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

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(A) Given:Radius $R = 1\, m$Number of revolutions $n = 22$Time $t = 44\, s$Frequency $f = \frac{n}{t} = \frac{22}{44} = 0.5\, Hz$Angular velocity $\om

Vedclass · Accepted Answer

${\pi ^2}\,m\,s^{ - 2}$ and direction along the radius towards the centre.

Vedclass · Answer

(A) Given:Radius $R = 1\, m$Number of revolutions $n = 22$Time $t = 44\, s$Frequency $f = \frac{n}{t} = \frac{22}{44} = 0.5\, Hz$Angular velocity $\omega = 2\pi f = 2\pi 	imes 0.5 = \pi\, rad/s$The centripetal acceleration is given by $a_r = \omega^2 R$Substituting the values: $a_r = (\pi)^2 	imes 1 = \pi^2\, m/s^2$Since the speed is constant,the tangential acceleration $a_t = 0$. Thus,the net acceleration is the centripetal acceleration,which is directed towards the centre of the circle.

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

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