Final answer:
The minimum period of oscillation for a uniform rod with a movable pivot and a length of 0.310 m is approximately 0.56 seconds, which can be found using the formula for the period of a physical pendulum.
Step-by-step explanation:
To calculate the minimum period of oscillation of a uniform rod with a movable pivot point, we can use the formula for a physical pendulum:
T = 2π√(I/mgh)
where I is the moment of inertia, m is the mass of the rod, g is the acceleration due to gravity (9.81 m/s²), and h is the distance from the pivot to the center of mass. For a uniform rod of length L, the moment of inertia about an axis through the end is I = (1/3)mL². The distance h in this case would be L/2 since the rod is uniform and the center of mass would be at the midpoint.
The mass of the rod is not provided in the question, but it will cancel out in the equation because it appears in both the numerator and denominator. Thus, the period T depends only on the length of the rod and the acceleration due to gravity. Substituting these values we get:
T = 2π√((1/3)L²/(g(L/2)))
T = 2π√((2/3)L/g)
Plug in the provided length of L = 0.310 m, we find the minimum period of oscillation.
T = 2π√((2/3)×0.310 m / 9.81 m/s²)
T ≈ 0.56 seconds
The minimum period of oscillation for the uniform rod is therefore approximately 0.56 seconds.