The moment of inertia of a circular disc of radius 2 m and mass 1 kg about an axis passing through the centre of mass but perpendicular to the plane of the disc is 2 kg m2. Its moment of inertia about an axis parallel to this axis but passing through the edge of the disc is............. (See the given figure).

Question

The moment of inertia of a circular disc of radius 2 m and mass 1 kg about an axis passing through the centre of mass but perpendicular to the plane of the disc is 2 kg m2. Its moment of inertia about an axis parallel to this axis but passing through the edge of the disc is............. (See the given figure). (A) 10 kg m2 (B) 6 kg m2 (C) 8 kg m2 (D) 4 kg m2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f4ccc58a04cbb7386ba55e00e81619f4-.png" /> According to parallel axis theorem IX.Y' =IX Y+M R2 =2+(1)(2)2 =6 kg - m 2

$10 k g m^{2}$

$6 k g m^{2}$

$8 k g m^{2}$

$4 k g m^{2}$

According to parallel axis theorem
$I_{X . Y^{'}} = I_{X Y} + M R^{2}$
$= 2 + (1) (2)^{2}$
$= 6 k g - m^{2}$

10kgm2

6kgm2

8kgm2

4kgm2

According to parallel axis theorem IX.Y′​=IXY​+MR2 =2+(1)(2)2 =6kg−m2

$10 k g m^{2}$

$6 k g m^{2}$

$8 k g m^{2}$

$4 k g m^{2}$

According to parallel axis theorem
$I_{X . Y^{'}} = I_{X Y} + M R^{2}$
$= 2 + (1) (2)^{2}$
$= 6 k g - m^{2}$