Homework 2
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Physics 3309 Homework 2

Chapter 1

1-31.
a)
                                                      (1)
Therefore,
                                                                        (2)
b)
                                                      (3)
Therefore,
                                                                      (4)
c)
                                                      (5)
or,
                                                                                (6)

1-33. Since
                                                                (1)
we have
                                                              (2)
from which
                                                                    (3)
where C is the integration constant (a vector).

1-37. 
                                          
To do the integral directly, note that , on the surface, and that .
                                                      (1)
To use the divergence theorem, we need to calculate . This is best done in spherical coordinates, where . Using Appendix F, we see that
                                                              (2)
Therefore,
                                                      (3)
Alternatively, one may simply set  in this case.

Chapter 2

2-1. The basic equation is
                                                                                           (1)
a) : Not integrable                                                          (2)
b) 
                                             
                                                    : Integrable        (3)
c) : Not integrable                                                          (4)

2-5.
                          
a) From the force diagram we have . The acceleration that the pilot feels is , which has a maximum magnitude at the bottom of the maneuver.
b) If the acceleration felt by the pilot must be less than 9g, then we have
                                                      (1)
A circle smaller than this will result in pilot blackout.

2-9.
a) Zero resisting force ():
The equation of motion for the vertical motion is:
                                                                    (1)
Integration of (1) yields
                                                                                     (2)
where  is the initial velocity of the projectile and t = 0 is the initial time.
The time  required for the projectile to reach its maximum height is obtained from (2). Since  corresponds to the point of zero velocity,
                                             ,                          (3)
we obtain
                                                                                          (4)
b) Resisting force proportional to the velocity :
The equation of motion for this case is:
                                                                  (5)
where –kmv is a downward force for  and is an upward force for . Integrating, we obtain
                                                                 (6)
For , v(t) = 0, then from (6),
                                                                              (7)
which can be rewritten as
                                                                           (8)
Since, for small z (z = 1) the expansion
                                                              (9)
is valid, (8) can be expressed approximately as
                                                      (10)
which gives the correct result, as in (4) for the limit k ® 0.

                                         
2-13. The equation of motion of the particle is
                                                                    (1)
Integrating,
                                                                     (2)
and using Eq. (E.3), Appendix E, we find
                                                            (3)
Therefore, we have
                                                                             (4)
where  and where C¢ is a new constant. We can evaluate C¢ by using the initial condition,  at t = 0:
                                                                                    (5)
Substituting (5) into (4) and rearranging, we have
                                                                 (6)
Now, in order to integrate (6), we introduce  so that du = –Au dt. Then,
                                                      (7)
Using Eq. (E.8c), Appendix E, we find
                                                           (8)
Again, the constant C² can be evaluated by setting x = 0 at t = 0; i.e., x = 0 at u = 1:
                                                                 (9)
Therefore, we have
                          
Using (4) and (5), we can write
                                                      (10)
From (6) we see that v ® 0 as t ® ¥. Therefore,
                                                      (11)
Also, for very large initial velocities,
                                                      (12)
Therefore, using (11) and (12) in (10), we have
                                                                             (13)

2-17. 
                          
The setup for this problem is as follows:
                                                                                   (1)
                                                                (2)
where  and . The ball crosses the fence at a time , where R = 60 m. It must be at least h = 2 m high, so we also need . Solving for , we obtain
                                                      (3)

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