A stone tied to the end of a string 1,m long | vedclass

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(C) The length of the string is the radius of the circular path,$r = 1\,m$.The frequency of revolution $n$ is given by $n = \frac{	ext{number of revo

Vedclass · Accepted Answer

$\pi^2\,m/s^2$ and direction along the radius towards the centre

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(C) The length of the string is the radius of the circular path,$r = 1\,m$.The frequency of revolution $n$ is given by $n = \frac{	ext{number of revolutions}}{	ext{time}} = \frac{22}{44} = 0.5\,Hz$.The angular velocity $\omega$ is given by $\omega = 2\pi n = 2\pi(0.5) = \pi\,rad/s$.The centripetal acceleration $a$ is given by $a = \omega^2 r$.Substituting the values,$a = (\pi)^2 	imes 1 = \pi^2\,m/s^2$.In uniform circular motion,the acceleration is centripetal,meaning its direction is always along the radius and towards the centre.

$A$ stone tied to the end of a string $1\,m$ 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 the acceleration of the stone?

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