A stone tied to the end of a string 80; cm lo | vedclass

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Question

(A) Length of the string,$r = 80\; cm = 0.8\; m$.Number of revolutions $= 14$,Time taken $= 25\; s$.Frequency,$
u = \frac{	ext{Number of revolutions

Vedclass · Accepted Answer

$9.91\; m/s^2$ directed towards the center

Vedclass · Answer

(A) Length of the string,$r = 80\; cm = 0.8\; m$.Number of revolutions $= 14$,Time taken $= 25\; s$.Frequency,$
u = \frac{	ext{Number of revolutions}}{	ext{Time taken}} = \frac{14}{25}\; Hz$.Angular frequency,$\omega = 2\pi
u = 2 	imes \frac{22}{7} 	imes \frac{14}{25} = \frac{88}{25}\; rad/s = 3.52\; rad/s$.Centripetal acceleration,$a_c = \omega^2 r = (3.52)^2 	imes 0.8$.$a_c = 12.3904 	imes 0.8 = 9.91232\; m/s^2 \approx 9.91\; m/s^2$.The direction of centripetal acceleration is always directed along the radius towards the center of the circular path.

$A$ 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?

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