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A Stone Tied to the End of a String 80 Cm Long is Whirled in a Horizontal Circle with a C Magnitude and Direction of Acceleration of the Stone - Physics

Numerical

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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Solution 1

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`

=`(88/22)^2 xx 0.8`

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

=9.90 `ms^(-2)`

The0direction of centripetal acceleration is always directed along the string, toward the centre, at all points.

Solution 2

Here r = 80 cm = 0.8 m;

v = 14/25 rev/s

`omega = 2pi v = 2xx22/7xx14/25 "rad/s" =88/25 "rad s"^(-1)`

The centripetal acceleration

`a = omega^2r = (88/25)^2 xx 0.80 = 9.90 ms^(-2)`

The direction of centripetal accleration is along the string directed towards the centre of cicrular path.

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APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 4 Motion in a Plane
Q 17 | Page 87
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