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Calculate the angular velocity of the Earth \((a)\) in its orbit around the Sun, and \((b)\) about its axis.

Short Answer

Expert verified
(a) Earth's orbital angular velocity is approximately \( 1.99 \times 10^{-7} \) radians/second. (b) Earth's rotational angular velocity is approximately \( 7.27 \times 10^{-5} \) radians/second.

Step by step solution

01

Understanding Angular Velocity

Angular velocity \( \omega \) is defined as the rate of change of angular displacement and is given by \( \omega = \frac{\theta}{t} \), where \( \theta \) is the angle in radians and \( t \) is the time in seconds. We need to find two separate angular velocities: (a) the Earth's orbit around the Sun, and (b) the Earth's rotation about its axis.
02

Calculate Earth's Orbital Angular Velocity

To find the Earth's angular velocity in its orbit around the Sun, we consider that it completes one revolution ( \( 2\pi \) radians) in approximately 365.25 days. Convert the time period into seconds: \[ t = 365.25 \times 24 \times 60 \times 60 = 31,557,600 \text{ seconds} \] Then, calculate the angular velocity: \[ \omega = \frac{2\pi}{31,557,600} \] This gives \( \omega \approx 1.99 \times 10^{-7} \text{ radians/second} \) as the Earth's orbital angular velocity.
03

Calculate Earth's Rotational Angular Velocity

For the Earth's rotation about its axis, it completes one full rotation ( \( 2\pi \) radians) in approximately 24 hours. Convert the time period into seconds: \[ t = 24 \times 60 \times 60 = 86,400 \text{ seconds} \] Then, calculate the angular velocity: \[ \omega = \frac{2\pi}{86,400} \] This gives \( \omega \approx 7.27 \times 10^{-5} \text{ radians/second} \) as the Earth's rotational angular velocity.

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Key Concepts

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

Earth's orbit
The Earth's orbit is the path along which it travels around the Sun. This path is nearly circular, with a slight elliptical shape, meaning the distance between the Earth and the Sun changes slightly as it moves along its orbit. This journey around the Sun is what defines a year, taking approximately 365.25 days to complete one full orbit.

When considering the concept of angular velocity, the Earth's orbit can be understood in terms of both rotation and revolution. The planet is not only spinning on its own axis but is also revolving around the Sun. This movement is what leads to the cycle of seasons as the Earth tilts its axis during its orbit, bringing different parts of the planet closer to or further from the sun at different times of the year.

In terms of angular displacement, the Earth completes a full circle around the Sun, which we can equate to an angular displacement of \(2\pi\) radians. To find the angular velocity \(\omega\), we divide this angular displacement by the time it takes for this journey, about 31,557,600 seconds, resulting in a small angular velocity of approximately \(1.99 \times 10^{-7}\) radians per second.

To summarize, the Earth's orbit may seem vast and slow from our point of view, but it plays a crucial role in creating the rhythms of our days and seasons.
Earth's rotation
Earth's rotation refers to the spinning motion of the planet around its own axis. This axis is an imaginary line that runs through the North and South Poles. The act of rotating gives us our daily cycle of day and night as different areas of the Earth's surface are exposed to sunlight.

The planet completes a full spin every 24 hours, almost like an invisible top spinning in space. This time period is a day as we know it. Scientific calculations show that this rotation is consistent, leading to an angular displacement of \(2\pi\) radians every day. In seconds, this duration is 86,400, which helps us determine the Earth's angular velocity of about \(7.27 \times 10^{-5}\) radians/second for rotation.

Understanding Earth's rotation is key to grasping aspects like the time zone divisions and even the Coriolis effect—an interesting phenomenon that influences weather patterns and ocean currents due to the rotation speed.
  • Creates time divisions - morning, afternoon, evening, night
  • Influences climate and weather through rotation-induced Coriolis effect
In essence, Earth's rotation is a fundamental process that defines not only our daily time structure but also has broader implications on the planet's environmental systems.
Radian measurement
Radian measurement is a way of measuring angles, essential for understanding rotational movements like Earth's orbit and rotation. Unlike degrees, radians provide a natural and mathematical approach to angular measurement. A radian relates the diameter of a circle to the length of the arc created by an angle. When the arc's length is equal to the circle's radius, the angle is one radian.

Since a complete circle revolves around \(2\pi\) radians, we use radians to calculate things like angular velocity with great precision. For instance, when we say the Earth rotates \(2\pi\) radians, we're acknowledging it makes one full turn.

Utilizing radians in computations offers a direct correlation between linear and angular speeds, simplifying the process of measuring angles and calculating pathways essential in the fields of physics, engineering, and astronomy.
  • Foundation for understanding circular motion
  • Simplifies mathematical calculations in relation to angular velocity
As a fundamental unit in trigonometry and calculus, radian measurement is not just a technical detail but an essential component of how we analyze and understand rotational dynamics, especially useful when computing Earth's movements in this exercise.

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Most popular questions from this chapter

On a 12.0 .0 -cm-diameter audio compact disc (CD), digital bits of information are encoded sequentially along an outward spiraling path. The spiral starts at radius \(R_{1}=2.5 \mathrm{cm}\) and winds its way out to radius \(R_{2}=5.8 \mathrm{cm} .\) To read the digital information, a CD player rotates the CD so that the player's readout laser scans along the spiral's sequence of bits at a constant linear speed of 1.25 \(\mathrm{m} / \mathrm{s}\) . Thus the player must accu- rately adjust the rotational frequency \(f\) of the \(\mathrm{CD}\) as the laser moves outward. Determine the values for \(f\) (in units of rpm) when the laser is located at \(R_{1}\) and when it is at \(R_{2}\) .

A marble of mass \(m\) and radius \(r\) rolls along the looped rough track of Fig. \(10-67 .\) What is the minimum value of the vertical height \(h\) that the marble must drop if it is to reach the highest point of the loop without leaving the track? (a) Assume \(r \ll R ;\) (b) do not make this assumption. Ignore frictional losses.

A 1.4 -kg grindstone in the shape of a uniform cylinder of radius \(0.20 \mathrm{~m}\) acquires a rotational rate of \(1800 \mathrm{rev} / \mathrm{s}\) from rest over a 6.0 -s interval at constant angular acceleration. Calculate the torque delivered by the motor.

(II) A sphere of radius \(r_{0}=24.5 \mathrm{cm}\) and mass \(m=1.20 \mathrm{kg}\) starts from rest and rolls without slipping down a \(30.0^{\circ}\) incline that is 10.0 \(\mathrm{m}\) long. (a) Calculate its translational and rotational speeds when it reaches the bottom. (b) What is the ratio of translational to rotational kinetic energy at the bottom? Avoid putting in numbers until the end so you can answer: \((c)\) do your answers in \((a)\) and \((b)\) depend on the radius of the sphere or its mass?

(II) A merry-go-round accelerates from rest to 0.68 \(\mathrm{rad} / \mathrm{s}\) in 24 \(\mathrm{s}\) . Assuming the merry-go-round is a uniform disk of radius 7.0 \(\mathrm{m}\) and mass \(31,000 \mathrm{kg}\) , calculate the net torque required to accelerate it.

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