Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 38

(a) What is the centripetal acceleration of an object on the Earth's equator due to the rotation of the Earth? (b) A \(25.0-\mathrm{kg}\) object hangs from a spring scale at the equator. If the free-fall acceleration due only to the Earth's gravity is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\), what is the reading on the spring scale?

Short Answer

Expert verified
For part (a) we find that the centripetal acceleration of an object on the Earth's equator due to the Earth's rotation is approximately \(0.034 m/s^2\). For part (b) the spring scale reading, after considering the net impact of both gravitational and centripetal forces, is about \(245 N\).

Step by step solution

01

Calculate the Earth's Rotational Speed

The velocity of the rotation on the equator, \(v\), can be calculated by using the formula \(v = 2\pi r/T\), where \(r\) is the Earth's radius and \(T\) is the period of rotation (24 hours or 86400 seconds). The Earth's radius can be approximated as 6.37 x \(10^6\) m.
02

Find the Centripetal Acceleration

Once we have the velocity, we can substitute its value into the formula for centripetal acceleration: \(a_c = \frac{v^2}{r}\). This will give us the centripetal acceleration at the Earth's equator due to its rotation.
03

Calculate the Effect on the 25.0 kg Object

The effect of this acceleration on the reading of the spring scale can be found by finding the force exerted by this mass. It should be noted that the net effect of centripetal acceleration and gravitational acceleration acting on the object at the equator should be considered. As the object moves in a circular path due to Earth's rotation, the net force will be the gravitational force minus the centrifugal force (mass * centripetal acceleration). So, by using the formula \(F= m(a_g - a_c)\) we can figure out the new weight of the object where \(a_g\) is the acceleration due to gravity and \(a_c\) is the centripetal acceleration.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If the pitcher's mound is \(1.25 \mathrm{ft}\) above the baseball field, can a pitcher release a fast ball horizontally at \(92.0 \mathrm{mi} / \mathrm{h}\) and still get it into the strike zone over the plate \(60.5 \mathrm{ft}\) away? Assume that, for a strike, the ball must fall at least \(1.30 \mathrm{ft}\) but no more than \(3.60 \mathrm{ft}\).

An \(8.5\) -kg object passes through the origin with a velocity of 42 \(\mathrm{m} / \mathrm{s}\) parallel to the \(x\) axis. It experiences a constant \(19-\mathrm{N}\) force in the direction of the positive \(y\) axis. Calculate \((a)\) the velocity and \((b)\) the position of the particle after \(15 \mathrm{~s}\) have elapsed.

A train travels due south at \(28 \mathrm{~m} / \mathrm{s}\) (relative to the ground) in rain that is blown to the south by the wind. The path of each raindrop makes an angle of \(64^{\circ}\) with the vertical, as measured by an observer stationary on the Earth. An observer on the train, however, sees perfectly vertical tracks of rain on the windowpane. Determine the speed of the drops relative to the Earth.

Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a star has a radius of \(20 \mathrm{~km}\) (a typical value), \((a)\) what is the speed of a point on the equator of the star, and \((b)\) what is the centripetal acceleration of this point?

A ball rolls off the top of a stairway with a horizontal velocity of magnitude \(5.0 \mathrm{ft} / \mathrm{s}\). The steps are \(8.0\) in. high and \(8.0\) in. wide. Which step will the ball hit first?
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Dynamics

Read Explanation

Quantum Physics

Read Explanation

Mechanics and Materials

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Oscillations

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free