Question

A stone of 0.5 kg mass is
attached to one end of a 0.8m
long aluminum wire of 0.7 mm
diameter and suspended
vertically. The stone is now
rotated in a horizontal plane at
a rate such that wire makes an
angle of 850 with the vertical. If
Y = 7 * 1010 Nm ?,
sin 850 = 0.9962 and
cos 850 = 0.0872, the increase
in length of wire is​

Answer 1

Given:

• Mass of stone $m = 0.5 kg$

• Angle  $\theta = 85^\circ$

• Length of wire $l = 0.8m$

• $Y = 7 \times 10^{10} N/m^2$

To Find:

• The increase  in length of wire

Solution:

Equations of motion

for horizontal line

        $Tsin \theta = \frac{mv^2}{r}$

and for vertical line

        $Tcos \theta = mg$

So,

           $T = \frac{mg}{cos \theta}$

So,

       $T = \frac{0.5 \times 9.81 }{cos 85^\cric}$

Thus, the tension in the rope as

       $T = F = 56.2 N$

Now, this is the force being exerted on the wire and would cause elongation $(dl)$  in it  

thus, the elongation will be given by

                 $dl = \frac{F.l }{Y.A}$

Area $(A) = \pi r^2$

              $= 3.14 \times (\frac{0.7 \times 10^{-3}}{2})^2$

        $A = 3.85 \times 10^{-6} m^2$

thus, we have

            $dl = \frac{(56.2 \times 0.8)}{ (7 \times 10^{10} \times 3.85 \times 10^{-6})}$

           $dl = 16.66 \times 10^{-4} m$

          $dl = 1.66 mm$

Thus, The increased length of the wire is    $dl = 1.66 mm$