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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 and Direction of Acceleration of the Stone - CBSE (Science) Class 11 - 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?

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`

=9.91 `m/s^2`

The direction 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.

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Physics Textbook for Class 11 (2018 to Current)
Chapter 4: Motion in a Plane
Q: 17 | Page no. 87

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Solution 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 Concept: Uniform Circular Motion.
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