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

The moment of inertia for the rotation of a solid disk about an axis through its center is given by I = (1/2)MR^2. The moment of inertia about an axis through the center of a circle with a hole is given by Iparallel-axis = (1/2)MR^2 + Md^2.

Step-by-step explanation:

The moment of inertia for the rotation of a solid disk about an axis through its center is given by the formula I = (1/2)MR^2, where M is the mass of the disk and R is the radius of the disk. In this case, since the center of the hole is a distance d from the center of the circle, we must use the parallel-axis theorem to find the moment of inertia. The moment of inertia of the disk about its center is given by Icenter = (1/2)MR^2, and the moment of inertia about the axis through the center of the circle is given by Iparallel-axis = Icenter + Md^2. Therefore, the moment of inertia for the rotation about an axis through the center, perpendicular to the plane of the plate, is Iparallel-axis = (1/2)MR^2 + Md^2.a

Answer 2

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

Step-by-step 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))]