A circular thin disc of mass 2 kg has a diameter 0.2 m. Calculate its moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc (. in . kg m 2)

Question

A circular thin disc of mass 2 kg has a diameter 0.2 m. Calculate its moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc (. in . kg m 2) (A) 0.01 (B) 0.03 (C) 0.02 (D) 3

Tardigrade Admin · Accepted Answer

I=(M R2/2) About axis passing through edge, moment of inertia =I prime ∴ I prime=I+M R2 (By theorem of parallel axes) or I prime=(M R2/2)+M R2=(3/2) M R2=(3/2) × 2 ×(0.1)2 or I prime=0.03 kg m 2

Q. A circular thin disc of mass $2 k g$ has a diameter $0.2 m$ . Calculate its moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc $($ in $k g m^{2})$

0.01

0.03

0.02

3

$I = \frac{M R ^{2}}{2}$
About axis passing through edge, moment of inertia $= I^{'}$
$∴ I^{'} = I + M R^{2}$ (By theorem of parallel axes)
or $I^{'} = \frac{M R ^{2}}{2} + M R^{2} = \frac{3}{2} M R^{2} = \frac{3}{2} \times 2 \times (0.1)^{2}$
or $I^{'} = 0.03 k g m^{2}$

Q. A circular thin disc of mass 2kg has a diameter 0.2m. Calculate its moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc ( in kgm2)

0.01

0.03

0.02

3

I=2MR2​ About axis passing through edge, moment of inertia =I′ ∴I′=I+MR2 (By theorem of parallel axes) or I′=2MR2​+MR2=23​MR2=23​×2×(0.1)2 or I′=0.03kgm2

Q. A circular thin disc of mass $2 k g$ has a diameter $0.2 m$ . Calculate its moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc $($ in $k g m^{2})$

$I = \frac{M R ^{2}}{2}$
About axis passing through edge, moment of inertia $= I^{'}$
$∴ I^{'} = I + M R^{2}$ (By theorem of parallel axes)
or $I^{'} = \frac{M R ^{2}}{2} + M R^{2} = \frac{3}{2} M R^{2} = \frac{3}{2} \times 2 \times (0.1)^{2}$
or $I^{'} = 0.03 k g m^{2}$