The maximum and minimum tension in the string | vedclass

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(A) In a vertical circular motion with constant speed $v$,the tension at the lowest point is $T_{\max} = \frac{mv^2}{r} + mg$ and the tension at the h

Vedclass · Accepted Answer

$\sqrt{98} \, m/s$

Vedclass · Answer

(A) In a vertical circular motion with constant speed $v$,the tension at the lowest point is $T_{\max} = \frac{mv^2}{r} + mg$ and the tension at the highest point is $T_{\min} = \frac{mv^2}{r} - mg$.Given the ratio $\frac{T_{\max}}{T_{\min}} = \frac{5}{3}$,we have:$\frac{\frac{mv^2}{r} + mg}{\frac{mv^2}{r} - mg} = \frac{5}{3}$Cross-multiplying gives: $3(\frac{mv^2}{r} + mg) = 5(\frac{mv^2}{r} - mg)$$3\frac{mv^2}{r} + 3mg = 5\frac{mv^2}{r} - 5mg$$8mg = 2\frac{mv^2}{r}$$4gr = v^2$$v = \sqrt{4gr} = \sqrt{4 	imes 9.8 	imes 2.5} = \sqrt{98} \, m/s$.

The maximum and minimum tension in the string whirling in a circle of radius $2.5 \, m$ with constant velocity are in the ratio $5 : 3$. Then its velocity is:

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