The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6.38 x 106 m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 76.0 ° north of the equator.
ω=2π/T= 2π/24•3600= 7.27•10⁻⁵rad/s
1) equator
v= ω•R=7.27•10⁻⁵•6.38•10⁶ = … m/s
a=ω²•R= … m/s²
2) at a latitude of 76.0 ° north of the equator
r=Rcos76.6⁰=6.378•10⁶•cos76.6⁰=1.48•10⁶ m
v= ω•r= … m/s
a=ω²•r= … m/s²
To determine the speed and centripetal acceleration of a person situated at different locations on Earth, we can use the formulas for rotational motion.
Let's start with the speed. The speed of an object moving in a circle can be calculated using the formula:
v = ω * r
Where:
v is the linear velocity (speed)
ω is the angular velocity (angular speed)
r is the radius of the circular path
The angular velocity ω can be found by dividing the angle through which an object rotates by the time it takes to rotate that angle. In this case, the Earth completes one full rotation in 24 hours, or 86400 seconds. Therefore, the angular velocity is:
ω = 2π / T
Where:
T is the period, which is the time it takes for one complete rotation.
Now, let's calculate the speed for each scenario:
(a) Equator:
At the equator, the radius of rotation is equal to the Earth's radius (6.38 x 10^6 m). We know that the period T is 86400 seconds. Plugging this information into the equations, we can calculate:
ω = 2π / 86400 = 7.27 x 10^-5 rad/s
v = (7.27 x 10^-5 rad/s) * (6.38 x 10^6 m) = 465.1 m/s
Therefore, a person at the equator is moving at a speed of approximately 465.1 m/s.
(b) Latitude of 76.0° North of the equator:
To calculate the speed and centripetal acceleration at a specific latitude, we need to consider the effective radius of rotation. This radius can be found using the formula:
R = r * cos(latitude)
Where:
R is the effective radius of rotation
r is the Earth's radius (6.38 x 10^6 m)
latitude is the angle of latitude in radians (76.0° converted to radians is 1.326 rad)
R = (6.38 x 10^6 m) * cos(1.326 rad) = 2.42 x 10^6 m
Now we can calculate the speed:
ω = 2π / 86400 = 7.27 x 10^-5 rad/s
v = (7.27 x 10^-5 rad/s) * (2.42 x 10^6 m) = 176.3 m/s
Therefore, a person at a latitude of 76.0° North of the equator is moving at a speed of approximately 176.3 m/s.
Now, let's move on to calculating the centripetal acceleration.
The centripetal acceleration of an object moving in a circle can be found using the formula:
a = ω^2 * r
Where:
a is the centripetal acceleration
r is the radius of the circular path
ω is the angular velocity
Let's calculate the centripetal acceleration for both scenarios:
(a) At the equator:
Using the radius of rotation (6.38 x 10^6 m) and the angular velocity (7.27 x 10^-5 rad/s) that we calculated earlier, we can find the centripetal acceleration:
a = (7.27 x 10^-5 rad/s)^2 * (6.38 x 10^6 m) = 0.034 m/s^2
Therefore, a person at the equator experiences a centripetal acceleration of approximately 0.034 m/s^2.
(b) Latitude of 76.0° North of the equator:
Using the effective radius of rotation (2.42 x 10^6 m) and the angular velocity (7.27 x 10^-5 rad/s) that we calculated earlier, we can find the centripetal acceleration:
a = (7.27 x 10^-5 rad/s)^2 * (2.42 x 10^6 m) = 0.005 m/s^2
Therefore, a person at a latitude of 76.0° North of the equator experiences a centripetal acceleration of approximately 0.005 m/s^2.