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?
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....
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