A solid circular disk has a mass of 1.67 kg and a radius of 0.144 m. Each of three identical thin rods has a mass of 0.351 kg. The rods are attached perpendicularly to the plane of the disk at its outer edge to form a three-legged stool (see the figure). Find the moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center. (Hint: When considering the moment of inertia of each rod, note that all of the mass of each rod is located at the same perpendicular distance from the axis.)
3 m r^2 + (1/2) M r^2
where M = 1.67
and
m = .351
and
r = .144
To find the moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center, we need to consider the contributions of both the disk and the rods.
The moment of inertia of the disk can be calculated using the formula for a solid disk:
I_disk = (1/2) * m * r^2,
where I_disk is the moment of inertia of the disk, m is the mass of the disk, and r is the radius of the disk.
Substituting the given values, we have:
I_disk = (1/2) * 1.67 kg * (0.144 m)^2
= 0.065 kg·m^2.
Next, we need to calculate the moment of inertia of each rod. Since each rod is attached perpendicularly to the plane of the disk at its outer edge, all of the mass of each rod is located at the same perpendicular distance from the axis (which is the radius of the disk, 0.144 m in this case). Thus, we can consider each rod as a point mass.
The moment of inertia of a point mass rotating about an axis at a perpendicular distance r from the mass is given by:
I_rod = m * r^2,
where I_rod is the moment of inertia of the rod and m is the mass of the rod.
Substituting the given values, we have:
I_rod = 0.351 kg * (0.144 m)^2,
= 0.007 kg·m^2.
Since there are three identical rods attached to the disk, we need to consider the total moment of inertia contributed by the rods:
I_total_rods = 3 * I_rod,
= 3 * 0.007 kg·m^2,
= 0.021 kg·m^2.
Finally, to find the moment of inertia of the stool, we add the moment of inertia of the disk and the moment of inertia contributed by the rods:
I_total = I_disk + I_total_rods,
= 0.065 kg·m^2 + 0.021 kg·m^2,
= 0.086 kg·m^2.
Therefore, the moment of inertia of the stool with respect to the given axis is 0.086 kg·m^2.