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  • Help me answer: An equilateral triangle ABC is cut from a thin solid sheet of wood. (see fig.) D, E and F are the mid - points of its sides as shown and G is the centre of the triangle . The moment of interia of triangle about an axis passing through G an

An equilateral triangle ABC is cut from a thin solid sheet of wood. (see fig.) D, E and F are the mid - points  of its sides as shown and G is the centre of the triangle . The moment of interia of triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. If the smaller triangle DEF is removed from ABC, the moment of interia of the remaining figure about the same axis is I. Then:

  • Option 1) I=(9/16)I0
  • Option 2) I=(3/4)I0 
  • Option 3)  I=(1/4)I0 
  • Option 4) I=(15/16)I0 

Answers (1)

best_answer

 

Moment of inertia for continuous body -

I= \int r^{2}dm

- wherein

r is perpedicular distance of a particle of mass dm of rigid body from axis of rotation

Let m is the mass and a is the sides of larger triangle

\Rightarrow \frac{m}{4}is mass & \frac{a}{2} is side of smaller triangle

\frac{I_{removed}}{I_{original}}=\frac{\frac{m}{4}(\frac{a}{2})^{2}}{m(a)^{2}}

I_{removed}=\frac{I_{original}}{16}

Lets say I_{original}=I_{o}

I = I_{o}\frac{-I_{o}}{16}=\frac{15I_{o}}{16}

 

 

 


Option 1)

I = \frac{9}{16}I_{0}

Option 2)

I = \frac{3}{4}I_{0}

Option 3)

I = \frac{I_{0}}{4}

Option 4)

I = \frac{15}{16}I_{0}

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