A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed . If the stone makes 14 revo-lutions is 25 s, what is the magnitude and direction of acceleration of the stone ?

Explanation

Here r = 80 cm = 0.8 m v `= 14//25s^(-1)` .
`:. Omega = 2 pi v = 2 xx (22)/(7) xx (14)/(25) = (88)/(25) red . S^(-1)`.
The centripetal acceleration ,
` a = omega^(2)v r = ((88)/(25))^(2) xx 0 . 80 = 9. 90 m// s^(2)`
The direction centripetal acceleration is along the string directed towards the centre of circular path .