# Moment of Inertia Problem?

A thin, flat, uniform disk has mass M and radius R. A circular hole of radius R/4, centered at a point R/2 from the disk’s center, is then punched in the disk. a.Find the moment of inertia of the disk with the hole about an axis through the original center of the disk, perpendicular to the plane of the disk. (Hint: Find the moment of inertia of the piece punched from the disk.) b.Find the moment of inertia of the disk with the hole about an axis through the center of the hole, perpendicular to the plane of the disk. I found the answer to part a (247/512*MR^2) but I can’t seem to find the answer to part b

### 1 Answer

• I get the same result for part a. I wonder why you can’t get the answer for part b because it solution employ the same physical concept: the parallel-axis theorem. For each part you need to apply this theorem to one of the two objects considered. a) The moment of inertia of the a disk of radius r and mass m about an axis through its center, perpendicular to the disk is: J = (1/2)∙m∙r² So the moment of inertia of the disk without the hole about this axis is: J₀ = (1/2)∙M∙R² Consider the circular piece punched from the disk. Assuming uniform density and thickness of the disk, its mass is proportional to the area of the piece. So m/M = (π∙(R/4)²) / (π∙R²) = (1/16) <=> m = (1/16)∙M The moment of inertia about an axis through its center is: J₁ = (1/2)∙((1/16)∙M)∙(R/4)² = (1/512)∙M∙R² The moment of inertia about the axis through the center the original disk, which is located at distance R/2 from the center of the hole, can be found from parallel-axis theorem [1]. J’ = J + m∙l² => J₁’ = J₁ + ((1/16)∙M)∙(R/2)² = (1/512)∙M∙R² + (1/64)∙M∙R² = (9/512)∙M∙R² Hence the moment of inertia of the disk with hole about that axis is J = J₀ – J₁’ = (1/2)∙M∙R² – (9/512)∙M∙R² = (247/512)∙M∙R² b) For this part you need compute the moment of inertia of the original disk about the center of the hole using the parallel axis theorem: J₀’ = J₀ + M∙(R/2)² = (1/2)∙M∙R² + (1/4)∙M∙R² = (3/4)∙M∙R² The moment of inertia of the piece punched out about the axis for this part was already computed in part a J₁ = (1/2)∙((1/16)∙M)∙(R/4)² = (1/512)∙M∙R² Hence, J = J₀’ – J₁ = (3/4)∙M∙R² – (1/512)∙M∙R²

= (383/512)∙M∙R²