A stone of mass 2 ,kg is tied at one end of a | vedclass

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(D) Given: Mass of stone, $m = 2 \,kg$.Length of the string (radius), $r = 2 \,m$.Maximum tension, $T_{\max} = 64 \,N$.The centripetal force required

Vedclass · Accepted Answer

$\frac{120}{\pi}$

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(D) Given: Mass of stone, $m = 2 \,kg$.Length of the string (radius), $r = 2 \,m$.Maximum tension, $T_{\max} = 64 \,N$.The centripetal force required for circular motion is provided by the tension in the string: $T_{\max} = \frac{m v_{\max}^2}{r}$.Substituting the values: $64 = \frac{2 	imes v_{\max}^2}{2}$.Solving for $v_{\max}$: $v_{\max}^2 = 64$, so $v_{\max} = 8 \,m/s$.We know that $v = r \omega$, where $\omega = 2 \pi f$ and $f$ is the frequency in rotations per second.So, $v_{\max} = r (2 \pi f) \implies 8 = 2 	imes 2 \pi f$.$4 \pi f = 8 \implies f = \frac{2}{\pi}$ rotations per second.To find the number of rotations per minute $(N)$, we multiply the frequency by $60$: $N = f 	imes 60 = \frac{2}{\pi} 	imes 60 = \frac{120}{\pi}$.Thus, the permissible maximum number of rotations per minute is $\frac{120}{\pi}$.

$A$ stone of mass $2 \,kg$ is tied at one end of a string of length $2 \,m$ and is whirled in a horizontal circle. If the string can withstand a maximum tension of $64 \,N$, then the permissible maximum number of rotations per minute is

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