Question
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 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
Solution 1
Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s
Frequency v = `"Number of revolution"/"Time taken" 14/25 Hz`
Angular frequency, ω = 2πν
`2xx22/7xx14/25 = 88/25 rads^(-1)`
Centripetal acceleration, `a_c = omega^2 r`
=`(88/22)^2 xx 0.8`
=9.91 `m/s^2`
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.
Solution 2
Here r = 80 cm = 0.8 m;
v = 14/25 rev/s
`omega = 2pi v = 2xx22/7xx14/25 "rad/s" =88/25 "rad s"^(-1)`
The centripetal acceleration
`a = omega^2r = (88/25)^2 xx 0.80 = 9.90 ms^(-2)`
The direction of centripetal accleration is along the string directed towards the centre of cicrular path.
