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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 - Physics

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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?

Numerical
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Solution

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

Number of revolutions = 14

Time taken = 25 s

Frequency v = `"Number of revolution"/"Time taken" = 14/25 Hz`

Angular frequency, ω = 2πν

`2xx22/7xx14/25 = 88/25 rads^(-1)`

Centripetal acceleration, `a_c = omega^2 r`

=`(88/25)^2 xx 0.8`

= `0.80 xx 88/25 xx 88/25`

= 9.90 `ms^(-2)`

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

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Chapter 3: Motion in a Plane - EXERCISE [Page 48]

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NCERT Physics [English] Class 11
Chapter 3 Motion in a Plane
EXERCISE | Q 3.14 | Page 48

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