A very light rigid rod (0.5m) is connected to a more massive (0.50 kg) rigid 0.65 m rod, connected end to end. At the free end of a light rod,it is attached to the ceiling and pivots at that connection. Assuming the system is in Simple Harmonic Motion, Calculate the period of the system’s oscillation.
To calculate the period of the system's oscillation, we need to consider the physical properties of the rods and apply the principles of simple harmonic motion.
The period of an oscillating system is defined as the time it takes for one complete cycle or one full oscillation. In the case of this system, we have two connected rods, and the oscillation is driven by the force of gravity acting on the more massive rod.
To begin, let's break down the system and analyze it step by step:
1. Moment of inertia (I):
The moment of inertia is a physical property that describes an object's resistance to changes in rotational motion. In this case, we have two rods rotating around a pivot point. For simplicity, we will treat the rods as slender rods rotating about their end points.
For the light rigid rod (0.5m), the moment of inertia can be calculated using the formula:
I1 = (1/3) * m1 * L1^2
where m1 is the mass of the rod and L1 is its length. Since the rod is light, we can assume its mass is negligible compared to the more massive rod.
2. Period formula:
The period (T) of an oscillating system can be calculated using the formula:
T = 2π * sqrt(I / mgd)
where I is the moment of inertia of the system, m is the mass of the more massive rod, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of mass of the more massive rod.
3. Center of mass (CM):
To calculate the distance (d), we need to find the center of mass of the system. Since the two rods are connected end to end, the center of mass will be at the midpoint between the two rods.
The distance from the pivot point to the center of mass can be calculated as:
d = (L1 + L2) / 2
where L1 is the length of the light rod and L2 is the length of the more massive rod.
4. Moment of inertia (I):
For the more massive rod (0.50 kg, 0.65 m), the moment of inertia can be calculated using the formula:
I2 = (1/3) * m2 * L2^2
where m2 is the mass of the more massive rod and L2 is its length.
Now let's plug in the values and calculate the period:
m1 = mass of the light rod = 0.00 kg
L1 = length of the light rod = 0.50 m
m2 = mass of the more massive rod = 0.50 kg
L2 = length of the more massive rod = 0.65 m
g = acceleration due to gravity = 9.8 m/s^2
d = (L1 + L2) / 2 = (0.50 m + 0.65 m) / 2 = 0.575 m
I1 = (1/3) * m1 * L1^2 = (1/3) * 0.0 kg * (0.50 m)^2 = 0.0 kgm^2
I2 = (1/3) * m2 * L2^2 = (1/3) * 0.50 kg * (0.65 m)^2 = 0.056875 kgm^2
I = I1 + I2 = 0.0 kgm^2 + 0.056875 kgm^2 = 0.056875 kgm^2
Now we can substitute the values into the period formula:
T = 2π * sqrt(I / mgd)
= 2π * sqrt(0.056875 kgm^2 / (0.50 kg * 9.8 m/s^2 * 0.575 m))
= 2π * sqrt(0.11607143)
= 2π * 0.3408100
≈ 2.13657 seconds
Therefore, the period of the system's oscillation is approximately 2.14 seconds.