A body weighing `0.4kg` is whirled in a verticle circle making `2rps` If the radius of the circle is `1.2m` find the tension in the stringat (i) top of the circle (ii) bottom of the circle .

Answer

`T_(H) = 71.94 N, T_(L) = 79.78 N` .

Explanation

Here, `m = 0.4 kg , n = 2 rps, r = 1.2 m`
`T_(H) = ? T _(L) = ? `
From `T_(H) + m g = m r omega^(2) = 4 pi^(2) m r n^(2)`
`T_(H) = 4 pi^(2) m r n^(2) - m g`
` = 4 xx (22)/(7) xx (22)/(7) xx 0.4 xx 1.2 xx 4 xx 9.8`
`T_(H) = 75.86 - 3.92 = 71.94 N`
Again `T_(L) = 4 pi^(2) m r n^(2) + mg`
`= 75.86 + 3.92 = 79.78 N`.