A stone of mass `100g` is suspended from the end of a weightless string of length `100cm` and is allowed to swing in a verticle plane The speed of the mass is `200cm^(-1)` when the string makes an angle of `60^(@)` Also calculate the speed of the stone when it is in the lowest position Given `g = 980cm s^(-2)` .

Answer

`8.9 xx10^(4) dyne, 3.7 ms^(-1)` .

Explanation

Here, `m = 100g, r = 100 cm, upsilon = 200 cm//s`
`theta =60^(@)`
`T = (m upsilon^(2))/(r) + mg cos theta`
`=(10^(2) xx(200)^(2))/(100) + 100 xx 980 cos 60^(@)`
`=8.9 xx10^(4) "dyne"`
At `theta = 60^(@),` height above the lowest point
`h = r - r cos 60^(@) = r - (r)/(2) = (r)/(2)`
As total energy at L= total energy at `60^(@)`
`:. (1)/(2) m upsilon_(L)^(2) =(1)/(2) m upsilon^(2) + mg (r //2)`
`upsilon_(L)= sqrt(upsilon^(2) + g r) = sqrt(2^(2) + 9.8 xx 1 ) = 3.7 m//s` .