Question:

Moment of inertia of an equilateral triangular lamina $ABC$, about the axis passing through its centre $O$ and perpendicular to its plane is $I_o$ as shown in the figure. A cavity $DEF$ is cut out from the lamina, where $D, E, F$ are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is :

Updated On: Sep 14, 2024
  • $\frac{7}{8} I_o$
  • $\frac{15}{16} I_o$
  • $\frac{3I_o}{4} $
  • $\frac{3 1I_o}{32} $
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The Correct Option is B

Solution and Explanation

$I_{0} = kml^{2}\quad\quad\quad BC = l$ $I_{DEF} = k \frac{m}{4} \left(\frac{l}{2}\right)^{2}$ $= \frac{k}{16}\,ml^{2}$ $I_{DEF} = \frac{I_{0}}{16}$ $I_{remain} = I_{0} = \frac{I_{0}}{16}$ $= \frac{15I_{0}}{16}$
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Concepts Used:

Moment of Inertia

Moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

Moment of inertia mainly depends on the following three factors:

  1. The density of the material
  2. Shape and size of the body
  3. Axis of rotation

Formula:

In general form, the moment of inertia can be expressed as, 

I = m × r²

Where, 

I = Moment of inertia. 

m = sum of the product of the mass. 

r = distance from the axis of the rotation. 

M¹ L² T° is the dimensional formula of the moment of inertia. 

The equation for moment of inertia is given by,

I = I = ∑mi ri²

Methods to calculate Moment of Inertia:

To calculate the moment of inertia, we use two important theorems-

  • Perpendicular axis theorem
  • Parallel axis theorem