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Home » 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 direction and magnitude of the 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 direction and magnitude of the acceleration of the stone?
CBSE | Class 11 | Motion in a Plane | NCERT | Physics
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 direction and magnitude of the acceleration of the stone?

Answer –

According to the question, the length of the string, l is 80 cm, or 0.8 m

and the number of revolutions is 14

And the time taken = 25 s

Expression for the frequency is given as follows –

Frequency (v) = Number of revolutions / Time taken

v = 14 / 25 Hz

Angular frequency is given as follows –

\omega =2\pi \nu

\omega =2\times \frac{22}{7}\times \frac{14}{25}

\omega =\frac{88}{25}rad/s

Expression for centripetal acceleration is as follows :

{{a}_{c}}={{\omega }^{2}}r={{\left( \frac{88}{25} \right)}^{2}}\times 0.8

{{a}_{c}}=9.91m{{s}^{-2}}

At all times, the direction of centripetal acceleration is always along the string, toward the centre.

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