A machine part is made from a uniform solid disk of radius R = 0.161 m and mass M = 7.75 kg. A hole of radius R/2 is drilled into the disk, with the center of the hole at a distance R/2 from the center of the disk (the diameter of the hole spans from the center of the disk to its outer edge). What is the moment of inertia of this machine part about the center of the disk?
To find the moment of inertia (I) of the machine part about the center of the disk, we need to consider the individual moments of inertia of the solid disk (I_disk) and the hole (I_hole), and then subtract the moment of inertia of the hole from the moment of inertia of the disk.
1. Moment of Inertia of the Solid Disk:
The moment of inertia of a solid disk about its center can be found using the formula:
I_disk = (1/2) * M * R^2
where M is the mass of the disk and R is the radius.
Substituting the given values:
I_disk = (1/2) * 7.75 kg * (0.161 m)^2
2. Moment of Inertia of the Hole:
The moment of inertia of the hole can be found using the parallel-axis theorem. Since the hole is drilled at a distance R/2 from the center of the disk, the moment of inertia of the hole can be calculated as:
I_hole = (1/2) * (1/2) * M * (R/2)^2
where M is the mass of the disk and R is the radius.
3. Subtract I_hole from I_disk:
Finally, we subtract the moment of inertia of the hole from the moment of inertia of the disk to get the moment of inertia of the machine part about the center of the disk:
I = I_disk - I_hole
Calculating the values:
I_disk = (1/2) * 7.75 kg * (0.161 m)^2
I_hole = (1/2) * (1/2) * 7.75 kg * (0.081 m)^2
I = I_disk - I_hole
Now, you can substitute the values into the formulae and calculate the moment of inertia of the machine part about the center of the disk.
To determine the moment of inertia of this machine part about the center of the disk, we can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia about any axis parallel to the center of mass can be calculated by adding the moment of inertia about the center of mass and the product of the mass and the square of the perpendicular distance between the two axes.
In this case, we have a disk with a hole drilled into it. The moment of inertia of the whole disk about its center of mass can be calculated using the formula for the moment of inertia of a solid disk:
I_disk = (1/2) * M * R^2
Given that the radius of the disk is R = 0.161 m and the mass is M = 7.75 kg, we can substitute these values into the formula to find the moment of inertia of the whole disk:
I_disk = (1/2) * 7.75 kg * (0.161 m)^2 = 0.124 kg m^2
Now, let's calculate the moment of inertia of the drilled hole. The moment of inertia of a solid disk can be expressed as:
I_hole = (1/2) * m * r_hole^2
Where m is the mass of the drilled hole and r_hole is the radius of the hole.
Since the drilled hole is a cylindrical shape, we can assume that its mass is the mass of the material that was removed from the disk. The volume of the removed material is given by:
V_hole = π * (r_hole^2) * h_hole
Where h_hole is the height of the removed material. Since the radius of the hole is R/2 and its height is also R/2, we can substitute these values into the formula to find the volume of the removed material:
V_hole = π * ((R/2)^2) * (R/2) = (π/8) * R^3
The mass of the drilled hole can be calculated using the density of the material, denoted by ρ:
m_hole = ρ * V_hole
Substituting the volume of the drilled hole into this equation, we get:
m_hole = ρ * (π/8) * R^3
Now, let's substitute this expression for the mass of the drilled hole and the formula for the moment of inertia of a solid disc into the parallel axis theorem. The moment of inertia of the drilled hole about the center of the disk is given by:
I_hole = (1/2) * (ρ * (π/8) * R^3) * (R/2)^2 = (ρ * π * R^5) / 128
Finally, we can add the moment of inertia of the whole disk and the moment of inertia of the drilled hole to obtain the moment of inertia of the machine part about the center of the disk:
I_total = I_disk + I_hole = (0.124 kg m^2) + ((ρ * π * R^5) / 128)