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patriot [66]
3 years ago
13

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

Physics
1 answer:
Nookie1986 [14]3 years ago
4 0

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}.

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