Moment of inertia Physical pendulum?
Consider the physical pendulum shown in the figure, which consists of a thin rod of length l. Its total mass m is distributed evenly along its length.
Calculate the moment of inertia with respect to the axis of rotation passing through the suspension point shown (distance l/3 from the upper end).
(b) Derive the equation of motion for the deflection angle α and, using an oscillation approach for α(t), determine the oscillation frequency ω0 of the pendulum. Neglect friction. Note for (b): sin(180◦ − α) = sin(α). Then use a small-angle approximation.
Jensek81 approach
a) J =∫ r² dm = m/l ∫(0 to l) dr r² = m/L * 1/3 |³ = 1 /3 ml²
b) M = L/2 * FG
M = Jw = Yes
=> -l/2 mg α = 1/3 m |² ider α'' = -3/2 g /l α = – ω0²α
SO ω0² = 3/2 g/l or ω0 = √3/2 g/l
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Isn’t your J just for a rod hanging out at the end? What did you do with the l/3?
I don’t know how icke (=I) should integrate it into the account.
At first, I would set the moment around the point of rotation:
You have an upper part where the center of gravity L/6 is remote from the center of gravity G/3, and a lower part with L/3 and the weight 2G/3.
The resulting moment is then
M = sin(α)*(l/3*2G/3 – l/6*G/3) = G*L/6*sin(α) ~ G*L/6*α
naja: the returning torque is not -mgl sin(α). The center of gravity has the distance L/6 from the center of rotation, so is the returning moment
M= -mgL/6 sin(α) (see above)
So J a” = -mgl sin (alpha)
ML2 / ) = – mg alppha
where2 = root from (9 mg / ML).
Can that be?
The moment of inertia around the center is Js=ML2/12
The inertia monster around the pivot point, which is offset by L/6 with respect to the heavy powder, is then
J=ML2/12 + ML2/36 = ML2/9
Now you can bring everything together…
And for the moment of inertia around the point of rotation, you have to work with the Steiner’s set.