Consider the rods plus disc system shown in the Figure. The rod has a length L =0.50 m, and mass M = 0.270kg. The large disc has a radius of R = 0.130m and mass m=0.810kg. The second disk should be treated as a point mass of 0.405kg. It is glued to the rod at position X = 0.22m. The system behaves as a compound pendulum that oscillates about pivot point P. The pivot is at P where the tip of rod rod meets with one end of the disk. Assume that the disc and rod have uniform mass distributions.
a) Find how far below P is the center of mass of the rods and disks system. Answer in units of m.
b) Find the moment of inertia of one of the rod, in units of kgm2, as it freely rotates about the pivot point P.
c) Find the moment of inertia of the disk (the large disk), in units of kgm2, about its center of mass.
d) Find the moment of inertia of the disk (the large disk), in units of kgm2, as it freely rotates about the pivot point P.
e) Find the moment of inertia of the whole system, in units of kgm2, as it freely rotates about the pivot point P.
exam question
a) To find the center of mass of the rods and disks system, we need to consider the masses and positions of each component.
The center of mass of the rod can be found using the formula:
x_cm_rod = (1/2) * L
The center of mass of the large disk can be found using the formula:
x_cm_disk_large = R
The center of mass of the small disk can be found at the glue point position, given as:
x_cm_disk_small = X
The total mass of the system is the sum of the masses of the rod and the two disks:
M_total = M + m_large_disk + m_small_disk
To find the center of mass of the system, we can use the formula:
x_cm_system = (M * x_cm_rod + m_large_disk * x_cm_disk_large + m_small_disk * x_cm_disk_small) / M_total
Plugging in the values given:
x_cm_system = (0.270kg * 0.50m/2 + 0.810kg * 0.130m + 0.405kg * 0.22m) / (0.270kg + 0.810kg + 0.405kg)
Simplifying the equation:
x_cm_system = (0.135kg*m + 0.1053kg*m + 0.0891kg*m) / 1.485kg
x_cm_system = 0.3294kg*m / 1.485kg
x_cm_system ≈ 0.2218m
So, the center of mass of the rods and disks system is approximately 0.2218 meters below point P.
b) The moment of inertia of a rod rotating about an axis passing through one end perpendicular to its length is given by the formula:
I_rod = (1/3) * M * L^2
Plugging in the values given:
I_rod = (1/3) * 0.270kg * (0.50m)^2
Simplifying the equation:
I_rod = 0.005625kg*m^2
So, the moment of inertia of one of the rods about the pivot point P is approximately 0.005625 kgm^2.
c) The moment of inertia of a disk about its center of mass can be calculated using the formula:
I_disk_center = (1/2) * m * R^2
Plugging in the values given:
I_disk_center = (1/2) * 0.810kg * (0.130m)^2
Simplifying the equation:
I_disk_center = 0.065835kg*m^2
So, the moment of inertia of the large disk about its center of mass is approximately 0.065835 kgm^2.
d) To find the moment of inertia of the large disk as it freely rotates about the pivot point P, we need to use the parallel axis theorem. The parallel axis theorem states that the moment of inertia about an axis parallel to and a distance d away from an axis through the center of mass is given by:
I_disk_pivot = I_disk_center + m * d^2
In this case, the distance from the center of mass of the disk to the pivot point P is the radius of the disk R, so we can substitute that in the formula:
I_disk_pivot = I_disk_center + m * R^2
Plugging in the values given:
I_disk_pivot = 0.065835kg*m^2 + 0.810kg * (0.130m)^2
Simplifying the equation:
I_disk_pivot = 0.065835kg*m^2 + 0.810kg * 0.0169m^2
I_disk_pivot = 0.065835kg*m^2 + 0.013779kg*m^2
I_disk_pivot ≈ 0.079614kg*m^2
So, the moment of inertia of the large disk as it freely rotates about the pivot point P is approximately 0.079614 kgm^2.
e) To find the moment of inertia of the whole system as it freely rotates about the pivot point P, we need to sum the individual moments of inertia of the rod and the two disks.
I_system = I_rod + I_disk_large + I_disk_small
Plugging in the values calculated in previous parts:
I_system = 0.005625kg*m^2 + 0.065835kg*m^2 + 0.079614kg*m^2
I_system ≈ 0.151074kg*m^2
So, the moment of inertia of the whole system as it freely rotates about the pivot point P is approximately 0.151074 kgm^2.