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InterviewSolution
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A ring of radius `r` is suspended from a point on its circumference. Determine its angular frequency of small oscillations. |
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Answer» Correct Answer - B It is a physical pendulum, the time period of which is, `T = 2pi sqrt((I)/(mgl))` Here, `I` = moment of inertia of the ring about point of suspension ` = mr^(2) + mr^(2) = 2mr^(2)` and `l =` distance of point of suspension from center of gravity `= r` `:. T = 2pi sqrt ((2mr^(2))/(mgr))` `= 2pi sqrt ((2 r)/(g))` `:.` Angular frequency `omega = (2pi)/(T)` or `omega = sqrt ((g)/(2r))`. |
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