A sphere of mass 200 g is attached to an inextensible string of length 130 cm whose upper end is fixed to the ceilling . The sphere is made to describe a horizontal circle of radius 50 cm Calculate the periodic time of this conical pendulum and the tension in the string .

Explanation

Here , ` m = 200 g , OB = l = 130 cm , `
` r = AB = 50 cm , t = ? T = ? `
` O A = sqrt(l^(2) - r^(2) )= sqrt((130)^(2) - (50)^(2))= 120 cm `
As is clear from
` T cos theta = mg `
` T sin theta = (m upsilon^(2))/(r) = m r omega^(2) `
Dividing , we get `tan theta= (m r omega^(2))/(mg)=(r omega^(2))/(g)`
`omega = sqrt(g tan theta )/(r) = (2 pi)/(r)`
` t = 2 pi sqrt((r)/(g tan theta)) = 2 xx (22)/(7) sqrt((0.5)/9.8 xx 5 //12) = 2.19 s`
From (i) ` T = (mg)/(cos theta) = (200)/(1000) xx (9.8)/(12 // 13)= 2.12 N `
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