First slide

Moment of intertia

Question

With reference to figure of a cube of edge a and mass m, which of the following is the correct statement? (O is the centre of the cube)

Moderate

A

The moment of inertia of cube and z' is
${I_{z}}^{'} = I_{Z} + \frac{{ma}^{2}}{2}$
B

The moment of inertia of cube about $z^{''}$
$I_{Z} + \frac{m a^{2}}{4}$
C

$I_{x} = 2 I_{y}$
D

$2 I_{x} = I_{Z}$

Solution

As z axis is parallel to $z^{'}$ axis and distance between them = $\frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$
By the theorem of parallel axis,
${I_{z}}^{'} = I_{z} + m {(\frac{a}{\sqrt{2}})}^{2} = I_{Z} + \frac{{ma}^{2}}{2}$
Hence, option (1) correct.
z and $z^{''}$ are not parallel axes. Hence, option (2) is incorrect.
By symmetry, $I_{x} = I_{y}$ . Hence, option(3) is not correct. Again $I_{X} = 1 / 6 m a^{2} + m {(a / 2)}^{2} = 5 / 12 m a^{2} a n d I_{z} = 1 / 6 m a^{2} . S o o p t i o n (4) i s w r o n g .$

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