<span>1.57 seconds.
The rod hanging from the nail constructs a physical pendulum. The period of such a pendulum follows the formula
T = 2*pi*sqrt(L/g)
where
T = time
L = length of pendulum
g = local gravitational acceleration
So the problem becomes one of determining L. It's tempting to consider L to be the distance between the center of mass and the pivot, but that isn't the right value. The correct value is the distance between the pivot and the center of percussion. So let's determine what that is. We can treat the uniform thin rod as an uniform beam and for an uniform beam the distance between the center of mass and the center of percussion is expressed as
b = L^2/(12A)
where
b = distance between center of mass and center of percussion
L = length of beam
A = distance between pivot and center of mass
Since the rod is uniform, the CoM will be midway from either end, or 0.962 m / 2 = 0.481 m from the end. The pivot will therefore be 0.481 m - 0.048 m = 0.433 m from the CoM
Now let's calculate the distance the CoP will be from the CoM:
b = L^2/(12A)
b = (0.962 m)^2/(12 * 0.433 m)
b = (0.925444 m^2)/(5.196 m)
b = 0.178107005 m
With the distance between the CoM and CoP known, we can now calculate the effective length of the pendulum. So:
0.433 m + 0.178107005 m = 0.611107005 m
And finally, with the effective length known, let's calculate the period.
T = 2*pi*sqrt(L/g)
T = 2*pi*sqrt((0.611107005 m)/(9.8 m/s^2))
T = 2*pi*sqrt(0.062357858 s^2)
T = 2*pi*0.249715554 s
T = 1.569009097 s
Rounding to 3 significant figures gives 1.57 seconds.
Let's check if this result is sane. Looking up "Seconds Pendulum", I get a length of 0.994 meters which is longer than the length of 0.611 meters calculated. But upon looking closer at the "Seconds Pendulum", you'll realize that it's period is actually 2 seconds, or 1 second per swing. So the length of the calculated pendulum is sane.</span>
