Interestingly, there have been several studies using cadavers to determine the moment of inertia of human body parts by letting them swing as a pendulum about a joint. In one study, the center of gravity of a 5.0 kg lower leg was found to be 18 cm from the knee. When pivoted at the knee and allowed to swing, the oscillation frequency was 1.6 Hz.What was the moment of inertia of the lower leg?
PLEASE WORK IT OUT, I HAVE TRIED DO IT, BUT AM NOT GETTING THE RIGHT ANSWER, PLEASE HELP
THANKS
period= 2PI sqrt (I/mg*.18)
I= mg*.18*Period^2/(2PI)^2
check that. Period= 1/1.6 sec
To find the moment of inertia of the lower leg, you can use the formula for the period of a simple pendulum:
T = 2π√(I / mg),
where T is the period, I is the moment of inertia, m is the mass, and g is the acceleration due to gravity.
In this case, you are given the oscillation frequency (f), not the period (T). The period (T) is the inverse of the frequency (f), so you can calculate the period using:
T = 1 / f.
In the given problem, the frequency (f) is 1.6 Hz, so the period (T) would be:
T = 1 / 1.6 = 0.625 s.
Next, you need to convert the period (T) into angular frequency (ω) using the formula:
T = 2π / ω.
Rearranging the formula, you can solve for angular frequency (ω):
ω = 2π / T.
Plugging in the value for T:
ω = 2π / 0.625 = 10.043 s^(-1).
Now, you can use the formula for angular frequency (ω) in a pendulum:
ω = √(g / L),
where L is the length from the axis of rotation (knee) to the center of gravity of the lower leg (18 cm = 0.18 m).
Rearranging the formula, you can solve for g (acceleration due to gravity):
g = ω^2 * L.
Plugging in the values, you get:
g = (10.043 s^(-1))^2 * 0.18 m = 1.808 m/s^2.
Finally, you can use the formula for moment of inertia (I) of a pendulum:
I = m * L^2 / g.
Plugging in the values, you get:
I = 5.0 kg * (0.18 m)^2 / 1.808 m/s^2 ≈ 0.464 kg·m².
Therefore, the moment of inertia of the lower leg is approximately 0.464 kg·m².