Rotational Inertia of a block

[seraphimhouse]

Homework Statement

The uniform solid block in Fig. 10-37 has mass .172 kg and edge lengths a = 0.035 m, b = .084 m, and c = .014 m. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c10/fig10_37.gif

Homework Equations

I = Icom + M(h^2)
I = 1/12M(a^2+b^2)

The Attempt at a Solution

With the pythagorean theorem, I found the distance it has shifted which was [1/2(a^2+b^2)]^1/2 = h so,

I = 1/12M(a^2+b^2) + M(h^2) substituting h with the previous discovery giving:

I = 1/12M(a^2+b^2) + M[1/2(a^2+b^2)] I still come out with an incorrect answer.

I'm wondering if the c should be included in the problem or not, but when I do input it in the Icom, it still does not work. I must be missing something important

The answer should be fitting with 4.7 x 10^-4 kg.m^2

 

[LowlyPion]

I = Icom + M(h^2)
I = 1/12M(a^2+b^2)

With the pythagorean theorem, I found the distance it has shifted which was [1/2(a^2+b^2)]^1/2 = h

I think your 1/2 is misplaced.

1/2*(a² + b²)^1/2

Squaring that for your || axis displacement yields

(1/12 + 1/4)*(a² + b²) = 1/3*(a² + b²)

 

[nlightner]

I would recommend double-checking your calculations and making sure you are using the correct equations for rotational inertia. It may also be helpful to draw a diagram and label all the variables before plugging them into the equations. Additionally, make sure you are using the correct units for each variable. If you are still having trouble, it may be helpful to seek assistance from a classmate or your instructor. Remember, it is important to understand the concept behind the calculation, not just the final answer. Keep practicing and don't be afraid to ask for help!