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A $14.5 \mathrm{~kg}$ mass, fastened to the end of a steel wire of unstretched length $1 \mathrm{~m}$, is whirled in a vertical circle with an angular velocity of $2 \mathrm{rev} / \mathbf{s}$ at the bottom of the circle. The cross-sectional 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.
$$
Y_{\text {steel }}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2
$$
$$
Y_{\text {steel }}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2
$$
Solution:
1457 Upvotes
Verified Answer
Here,
$$
\begin{aligned}
&m=14.5 \mathrm{~kg}, l=r=1 \mathrm{~m} \\
&\omega=2 r p s=2 \times 2 \pi \mathrm{rad} / \mathrm{s} \\
&A=0.065 \times 10^{-4} \mathrm{~m}^2
\end{aligned}
$$
Tension in the wire at the lowest position on the vertical
$$
\begin{aligned}
&\text { circle }=F=m g+m r \omega^2 \\
&=14.5 \times 9.8+14.5 \times 1 \times 4 \times\left(\frac{22}{7}\right)^2 \times 4 \\
&=142.1+2291.6=2433.7 \mathrm{~N} \\
&\begin{aligned}
Y=\frac{F l}{A \Delta l} \Rightarrow \Delta l &=\frac{F l}{A Y}=\frac{2433.7 \times 1}{0.065 \times 10^{-4} \times 2 \times 10^{11}} \\
&=1.87 \times 10^{-3} \mathrm{~m}=1.87 \mathrm{~mm}
\end{aligned}
\end{aligned}
$$
$$
\begin{aligned}
&m=14.5 \mathrm{~kg}, l=r=1 \mathrm{~m} \\
&\omega=2 r p s=2 \times 2 \pi \mathrm{rad} / \mathrm{s} \\
&A=0.065 \times 10^{-4} \mathrm{~m}^2
\end{aligned}
$$
Tension in the wire at the lowest position on the vertical
$$
\begin{aligned}
&\text { circle }=F=m g+m r \omega^2 \\
&=14.5 \times 9.8+14.5 \times 1 \times 4 \times\left(\frac{22}{7}\right)^2 \times 4 \\
&=142.1+2291.6=2433.7 \mathrm{~N} \\
&\begin{aligned}
Y=\frac{F l}{A \Delta l} \Rightarrow \Delta l &=\frac{F l}{A Y}=\frac{2433.7 \times 1}{0.065 \times 10^{-4} \times 2 \times 10^{11}} \\
&=1.87 \times 10^{-3} \mathrm{~m}=1.87 \mathrm{~mm}
\end{aligned}
\end{aligned}
$$
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