Answer
(a) $1.6\times 10^{-4}m$
(b) $2.9\times 10^{-4}m$
Work Step by Step
(a) We know that at the top of the circle, the tension in the wire is given as
$F=m(\frac{v^2}{r}-g)$
$\implies F=1.078Kg(\frac{(7.8m/s)^2}{0.852m}-9.8m/s^2)$
$\implies F=66.4Kg.m^2/s^2$
Now the stretch in the wire can be determined as
$\Delta L=\frac{FL_{\circ}}{YA}$
We plug in the known values to obtain:
$\Delta L=\frac{(66.4N)(0.82m)}{(6.9\times 10^{10}N/m^2)(\frac{\pi}{4}d^2)}$
$\implies \Delta L=1.6\times 10^{-4}m$
(b) We know that the tension at the bottom of the circle is given as
$F^{\prime}=m(\frac{v^2}{r}+g)$
We plug in the known values to obtain:
$F^{\prime}=(1.078Kg)(\frac{(9.3m/s)^2}{0.852m}+9.8m/s^2)$
$F^{\prime}=120N$
Now $\Delta L^{\prime}=\frac{F^{\prime}L_{\circ}}{YA}$
We plug in the known values to obtain:
$\Delta L^{\prime}=\frac{(120N)(0.82m)}{(6.9\times 10^{10}N/m^2)(4.9\times 10^{-6}m^2)}=2.9\times 10^{-4}m$
