Respuesta :
Answer:
a) [tex] v = R w = 6400000 m *7.27 x10^{-5} \frac{rad}{s}= 465.421 m/s[/tex]
b) [tex] F_c = 98 kg* (6400000 m) (7.27 x10^{-5} \frac{rad}{s})^2 = 3.316N[/tex]
c) [tex] W = 98kg * 9.8 m/s^2 = 960.4 N[/tex]
d) [tex] W_a = 960.4 N -3.316 N = 957.04 N[/tex]
Explanation:
For this case we have the following data:
[tex] R= 6.4x10^6 m = 6400000 m[/tex] represent the radius of the Earth
The gravity for this case is assumed [tex] g = 9.8 m/s^2[/tex]
m =98 kg represent the mass of the person
[tex] w=\frac{2\pi}{T} =\frac{2\pi}{24 hr *\frac{3600 s}{1hr}}= 7.27 x10^{-5} \frac{rad}{s}[/tex] represent the angular velocity for the Earth
Part a
Since we need to find the speed of the person at the equator we can use the following formula:
[tex] v = R w = 6400000 m *7.27 x10^{-5} \frac{rad}{s}= 465.421 m/s[/tex]
Part b
We can use the formula for the centripetal force for the person at the equator to allow the rotation like this:
[tex] F_c = m R w^2[/tex]
And if we replace we got:
[tex] F_c = 98 kg* (6400000 m) (7.27 x10^{-5} \frac{rad}{s})^2 = 3.316N[/tex]
Part c
The weight of the person is just defined as:
[tex] W = mg[/tex]
And if we replace we got:
[tex] W = 98kg * 9.8 m/s^2 = 960.4 N[/tex]
Part d
The normal force of Earth on the person represent the apparent weight and is given by:
[tex] W_a = W - F_c[/tex]
And since we have the weight and the centripetal forc we can replace like this:
[tex] W_a = 960.4 N -3.316 N = 957.04 N[/tex]
The Earth is rotating at a speed, that is slow enough for an object to
require only a small fraction of its weight remain on the Earth's surface.
These are the correct responses;
(a) The speed of the person at the equator is approximately 465.42 m/s.
(b) The force required to maintain circular motion is approximately 3.317 N.
(c) Weight of the person is the person is approximately 961.38 N.
(d) The normal force, of Earth on the person, is approximately 958.06 N.
Reasons:
(a) Linear speed at the equator is given by the equation;
[tex]v = \dfrac{2 \cdot \pi}{T} \times R[/tex]
Where;
v = The linear speed at the equator
T = The time it takes the Earth to complete rotation about its axis = 24 hours
R = The radius of the Earth = 6,400 km = 6,400,000 m
24 hours = 24 hr. × 60 min/hr. × 60 s/min = 86,400 seconds
[tex]v = \dfrac{2 \cdot \pi}{86,400} \times 6,400,000 = 465.42[/tex]
The linear speed of the Earth at the equator, is the speed of the 98 kg
person at the equator, v = 465.42 m/s
(b) The centripetal force required to maintain a circular motion is, [tex]F = \dfrac{m \cdot v^2}{R}[/tex]
Therefore;
[tex]\mathrm{The \ force \ requied , \ F} \approx \dfrac{98 \times 465.42^2}{6,400,000} \approx 3.317 \, N[/tex]
(c) The weight of the person, W = m·g
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
∴ W ≈ 98 × 9.81 = 961.38
Weight of the person, W ≈ 961.38 N
(d) The normal force is the reaction of the apparent (net downward) weight of the person.
The normal force reaction of the Earth, [tex]F_N[/tex] = Weight - Centrifugal force
Centrifugal force = -Centripetal force
∴ [tex]F_N[/tex] ≈ 961.38 N - 3.317 N = 958.06 N
The normal force, of Earth on the person [tex]F_N[/tex] ≈ 958.06 N
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