4.17 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?

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Vishal Baghel | Contributor-Level 10

5 months ago

4.17 Given, the length of the string, l = 80 cm, No. of revolution = 14, Time taken = 25 s

We know Frequency, v = $\frac{N o . o f r e v o l u t i o n}{t i m e t a k e n}$ = $\frac{14}{25}$ Hz

Angular frequency $ω$ = 2?v = 2 $\times$ $\frac{22}{7}$ $\times \frac{14}{25}$ rad/s = 3.52 rad/s

Centripetal acceleration ac = $ω^{2}$ r = ${3.52}^{2} \times \frac{80}{100}$ m/s2 = 9.91 m/s2

The direction of acceleration is towards the centre.

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4.17 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?

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