- #1
andyonassis
- 4
- 1
- Homework Statement
- A plane flies across the north pole at 223.52 m/s, following a meridian of longitude (which rotates around the earth). Find the angle between the direction of a plumb line hanging freely as it passes over the pole and one hanging freely at the surface of the earth over the pole.
- Relevant Equations
- F = ma
$$\omega = \frac{2\pi}{T} $$
From Newton's second law:
$$T_{x} = F_{turn}$$
So
$$T \sin \theta = ma$$
$$T_{y} = F_{y}$$
so
$$T \cos \theta = mg$$
Equate the two equations to get:
$$ \frac{T \sin \theta}{a} = \frac{T \cos \alpha}{g} $$
and the angle is given by:
$$tan (\theta) = \frac{a}{g} $$
where ##r = \frac{v}{w}## and ##a = r\omega^{2}##
From this we get:
$$\theta = arctan(\frac{v\omega}{g}) = arctan(\frac{223.52}{9.8} \times \frac{2\pi}{24\times 60 \times 60}) = 1.65 \times 10^{-3} radians$$
However, the book says ##\omega = 2.31 \times 10^-4 rad/s## so ##\theta = 5.25 \times 10^{-4}##radians.
$$T_{x} = F_{turn}$$
So
$$T \sin \theta = ma$$
$$T_{y} = F_{y}$$
so
$$T \cos \theta = mg$$
Equate the two equations to get:
$$ \frac{T \sin \theta}{a} = \frac{T \cos \alpha}{g} $$
and the angle is given by:
$$tan (\theta) = \frac{a}{g} $$
where ##r = \frac{v}{w}## and ##a = r\omega^{2}##
From this we get:
$$\theta = arctan(\frac{v\omega}{g}) = arctan(\frac{223.52}{9.8} \times \frac{2\pi}{24\times 60 \times 60}) = 1.65 \times 10^{-3} radians$$
However, the book says ##\omega = 2.31 \times 10^-4 rad/s## so ##\theta = 5.25 \times 10^{-4}##radians.
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