Consider the rods plus disc system shown in the Figure. The rods are identical. They have a length L =0.50 m, and mass M = 0.250kg. The disc has a radius of R = 0.130m and mass m=0.750kg. Both rods are glued to the disk as shown with X = 0.25m. The system behaves as a compound pendulum that oscillates about pivot point P. The pivot is at P where the tip of both rods meet with one end of the disk. Assume that the disc and rod have uniform mass distributions
Find the moment of inertia of one of the rods (they should be the same), in units of kgm2, as it freely rotates about the pivot point P.
To find the moment of inertia of one of the rods, we can use the formula for the moment of inertia of a uniform rod rotating about one end.
The moment of inertia of a uniform rod rotating about one end is given by the formula:
I = (1/3) * m * L^2
where I is the moment of inertia, m is the mass of the rod, and L is the length of the rod.
In this case, the mass of the rod is given as M = 0.250 kg, and the length of the rod is given as L = 0.50 m.
Substituting the given values into the formula, we have:
I = (1/3) * (0.250 kg) * (0.50 m)^2
I = (1/3) * (0.250 kg) * (0.25 m^2)
I = (1/3) * (0.250 kg) * (0.0625 m^2)
I = 0.015625 kgm^2
Therefore, the moment of inertia of one of the rods is 0.015625 kgm^2.
To find the moment of inertia of one of the rods about the pivot point P, we'll use the parallel axis theorem. The parallel axis theorem states that the moment of inertia of an object about an axis parallel to and a distance "d" away from its center of mass is given by the sum of its moment of inertia about its center of mass and the product of its mass (m) and the square of the distance (d) between the two axes.
First, let's calculate the moment of inertia of one rod about its center of mass. Since the rod has a uniform mass distribution, we can use the formula for the moment of inertia of a uniform rod about its center:
I_rod = (1/12) * M * L^2
where M is the mass of the rod and L is its length.
Second, let's calculate the moment of inertia of one rod about the pivot point P. Since the rod is rotating about an axis parallel to and a distance X away from its center of mass, we can use the parallel axis theorem:
I_pivot = I_center + M * X^2
where I_center is the moment of inertia about the center of mass, M is the mass of the rod, and X is the distance between the center of mass and the pivot point.
Now, let's substitute the given values into the formulas:
M = 0.250 kg (mass of the rod)
L = 0.50 m (length of the rod)
X = 0.25 m (distance between the center of mass and pivot point)
I_rod = (1/12) * 0.250 kg * (0.50 m)^2
= 0.005208333 kgm^2
I_pivot = I_center + (0.250 kg) * (0.25 m)^2
= 0.005208333 kgm^2 + 0.005208333 kgm^2
= 0.010416667 kgm^2
So, the moment of inertia of one of the rods, as it freely rotates about the pivot point P, is 0.010416667 kgm^2.