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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 - CBSE (Science) Class 11 - Physics

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 Physics Textbook for Class 11 Part 1 (with solutions)
Chapter 4: Motion in a Plane
Q: 17 | Page no. 87

Video TutorialsVIEW ALL [1]

Reference Material

Solution for 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 concept: Uniform Circular Motion. For the courses CBSE (Science), CBSE (Arts), CBSE (Commerce)
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