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?

Asked by Abhisek | 1 year ago |  187

1 Answer

Solution :-

Length of the string, l = 80 cm = 0.8 m

No. of revolutions = 14

Time taken = 25 s

Frequency, v = \(\dfrac{No. \; of \; revolution}{Time \; taken} = \dfrac{14}{25} Hz\)

Angular frequency \(\omega\),

=\(2\pi v\)

=\(2 \times \dfrac{22}{7} \times \dfrac{14}{25}\)

\(\dfrac{88}{25} \; rad \; s^{-1}\)

Centripetal acceleration:

\(a_{c} = \omega ^{2}r= \left (\frac{88}{25} \right )^{2} \times 0.8= 9.91 ms^{-2}\)

The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.

Answered by Pragya Singh | 1 year ago

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