `R` has a light pole fixed perpendicular to the disc at its periphery which in turn has a pendulum of length `R` attached to its other end as shown in figure. The disc is rotated with a constant angular velocity `omega` The string is making an angle `45^(@)` with the rod. Then the angular velocity `omega` of disc is A. `((sqrt3g)/(R))^(1//2)`B. `((sqrt3g)/(2R))^(1//2)`C. `((g)/(sqrt3R))^(1//2)`D. `((sqrt2 g)/((sqrt2 + 1)R))^(1//2)` |
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Answer» Correct Answer - D The bob of the pendulum moves in a circle of radius `(R + R "sin" 45^(@)) = ((sqrt2 + 1)/(sqrt2))R` T sin `45^(@) m ((sqrt2 + 1)/(sqrt2)) R omega^(2)` T cos `45^(@) = "mg" " " omega = sqrt((sqrt2 g)/((sqrt2 + 1) R))` |
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