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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

Engineers are designing a system by which a falling mass \(m\) imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (Fig. \(\mathbf{P 9 . 6 4}\) ). There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is \(3.71 \mathrm{~m} / \mathrm{s}^{2} .\) In the earth tests, when \(m\) is set to \(15.0 \mathrm{~kg}\) and allowed to fall through \(5.00 \mathrm{~m},\) it gives \(250.0 \mathrm{~J}\) of kinetic energy to the drum. (a) If the system is operated on Mars, through what distance would the \(15.0 \mathrm{~kg}\) mass have to fall to give the same amount of kinetic energy to the drum? (b) How fast would the \(15.0 \mathrm{~kg}\) mass be moving on Mars just as the drum gained \(250.0 \mathrm{~J}\) of kinetic energy?

The angular velocity of a flywheel obeys the equation \(\omega_{z}(t)=A+B t^{2},\) where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75 (for \(A\) ) and 1.50 (for \(B\) ). (a) What are the units of \(A\) and \(B\) if \(\omega_{z}\) is in \(\mathrm{rad} / \mathrm{s} ?\) (b) What is the angular acceleration of the wheel at (i) \(t=0\) and (ii) \(t=5.00 \mathrm{~s} ?\) (c) Through what angle does the flywheel turn during the first 2.00 s? (Hint: See Section \(2.6 .)\)

\(\mathrm{At} t=0\) a grinding wheel has an angular velocity of \(24.0 \mathrm{rad} / \mathrm{s}\) It has a constant angular acceleration of \(30.0 \mathrm{rad} / \mathrm{s}^{2}\) until a circuit breaker trips at \(t=2.00 \mathrm{~s}\). From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t=0\) and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

A thin, light wire is wrapped around the rim of a wheel (Fig. E9.49). The wheel rotates without friction about a stationary horizontal axis that passes through the center of the wheel. The wheel is a uniform disk with radius \(R=0.280 \mathrm{~m}\). An object of mass \(m=4.20 \mathrm{~kg}\) is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration. If the suspended object moves downward a distance of \(3.00 \mathrm{~m}\) in \(2.00 \mathrm{~s},\) what is the mass of the wheel?

A fan blade rotates with angular velocity given by \(\omega_{z}(t)=\gamma-\beta t^{2}, \quad\) where \(\quad \gamma=5.00 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.800 \mathrm{rad} / \mathrm{s}^{3}\) (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration \(\alpha_{z}\) at \(t=3.00 \mathrm{~s}\) and the average angular acceleration \(\alpha_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=3.00 \mathrm{~s}\). How do these two quantities compare? If they are different, why?

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