Question

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

Find 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.

A) 7/12 mb^2

B)1/3 mb^2

C) 1/2 mb^2

D) 2/3 mb^2

E) 7/4 mb^2

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