Polar moment of inertia of a circular area of dia ‘D’ is 

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  1. \(\rm \frac{\pi D^4}{8}\)
  2. \(\rm \frac{\pi D^4}{32}\)
  3. \(\rm \frac{\pi D^4}{64}\)
  4. \(\rm \frac{\pi D^4}{128}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{\pi D^4}{32}\)
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Detailed Solution

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CONCEPT:

  • Polar Moment of Inertia is a measure of an object’s capacity to oppose or resist torsion when some amount of torque is applied to it on a specified axis.
  • Polar Moment of Inertia also known as the second polar moment of area is a quantity used to describe resistance to torsional deformation.
  • It is denoted as Iz or J.

EXPLANATION:

  • The moment of inertia for a solid circular shaft with diameter d is

  \(⇒ I_{xx}=I_{yy}= \frac{{\pi \left( {d^4} \right)}}{{64}}\) 

Where Ix-x and Iyy are moments of inertial w.r.t. x-x axis and y-axis respectively.

We know that,

Polar moment of inertia. For objects that have rotational symmetry, such as a cylinder or hollow tube, the equation can be simplified to

⇒ Jz = Ixx + Iyy = 2Ixx = 2Iyy

\(\Rightarrow J_z = 2 \times \frac{{{\rm{\pi }}\left({{\rm{d}}^4} \right)}}{{64}} = \frac{{{\rm{\pi }}\left({{\rm{d}^4}} \right)}}{{32}}\)

Important Points Polar moment of inertia of some important sections are as follows:

F1 N.M. Nita 12.11.2019 D 8

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