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Physics 3309 Homework 10

Chapter 8

8-37. From the equations in Section 8.8 regarding Hohmann transfers:
                                          
                                                     
                                                      (1)
Substituting
                          
gives

                          

                          

 

8-41. From the equations in Section 8.8 regarding Hohmann transfers
                                         
where
                          
                          
Substituting
                          
gives
                                                 
From Eq. (8.58), the time of transfer is given by
                          
Substituting gives
                               

 

Chapter 9

9-3. 
                                        
By symmetry, .
From problem 9-2, the center of mass of the cone is at .
From problem 9-1, the center of mass of the hemisphere is at
                                       
So the problem reduces to
                              
                              
                          
for

                          

9-5.                   
                                         
                          
Calculate the torque about
                          
Now if the total torque is zero, we must have
                                                      
or
                                                    
which is the definition of the center of mass. So

                       

9-9. 
                          
Let the axes be as shown with the projectile in the y-z plane. At the top just before the explosion, the velocity is in the y direction and has magnitude .
                          
where  and  are the masses of the fragments. The initial momentum is
                          
The final momentum is
                                                         
                                                  
                                              
The conservation of momentum equations are
                                                    
                          
                                      
The energy equation is
                          
or
                                  
Substituting for  and  gives
                                    
                                                gives
                                      
So
                                                      travels straight down with speed =
                                                      travels in the y-z plane
                          
The mass  is the largest it can be when , meaning  and the mass ratio is

                          

9-13. From Eq. (9.9), the total force is given by
                                              
As shown in Section 9.3, the second term is zero. So the total force is
                                                                  
It is given that this quantity is zero.
Now consider two coordinate systems with origins at 0 and 0¢
                                     
where
                                                      is a vector from 0 to 0¢
                                                      is the position vector of  in 0
                                                      is the position vector of  in 0¢
We see that
The torque in 0 is given by
                                                    
The torque in 0¢ is
                              
But it is given that
Thus

                          

9-19. 
                            
The force that the tabletop exerts on the chain counteracts the force due to gravity, so that we may write the change in momentum of the center of the mass of the chain as
                                                                                   (1)
We can write out what the momentum is, though:
                                                                                  (2)
which has a time derivative
                                                      (3)
where we have used  and . Setting this last expression equal to (1) gives us
                                                                                         (4)

Although M. G. Calkin (personal communication) has found that experimentally the time of fall for this problem is consistently less than the value one would obtain in the above treatment by about 1.5%, he also finds evidence that suggests the free fall treatment is more valid if the table is energy absorbing.

9-21. Let us call  the length of rope hanging over the edge of the table, and  the total length of the rope. The equation of motion is
                               
Let us look for solution of the form
                                                  
Putting this into equation of motion, we find
                                                               
Initial conditions are ; .
From these we find .
Finally . When , the corresponding time is

                          

 


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Edgar A. Bering, III ,

Edgar A. Bering, III , <eabering@uh.edu>