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(A) For a stone rotating in a horizontal circle (conical pendulum),the forces acting on the stone are the tension $T$ in the string and the gravitatio

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$20$

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(A) For a stone rotating in a horizontal circle (conical pendulum),the forces acting on the stone are the tension $T$ in the string and the gravitational force $mg$. Resolving the tension $T$ into components: Vertical component: $T \cos 	heta = mg$ $(1)$ Horizontal component providing centripetal force: $T \sin 	heta = \frac{mv^2}{r}$,where $r = l \sin 	heta$ is the radius of the circular path. So,$T \sin 	heta = \frac{mv^2}{l \sin 	heta}$ $(2)$ From $(1)$,$\cos 	heta = \frac{mg}{T} = \frac{1 	imes 10}{400} = 0.025$. Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin^2 	heta = 1 - (0.025)^2 = 1 - 0.000625 = 0.999375$. From $(2)$,$v^2 = \frac{T l \sin^2 	heta}{m} = \frac{400 	imes 1 	imes 0.999375}{1} = 399.75$. Taking the square root,$v = \sqrt{399.75} \approx 19.99 \approx 20\,ms^{-1}$.

$A$ stone of mass $1\,kg$ is tied to the end of a massless string of length $1\,m$. If the breaking tension of the string is $400\,N$,then the maximum linear velocity the stone can have without breaking the string,while rotating in a horizontal plane,is $.......\,ms^{-1}$.

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