What Is the Correct Distance d in the Parallel Axis Theorem for a Cube's Edge?

[~angel~]

Could someone please help me with a particular question to this problem?

Consider a cube of mass m with edges of length a. The moment of inertia I_0 of the cube about an axis through its center of mass and perpendicular to one of its faces is given by I_0 = 1/6ma^2. Find I_edge, the moment of inertia about an axis p through one of the edges of the cube.
Express I_edge in terms of m and a. Use fractions rather than decimal numbers in your answer.

You know how the parallel axis theorem is I_edge = I_0 +md^2, what exactly is d in this particular problem. I tried a/2, and I tried using pythagoras' theorem to determine one of the diagonals, getting an answer of d= sqrt(2a^2)/2. Both answers where wrong when i substituted it into the equation.

A picture is available below.

Any help would be appreciated.

 

[ramollari]

what exactly is d in this particular problem

d is the displacement of the axis from the center of mass.

Now,

I_{edge} = I_0 + md^2 = \frac{1}{6}ma^2 + \frac{1}{2}ma^2 = \frac{2}{3}ma^2.

That should be the answer.

 

[thepatientmental]

Hi there,

To find the moment of inertia about an axis through one of the edges of the cube, we can use the parallel axis theorem, which states that the moment of inertia about any axis parallel to the original axis is equal to the moment of inertia about the original axis plus the product of the mass and the square of the distance between the two axes.

In this problem, the original axis is through the center of mass and perpendicular to one of the faces, and the parallel axis is through one of the edges. So, we can use the equation I_edge = I_0 + md^2, where d is the distance between the two axes.

To find the value of d, we can use the Pythagorean theorem. If we draw a diagonal from one corner of the cube to the opposite corner, we can see that the distance between the two axes is equal to the length of this diagonal. So, d = √(a^2 + a^2) = √2a.

Substituting this value into the equation, we get:

I_edge = 1/6ma^2 + m(√2a)^2

= 1/6ma^2 + 2ma^2

= 1/6ma^2 + 12/6ma^2

= 13/6ma^2

So, the moment of inertia about an axis through one of the edges of the cube is 13/6ma^2. I hope this helps! Let me know if you have any further questions.