# Finding the Moment of Inertia for a Rectangular Sheet

## Homework Statement

The horizontal side (x-axis) is length 2A and the vertical side (y-axis at x=+-A) is length B. The mass is uniform throughout the sheet so that the center of mass is at the center of the rectangle. What is the moment of inertia for. the rotation around the z-axis at the midpoint of the horizontal side, coordinate (0,0) in terms of A,B and/or mass? The z-axis at the center of mass (hint: use parallel axis)?

## Homework Equations

I=Icm+mr^2 where Icm is the center of mass moment and r is the distance to center of mass.

## The Attempt at a Solution

For part a, the moment of inertia for a point mass is just I=mr^2, so since the center of mass is (B/2) away from the axis, then I=m*(B/2)^2. However, shouldn't we use parallel axis for part a and not part b, in contrary to the hint for part b?
For part b, the moment of inertia would just be zero? This is because lcm=0 as the axis is at the center of mass and r=0 as the point is also at the center of mass, so I=Icm+mr^2=0+m(0)=0.

Is this the correct way to get both or did I miss something?

Related Introductory Physics Homework Help News on Phys.org
gneill
Mentor
For moments of inertia the geometry of the object is important. Mass elements that are further from the axis of rotation have a greater impact on the total, and elements closer have less of an impact. Because the contributions vary as the square of the distance you can't just lump the mass at the center of gravity and use the point mass formula on it. You'll have to do the integration over the object.

Once you've established the MOI about one axis though, you can use the parallel axis theorem to "move it" to any parallel axis.

• reslion