Respuesta :
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
a) The earth's angular speed is, [tex]w = 7.27 * 10^{-5} rad/s[/tex]
b) The speed of a point on the equator, v₁ = 463.1 m/s
c) The speed of a point, v₁ = 440.433 m/s
Given:
The radius of the earth, [tex]R = 6.37 * 10 ^6 m[/tex]
The time period for 1 revolution T = 24 hrs
Calculation of Angular speed:
The angular speed w of the earth can be related to the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
[tex]w = 7.27 * 10^{-5} rad/s[/tex]
The speed of the point on the equator v₁ can be determined from the linear and rotational motion kinematic relation.
v₁ = R*w
[tex]v_1 = (6.37 * 10 ^6)*(7.27 * 10^{-5})[/tex]
v₁ = 463.1 m/s
The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
The radius of the earth R' at a point where θ = 18° from the equator is:
R' = R*cos(18)
[tex]R' = (6.37 * 10 ^6)*cos(18)[/tex]
R' = 6058230.0088 m
The speed of the point where θ = 18° from the equator v₂ can be determined from the linear and rotational motion kinematic relation.
v₂ = R'*w
[tex]v_2 = (6058230.0088)*(7.27 * 10^{-5})[/tex]
v₂ = 440.433 m/s
Find more information about Angular speed here:
brainly.com/question/6860269
