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The earth is approximately spherical, with a diameter of \(1.27 \times 10^{7} \mathrm{~m} .\) It takes 24.0 hours for the earth to complete one revolution. What are the tangential speed and radial acceleration of a point on the surface of the earth, at the equator?

Short Answer

Expert verified
The tangential speed of a point on the surface of the Earth, at the equator, is approximately \(465 m/s\), and the radial acceleration is about \(0.0339 m/s^2\).

Step by step solution

01

Calculation of the radius of the Earth

We know that diameter \(D = 1.27 \times 10^{7} m\). The radius \(r\) can be calculated from the diameter using the formula \(r = D/2\).
02

Calculation of the tangential speed of a point on the surface of the Earth

The tangential speed \(v\) can be calculated using the formula for the circumference of a circle divided by the time period. The circumference of a circle is \(2\pi r\), and the time period for one complete revolution of the Earth is 24 hours, which is equivalent to 86400 seconds. Therefore, \(v = 2\pi r / T\), where \(T\) is the time period.
03

Calculation of the radial acceleration of a point on the surface of the Earth

The radial acceleration can be found by using the formula \(a = v^2/r\), where \(v\) is the speed and \(r\) is the radius.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangential Speed
Tangential speed refers to how fast a point on the edge of a rotating object moves. For the Earth, this is particularly important at the equator, where the journey around the Earth's circumference is longest. Imagine Earth spinning like a giant merry-go-round. A person standing at the equator moves through more space in a set time compared to someone standing closer to one of the poles, and thus has the highest tangential speed on Earth.

To find this speed, you use the formula for tangential speed, which involves dividing the circumference of Earth (a big circle, after all) by the time Earth takes to make one full spin (a day, or 86400 seconds). The formula is given by:
  • Formula: \[ v = \frac{2\pi r}{T} \]
  • Where:
    • \(v\) is the tangential speed.
    • \(r\) is Earth's radius, derived from its diameter.
    • \(T\) is the time period in seconds.
By understanding the tangential speed, you get a better grasp of the dynamics of rotating systems like Earth, especially useful in fields related to geophysics and meteorology.
Radial Acceleration
Radial acceleration is the acceleration directed towards the center of the circle along which a rotating point moves, also known as centripetal acceleration. At the Earth's surface, particularly at the equator, this force is crucial in keeping you from flying off into space due to the planet's rotation.

For a point on the Earth’s surface, radial acceleration can be found using the formula:
  • Formula: \[ a = \frac{v^2}{r} \]
  • Where:
    • \(a\) is the radial acceleration.
    • \(v\) is the tangential speed we just calculated.
    • \(r\) is the radius of the Earth.
Understanding radial acceleration helps in grasping why the sensation of gravity is slightly less at the equator compared to the poles. This difference is due to the slight but significant centrifugal force effect experienced due to Earth's rotation.
Earth's Rotation
Earth's rotation is the spinning of the planet around its axis. One complete rotation takes approximately 24 hours, giving us the cycle of day and night. This rotation plays a crucial role in defining climatic patterns and the apparent movement of the sun across the sky.

A few important points about Earth's rotation include:
  • It results in tangential speed and radial acceleration, affecting everything on the surface.
  • The rotation is responsible for the Coriolis Effect, which influences currents in ocean and air, steering weather patterns.
  • Equatorial regions spin faster than polar regions, affecting the shape of Earth slightly, making it an oblate spheroid.
By understanding Earth's rotation, you see its influence on weather, tides, and even technological systems like GPS, which must account for these motions to maintain accuracy. Recognizing these concepts in everyday phenomena enriches your comprehension of natural sciences.

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Most popular questions from this chapter

At \(t=0\) the current to a dc electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t)=(250 \mathrm{rad} / \mathrm{s}) t-\left(20.0 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}-\left(1.50 \mathrm{rad} / \mathrm{s}^{3}\right) t^{3} .\) (a) At what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at \(t=0,\) when the current was reversed? (e) Calculate the average angular velocity for the time period from \(t=0\) to the time calculated in part (a).

\(9.88^{\circ}\) DATA You are analyzing the motion of a large flywheel that has radius \(0.800 \mathrm{~m}\). In one test run, the wheel starts from rest and turns in a horizontal plane with constant angular acceleration. An accelerometer on the rim of the flywheel measures the magnitude of the resultant acceleration \(a\) of a point on the rim of the flywheel as a function of the angle \(\theta-\theta_{0}\) through which the wheel has turned. You collect these results: $$ \begin{array}{l|cccccccc} \boldsymbol{\theta}-\boldsymbol{\theta}_{\mathbf{0}}(\mathbf{r a d}) & 0.50 & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 3.50 & 4.00 \\ \hline \boldsymbol{a}\left(\mathbf{m} / \mathbf{s}^{\mathbf{2}}\right) & 0.678 & 1.07 & 1.52 & 1.98 & 2.45 & 2.92 & 3.39 & 3.87 \end{array} $$ Construct a graph of \(a^{2}\left(\right.\) in \(\left.\mathrm{m}^{2} / \mathrm{s}^{4}\right)\) versus \(\left(\theta-\theta_{0}\right)^{2}\) (in rad \(^{2}\) ). (a) What are the slope and \(y\) -intercept of the straight line that gives the best fit to the data? (b) Use the slope from part (a) to find the angular acceleration of the flywheel. (c) What is the linear speed of a point on the rim of the flywheel when the wheel has turned through an angle of \(135^{\circ} ?\) (d) When the flywheel has turned through an angle of \(90.0^{\circ},\) what is the angle between the linear velocity of a point on its rim and the resultant acceleration of that point?

You need to design an industrial turntable that is \(60.0 \mathrm{~cm}\) in diameter and has a kinetic energy of \(0.250 \mathrm{~J}\) when turning at \(45.0 \mathrm{rpm}\) (rev/min). (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?

An airplane propeller is rotating at 1900 rpm (rev/min). (a) Compute the propeller's angular velocity in rad/s. (b) How many seconds does it take for the propeller to turn through \(35^{\circ} ?\)

A hollow spherical shell has mass \(8.20 \mathrm{~kg}\) and radius \(0.220 \mathrm{~m} .\) It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of \(0.890 \mathrm{rad} / \mathrm{s}^{2}\). What is the kinetic energy of the shell after it has turned through 6.00 rev?

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