- Home
- Standard 11
- Physics
3-2.Motion in Plane
medium
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 ?
Option A
Option B
Option C
Option D
Solution
Length of the string, $l=80\, cm =0.8\, m$ Number of revolutions $=14$ Time taken $=25 \,s$
Frequency, $ v=\frac{\text { Number of revolutions }}{\text { Time taken }}=\frac{14}{25} Hz$
Angular frequency, $\omega=2 \pi v=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} rad s ^{-1}$
Centripetal acceleration, $ a_{\epsilon}=\omega^{2} r \quad=\left(\frac{88}{25}\right)^{2} \times 0.8$
$=9.91 \,m / s ^{2}$
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.
Frequency, $ v=\frac{\text { Number of revolutions }}{\text { Time taken }}=\frac{14}{25} Hz$
Angular frequency, $\omega=2 \pi v=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} rad s ^{-1}$
Centripetal acceleration, $ a_{\epsilon}=\omega^{2} r \quad=\left(\frac{88}{25}\right)^{2} \times 0.8$
$=9.91 \,m / s ^{2}$
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.
Standard 11
Physics
Similar Questions
hard
