A weightless thread can bear tension up to 3. | vedclass

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(B) Given: Tension $T_{max} = 3.7 \ kg \ wt = 3.7 	imes 10 \ N = 37 \ N$. Mass $m = 500 \ g = 0.5 \ kg$. Radius $r = 4 \ m$. $g = 10 \ m/s^2$. In a v

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$4$

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(B) Given: Tension $T_{max} = 3.7 \ kg \ wt = 3.7 	imes 10 \ N = 37 \ N$. Mass $m = 500 \ g = 0.5 \ kg$. Radius $r = 4 \ m$. $g = 10 \ m/s^2$. In a vertical circular motion,the tension is maximum at the lowest point of the path. The formula for tension at the lowest point is $T = mg + \frac{mv^2}{r}$. Substituting the values: $37 = (0.5 	imes 10) + \frac{0.5 	imes v^2}{4}$. $37 = 5 + \frac{0.5 	imes v^2}{4}$. $32 = \frac{0.5 	imes v^2}{4}$. $128 = 0.5 	imes v^2$. $v^2 = 256$. $v = 16 \ m/s$. Since $v = r\omega$,we have $\omega = \frac{v}{r} = \frac{16}{4} = 4 \ rad/s$.

$A$ weightless thread can bear tension up to $3.7 \ kg \ wt$. $A$ stone of mass $500 \ g$ is tied to it and revolved in a circular path of radius $4 \ m$ in a vertical plane. The maximum angular velocity of the stone will be (acceleration due to gravity,$g = 10 \ m/s^2$) (in $rad/s$)

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