Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
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Answer:
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A type O star is likely to appear blue.
Answer:
the position of the wood below the interface of the two liquids is 2.39 cm.
Explanation:
Given;
density of oil, 
= 926 kg/m³
density of the wood, 
= 974 kg/m³
density of water, 
= 1000 kg/m³
height of the wood, h = 3.69 cm
Based on the density of the wood, it will position across the two liquids.
let the position of the wood below the interface of the two liquids = x
Let the wood be in equilibrium position;
![F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm](https://tex.z-dn.net/?f=F_%7Bwood%7D%20-%20F_%7Boil%7D%20-%20F_%7Bwater%7D%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.gh%20-%20%5Crho%20_o%20.g%28h-x%29%20-%20%5Crho_w%20.gx%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.h%20-%20%5Crho%20_o%20%28h-x%29%20-%20%5Crho_w%20.x%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.h%20-%5Crho%20_o%20h%20%2B%20%5Crho%20_o%20x%20-%20%5Crho_w%20.x%20%3D0%5C%5C%5C%5Ch%20%28%5Crho%20_%7Bwood%7D%20%20-%5Crho%20_o%20%29%20%3D%20x%28%20%5Crho_w%20-%20%5Crho%20_o%29%5C%5C%5C%5Cx%20%3Dh%5B%5Cfrac%7B%20%5Crho%20_%7Bwood%7D%20%20-%5Crho%20_o%20%7D%7B%5Crho_w%20-%20%5Crho%20_o%7D%20%5D%5C%5C%5C%5Cx%20%3D%203.69%5C%20cm%20%5Ctimes%20%5B%5Cfrac%7B974%20-%20926%7D%7B1000-926%7D%20%5D%5C%5C%5C%5Cx%20%3D%202.39%20%5C%20cm)
Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.
