Answer:
it will be 1/√2 of its original period.
Explanation:
I think it is C) the resting position of the wave.
In Newton's cannonball experiment, if the velocity is equal to the orbital velocity then the cannonball will stay in Orbit.
Newtons cannonball experiment stated that the distance that a cannonball will travel, before being drawn into the Earth by the forces of gravity, is dependent on the initial velocity.
Therefore, if the cannonball is launched at a velocity that matches the orbital velocity, then it will not be able to be drawn in by gravity due to the Earth moving away from the cannonball at the same speed at which the cannonball itself is falling.
This means that the cannonball will continue to fall without reaching the Earth, therefore staying in orbit, much like that of the moon or planets around the sun.
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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
