A stone tied to the end of a string 80 c | Class 11 Physics Chapter Motion in a Plane, Motion in a Plane NCERT Solutions

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?

Answer:

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

Number of revolutions = 14

Time taken = 25 s

Frequency, v  =  Number of revolutions / time taken = 14 / 25 Hz

Angular frequency, ω = 2πν

= 2 x 22/7 x 14/25 = 88 / 25 rad s-1

Centripetal acceleration, ac  =  ω2 r

= (88 / 25) 2  x 0.8

= 9.91 m/s2

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


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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 17: A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant spe....