Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 10, Problem 83P
To determine
ToCalculate: The approximate value for this torque.
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Question 1
a. Suppose the Sun contracts (collapses) to a pulsar with radius 15 km. Estimate the period of rotation
of the pulsar. The period of revolution of the Sun about its axis is 25.38 days (1 day = 24 hours).
Compare the kinetic energies of rotation of the pulsar and of the Sun.
Figure 1: Question 1(a) A pulsar star system
b. A wheel on an extremely low friction slanted surface will slide without rotating. However, if the
same surface is rough and having some friction the wheel will roll down (with rotation). Explain
why, using free body diagram.
The rate at which a nebular cloud rotates increases as the cloud collapses to form systems of stars and planets. Consider a small
segment of a nebular cloud with a mass m of 1.9 x 102' kg, tangential velocity vinitial equal to 6.8 km s located at an orbital
distance rinitial = 2.5 x 104 km. After the cloud collapses, the same small segment is located at an orbital distance
rfinal = 3.2 x 10³ km. Calculate the change of the rotational velocity, Aw, for the cloud segment, assuming perfectly circular
orbits. Perform your work and report your solution using two significant figures.
Δω
rad s-1
The term precession of the equinoxes refers to the fact that Earth's spin axis does not stay fixed, but instead sweeps out a cone
once every 26,000 years. (This explains why our pole star, Polaris, will not remain the pole star forever.) The reason for this
instability is that Earth is a giant gyroscope. The spin axis of Earth precesses because of the torques exerted on it by the
gravitational forces of the Sun and moon. The angle between the direction of Earth's spin axis and the normal to the ecliptic
plane (the plane of Earth's orbit) is 22.5 degrees.
What is the approximate value for this torque t, given that the
period of rotation of Earth is 1.00 day and its moment of
inertia is 8.03 × 10³7 kg⋅m²?
T =
4.5 ×1023
Incorrect
N.m
Chapter 10 Solutions
Physics for Scientists and Engineers
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