Question

Moment of inertia of a cube of mass m and side a about one of its edges is equal to

Answer 1

Answer:

The moment of inertia of a cube of mass m and side a about one of its edges is equal to. I=Ic+m(a√2)2=(ma212+ma212)+ma22=23ma2.

Answer 2

The moment of inertia is $\dfrac{2}{3}ma^2$

Explanation:

Given that,

Mass of cube = m

Side of cube = a

We need to calculate the moment of inertia

Using theorem of perpendicular axes

$I=I_{cm}+mr^2$

Put the value into the formula

$I=\dfrac{ma^2}{12}+\dfrac{ma^2}{12}+m(\dfrac{a}{\sqrt{2}})^2$

$I=\dfrac{ma^2}{12}+\dfrac{ma^2}{12}+\dfrac{ma^2}{2}$

$I=\dfrac{2}{3}ma^2$

Hence, The moment of inertia is $\dfrac{2}{3}ma^2$

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Topic : moment of inertia

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