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?
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:
Length of the string, $I=80 \mathrm{~cm}=0.8 \mathrm{~m}$
Number of revolutions $=14$
Frequency, $v=\frac{\text { Number of revolutions }}{\text { Time taken }}=\frac{14}{25} \mathrm{~Hz}$
Angular frequency, $\omega=2 \pi \mathrm{V}$
$=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} \mathrm{rad} \mathrm{s}^{-1}$
Centripetal acceleration, $a_{\mathrm{c}}=\omega^{2} r$
$=\left(\frac{88}{25}\right)^{2} \times 0.8$
$=9.91 \mathrm{~m} / \mathrm{s}^{2}$
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.
Length of the string, $I=80 \mathrm{~cm}=0.8 \mathrm{~m}$
Number of revolutions $=14$
Frequency, $v=\frac{\text { Number of revolutions }}{\text { Time taken }}=\frac{14}{25} \mathrm{~Hz}$
Angular frequency, $\omega=2 \pi \mathrm{V}$
$=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} \mathrm{rad} \mathrm{s}^{-1}$
Centripetal acceleration, $a_{\mathrm{c}}=\omega^{2} r$
$=\left(\frac{88}{25}\right)^{2} \times 0.8$
$=9.91 \mathrm{~m} / \mathrm{s}^{2}$
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.