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
The height above the ground from where the skier start is 11.5 m.
<h3>
Conservation of energy</h3>
The height above the ground from where the skier start is determined by applying the principle of conservation of energy as shown below;
P.E = K.E
mgh = ¹/₂mv²
gh = ¹/₂v²

Thus, the height above the ground from where the skier start is 11.5 m.
Learn more about conservation of energy here: brainly.com/question/166559
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Answer:This also means that Mercury's surface gravity is 3.7 m/s2, which is the equivalent of 38% of Earth's gravity (0.38 g). This means that if you weighed 100 kg (220 lbs) on Earth, you would weigh 38 kg (84 lbs) on Mercury.
Explanation: