A hole of radius r has been drilled in a circular, flat plate of radius R. The center of the hole is a distance d from the center of the circle. You can assume that the mass of the solid disk before the hole is removed was M. Find the moment of inertia for the rotation about an axis through the center of the circle, perpendi

I_total = ½M[R² - ((r²/R²)(r² + 2d²))]

Explanation:

From the question, we have to consider the hole as an object of negative mass. Thus, the total inertia of the object will be;

I_total = I_plate - I_hole

Moment of inertia of a Disk is ½MR²

We are told it rotates around it's axis. Hence, we will use parallel axis theorem and we have;

I_plate = ½M_plate•R²

I_hole = ½M_hole•r² + M_hole•d²

I_hole = ½M_hole(r² + 2d²)

Thus;

I_total = ½M_plate•R² - ½M_hole(r² + 2d²)

Now,since the plate is uniform, then;

M_hole/M_plate = A_hole/M_plate

A_hole = πr²

A_plate = πR²

Thus;

M_hole/M_plate = πr²/πR²

M_hole/M_plate = r²/R²

M_hole = M_plate(r²/R²)

Let's put M_plate(r²/R²) for M_hole in the I_total equation to get;

I_total = ½M_plate•R² - (½M_plate(r²/R²)(r² + 2d²))

Factorizing, we have;

I_total = ½M_plate[R² - ((r²/R²)(r² + 2d²))]

We are told that mass of the solid disk before the hole is removed was M.

Thus;

I_total = ½M[R² - ((r²/R²)(r² + 2d²))]

the answer is 12 as i may remember

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i dont have a visual in front of me, but i can you reason to try to get the right one. the graph should look like this, 0 to 6 there is a spike, 12 to 14 there should be a very low point. slight increase in the graph at 12 to 14, and flat 14 to 18

explanation: