Question

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

Answer 1

Answer:

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))]