A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of accele…

Question

A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?

Accepted Answer

Correct answer: A. (a, b) a=r ω^2; ω=2 π v 22 revolution =4 4 sec. 1 revolution =4 4/2 2=2 sec. v=1/2 Hz a=r ω^2=1× 4 π^24=π^2m/s^2 Towards the centre, the centripetal acceleration =−ω^2R and away from the centre, the centrifugal acceleration is +ω^2R. (a) and (b) are correct as the directions are given.

$A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?$

Solution

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