Question

A piece of thin uniform wire of mass m and length 3b is bent into an equilaeral triangle.

Answer 1

Answer: Option (C) is the correct answer.

Explanation:

Formula for moment of inertia is as follows.

                 M.I = $\frac{mass \times (length)^{2}}{3}$

Hence, moment of inertia of two rods is as follows.

      M.I of two rods = $2 \times \frac{(\frac{m}{3} \times b^{2})}{3}$

                                = $\frac{2mb^{2}}{9}$    

As third rod have no connection with vertices. So, moment of inertia of a rod along an axis passing through its center is as follows.

           M.I = $\frac{mass \times (length)^{2}}{12}$

                 = $\frac{mb^{2}}{3 \times 12}$

                 = $\frac{mb^{2}}{36}$

Using parallel axis theorem moment of inertia through vertices is as follows.

  $\frac{mb^{2}}{36} + mass \times \text{distance between the two axes}$

       $h^{2} = b^{2} - \frac{b^{2}}{4}$

                  = $\frac{3b^{2}}{4}$

                h = $\frac{\sqrt{3b}}{2}$

Now, we will calculate the moment of inertia of third rod about vertices is as follows.

        $\frac{mb^{2}}{36} + [(\frac{m}{3}) \times 3\frac{b^{2}}{4}]$

            = $mb^{2}[\frac{1}{36} + \frac{1}{4}]$

            = 5 $\frac{mb^{2}}{18}$

Therefore, total moment of inertia is calculated as follows.

            Total M.I = $\frac{2mb^{2}}{9} + \frac{5mb^{2}}{18}$

                            = $\frac{mb^{2}}{2}$

Thus, we can conclude that the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices is $\frac{mb^{2}}{2}$.