Question

A $1kg$ mass is suspended by a rubber cord $2.00m$ long and of cross-section $0.5cm.$ It is made to describe a horizontal circle of radius $50cm$ in $4$ times a second. Find the extension $\left(incm\right)$ of the cord. $(\text{Young's modulus}Y=5\times {10}^{8}N/m^2)$

Answer 1

Step 1: Draw a labelled diagram.

Step 2: Find the extension of the cord.
Formula used: $Y=\frac{FL}{Al}$
given,
$m=1kg$
$r=0.5m$
$\omega =2\pi f=2\pi \times 4=8\pi rad/s$
$T\cos \theta =mg\cdots \left(i\right)$
$T\sin \theta =\frac{m{v}^2}{r}=m{\omega }^{2}r\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$

$T=\sqrt{{\left(T\sin \theta \right)}^2+{\left(T\cos \theta \right)}^2}$

$=\sqrt{{\left(m^{2}r\right)}^2{\left(mg\right)}^2}$

$T=\sqrt{{[1\times \left(8{\pi }^2\right)\times 0.5]}^2+{[1\times 10]}^2}$

$T=316N$
The elongation on the wire,
$\Delta L=\frac{TL}{AY}=\frac{316\times 2}{0.5\times {10}^{-4}\times 5\times {10}^8}$
$=252\times {10}^{-4}m$
$=2.52cm$
Final Answer: (2.52)