Moment of Inertia Problem?

Question

A thin, flat, uniform disk has a mass ( M ) and a radius ( R ). A circular hole with a radius of ( \frac{R}{4} ), centered at a point ( \frac{R}{2} ) from the disk’s center, is punched out of the disk. 
a. Calculate the moment of inertia of the disk with the hole about an axis that passes through the original center of the disk and is perpendicular to the plane of the disk. (Hint: First, find the moment of inertia of the portion removed from the disk.)
b. Determine the moment of inertia of the disk with the hole about an axis that passes through the center of the hole and is perpendicular to the plane of the disk. 
I have calculated the answer to part a as ( \frac{247}{512}MR^2 ), but I am unable to find the answer to part b.

Anonymous · Accepted 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∙r²So the moment of inertia of the disk without the hole about this axis is:J₀ = (1/2)∙M∙R²Consider 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. Som/M = (π∙(R/4)²) / (π∙R²) = (1/16)<=>m = (1/16)∙MThe moment of inertia about an axis through its center is:J₁ = (1/2)∙((1/16)∙M)∙(R/4)² = (1/512)∙M∙R²The 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∙R²Hence the moment of inertia of the disk with hole about that axis isJ = J₀ – J₁’ = (1/2)∙M∙R² – (9/512)∙M∙R²= (247/512)∙M∙R²b)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∙R²The moment of inertia of the piece punched out about the axis for this part was already computed in part aJ₁ = (1/2)∙((1/16)∙M)∙(R/4)² = (1/512)∙M∙R²Hence,J = J₀’ – J₁ = (3/4)∙M∙R² – (1/512)∙M∙R²

= (383/512)∙M∙R²

Anonymous

Moment of Inertia Problem?

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