A stone at the end of a string is whirled in a vertical circle of radius r=1.20 m at a constant speed u=1.50m/s. the centre of the string is 1.50m above the ground. What is the range of the stone if it is released when the string is inclined at `30^(@)` with the horizotnal (a) at A(b) at B? What is the accelration of the stone: (c) just before rleaseat A (d) just after release at A?

Answer

(a) 600m; (b)0.402m; (c)1.87`m//s^(2)` towards centre; (d) 8070 `m//s^(2)` down

Explanation

`h=1.5+0.6=v sin 60^(@)t-(1)/(2)g t^(2)`
`R_(1)=v cos 60^(@)t`
Solving `R_(1)=600m`

(b) `h=1.5+0.6=-v sin 60^(@)t-(1)/(2)g t^(2)`
`R_(1)=v cos 60^(@)t`
Solving `R_(2)=0.402m`

(c) `a=(v^(2))/(R)=((1.5)^(2))/(1.2)=1.87m//s^(2)` towards centre
(d) `a=g=9.8m//s^(2)` down.