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Home » A mass, fastened to the end of a steel wire of unstretched length , is whirled in a vertical circle with an angular velocity of at the bottom of the circle. The crosssectional area of the wire is Calculate the elongation of the wire when the mass is at the lowest point of its path.
A 14.5 \mathrm{~kg} mass, fastened to the end of a steel wire of unstretched length 1.0 \mathrm{~m}, is whirled in a vertical circle with an angular velocity of 2 \mathbf{r e v} / \mathrm{s} at the bottom of the circle. The crosssectional area of the wire is 0.065 \mathrm{~cm}^{2} . Calculate the elongation of the wire when the mass is at the lowest point of its path.
CBSE | Class 11 | Mechanical Properties of Solids | NCERT | Physics
A 14.5 \mathrm{~kg} mass, fastened to the end of a steel wire of unstretched length 1.0 \mathrm{~m}, is whirled in a vertical circle with an angular velocity of 2 \mathbf{r e v} / \mathrm{s} at the bottom of the circle. The crosssectional area of the wire is 0.065 \mathrm{~cm}^{2} . Calculate the elongation of the wire when the mass is at the lowest point of its path.

Mass is given as m=14.5 \mathrm{~kg}

Length of the steel wire is given as \mid=1 \mathrm{~m}

Angular velocity is given as v=2 rev/s

Cross-sectional area of the wire is given as A=0.065 \times 10^{-4} \mathrm{~m}^{2}

When the mass is at the lowest point of the vertical circle, total pulling force on the steel wire will be,

\mathrm{F}=\mathrm{mg}+ \operatorname{mr} \omega^{2}
=14.5 \times 9.8+14.5 \times 1 \times(12.56)^{2}
=2429.53 \mathrm{~N}

Young’s modulus = Stress / Strain

Y=\frac{F}{A} \frac{l}{\Delta l}

\Delta l=\frac{F}{A} \frac{l}{Y}

\Delta l=\frac{2429.53\times1}{0.065\times10^{-4}}\times(2\times1011)

As a result, the elongation of the wire when the mass is at the lowest is 1.87\times10^{-3}

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