A sphere of mass`0.1kg` is attached to an inextensible string of length `1.3m` whose upper end is fixed to the ceilling The sphere is made to describe a horizontal circle of radius `0.5m` Calculate time period of revolution and tension in the string .

Answer

`2.2 s ; 1.06 N` .

Explanation

Here, `m = 0.1kg, OP = l = 1.3m, CP = r = 0.5m`
time period `t =?` tension `T =?`
`sin theta = (r)/(t) =( 0.5)/(1.3) = 0.3846`
`theta = sin^(-1) (0.3846) = 22.62^(@)`
As is clear from
`T cos theta = mg`
`T = (mg)/(cos theta) = (0.1 xx 9.8)/(cos 22.62^(@)) = (0.98)/(0.9231) = 1.06 N`
Also `T sin theta = mr omega^(2) = (mr 4pi^(2))/(t^(2))`
`:. t^(2) = (4pi^(2)mr)/(T sin theta)`
`= 4 xx (22)/(7) xx (22)/(7) xx (0.1 xx 0.5)/(1.06 xx 0.3846) = 4.846`
`t = sqrt(4.846) = 2.2s`
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