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Week 7 - Rigid Body Motion II [6] [8]
Summary
This Monday (24 October 2005) is the midterm. It covers material up to 21 October 2005.
Rigid body motion using Euler angles.
Reading List
- ``Lagrangian Dynamics''
REF: Wells, D. A. 1967, Chapter 9.
- ``Classical Mechanics''
REF: Goldstein, H. 1980, Chapters 4-5.
- ``Mechanics''
REF: Landau, L. D., Lifshitz, E. M. 1988, Chapter 6.
Problem Set - 4 November 2005 Answers
Problem 1 - Like a rubber ball ...
When a rubber ball bounces, it rolls without slipping at the point of
contact between the ball and the floor. This means that the point of
contact is stationary at the moment of contact. Assume that energy is
conserved by the bounce.
Derive a relationship between the initial velocity and angular
velocity of the ball and the final velocities. What combination of
initial velocities result in an exact reversal of the velocities of
the ball (angular and linear)? If you sets things up like this the
ball will bounce back and forth between two fixed points.
Problem 2 - Flippin' coin
When you flip a coin such that the rotational axis almost
perpendicular to the face of the coin, how fast does the Queen's head
spin compared to rate at which the coin wobbles?
Problem 3 - Torquing Planet Earth
Derive an equation for the total potential energy of a body whose
centre of mass lies a distance R from a point mass m.
Expand this expression to second order in l/R where l is
the size of body. You should find a term proportional to
R-1 and a term proportional to
R-3. There is no R-2 term. The
first term should be familar, and you can write the second one in
terms of the moments of inertia of the body. Use this to estimate
the rate of precession of the Earth's rotation axis due the torques
exerted by the moon and sun. We are looking for an order of magnitude
estimate, so assume that the torque is about equal to the second-order
potential energy (the R-3 term) and that the angular
momentum of the Earth is simply Iz2π/(1 day).
The precession rate is the total torque (sun plus moon) divided by the
angular momentum; you will find that the answer only depends on the
ratios of the various moments of inertia of the Earth, the mass of the
sun and moon, and the distances to the sun and moon.
You may find the following Taylor series helpful.
(1+x)-1/2 = 1 - x/2 + 3/8 x2 + O(x3)
Also to translate your answer to some number of years, you will find the
following facts helpful.
G (M1 + M2)
R3
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= (The angular velocity of an orbit)2.
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We derived this in class. The mass of the Sun is 333,000 times that
of the Earth. The mass of the Earth is 81 times that of the Moon.
I got to within a factor of three of the actual precession rate, so this rough calculation is pretty good.
Last modified: Wednesday, 30 November 2005 12:14:26
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