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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

ConceptUniform Circular Motion

#### 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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