Finding moment of inertia

[demonelite123]

a uniform solid block has mass 0.172 kg and edge lengths a=3.5cm, b=8.4cm, and c=1.4cm (c is the height of the rectangular solid).Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.

i know the formula is integral of r^2 dm, but i have no idea what to do here. i have r^2 = a^2 + b^2. but i don't know how what to substitute for dm.

 

[tiny-tim]

Hi demonelite123!

(try using the X² tag just above the Reply box )

i know the formula is integral of r^2 dm, but i have no idea what to do here. i have r^2 = a^2 + b^2. but i don't know how what to substitute for dm.

If r² = x² + y², then dm = density times dxdydz (or, if you're only integrating over x and y, then dm = hdxdy).

 

[kerimek]

To calculate the moment of inertia for a solid block, we first need to determine the distribution of mass throughout the block. In this case, we can assume that the mass is evenly distributed throughout the block.

To find the rotational inertia about an axis through one corner and perpendicular to the large faces, we can use the parallel axis theorem. This theorem states that the moment of inertia about an 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 case, our original axis is through the center of mass of the block, which is located at (a/2, b/2, c/2). The distance between this axis and the desired axis through one corner is equal to the distance from the center of mass to the corner, which is given by (a/2, b/2, c/2) = (3.5/2, 8.4/2, 1.4/2) = (1.75, 4.2, 0.7).

Now, we can plug this distance into the parallel axis theorem and calculate the moment of inertia about the desired axis:

I = Icm + md^2 = (1/12)m(a^2 + b^2) + m(1.75^2 + 4.2^2 + 0.7^2)

= (1/12)(0.172)(3.5^2 + 8.4^2) + (0.172)(1.75^2 + 4.2^2 + 0.7^2)

= 0.0155 + 0.1596 = 0.1751 kgm^2

Therefore, the moment of inertia for the solid block about an axis through one corner and perpendicular to the large faces is 0.1751 kgm^2.