A steel wire 1.5,m long and of radius 1,mm is | vedclass

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Question

(A) At the lowest position,the tension $T$ in the wire is given by the sum of the gravitational force and the centripetal force: $T = Mg + M\omega^2\e

Vedclass · Accepted Answer

$1.77 	imes 10^{-3}\,m$

Vedclass · Answer

(A) At the lowest position,the tension $T$ in the wire is given by the sum of the gravitational force and the centripetal force: $T = Mg + M\omega^2\ell$.Here,$M = 3\,kg$,$\ell = 1.5\,m$,$g = 10\,m/s^2$,and frequency $f = 2\,Hz$.The angular velocity is $\omega = 2\pi f = 2 	imes 3.14 	imes 2 = 12.56\,rad/s$.The tension $T = 3 	imes 10 + 3 	imes (12.56)^2 	imes 1.5 = 30 + 3 	imes 157.75 	imes 1.5 = 30 + 710 = 740\,N$.The elongation $\Delta\ell$ is given by $\Delta\ell = \frac{T\ell}{AY}$,where $A = \pi r^2 = 3.14 	imes (10^{-3})^2 = 3.14 	imes 10^{-6}\,m^2$.$\Delta\ell = \frac{740 	imes 1.5}{3.14 	imes 10^{-6} 	imes 2 	imes 10^{11}} = \frac{1110}{6.28 	imes 10^5} \approx 1.77 	imes 10^{-3}\,m$.

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