Page 1 of 16
2/6/2016
Chapter 4 – Problems
4.7 Textbook, page 165
Particles of mud are thrown from the rim of a rolling
wheel. If the forward speed of the wheel is v0 and
the radius of the wheel is b, show that the greatest
height above the ground the mud can go is
v0 2 gb 2
b

2 g 2v0 2
At what point on the rolling wheel does this mud
leave?
(Note: It is necessary to assume that v0  bg . )
2

Let R be a vector from the center of the wheel to a
location where mud leaves the wheel. Let  be the
polar coordinate that also describes this point.
Then we have:

R  xi  yj  bcosi  sin j 

Velocity of mud = v0 kˆ  eR

v  v0 cosj  sin i 
at t=0, the initial conditions are:
Page 2 of 16
2/6/2016
x0  b cos
y0  b sin 
vxo  v0 sin 
v yo  v0 cos
ax  0
ay  g
Then we have the equation for motion when
a=constant.
x  x0  voxt  1 axt 2 becomes
2
x  b cos  v0 sin  t
Likewise for y motion:
y  y0  voyt  1 a yt 2 becomes
2
y  b sin   v0 cos t  1 gt 2
2
The velocity equations are:
vx  constant  v0 x  v0 sin 
v y  v0 y  a yt  v0 cos  gt
At the top point we have:
Page 3 of 16
2/6/2016
v y  0 This will give the time for mud to make it to
the top.
tTop  v0 cos 
g
y at the top is the evaluated using this time.
ytop
 v cos  1  v0 cos 
 b sin   v0 cos  0
  2 g

 g 
 g 
2
ytop  b sin   1 vo 2 cos2 
2
g
Now in order to maximize this with respect to  , let
dytop
dytop
d
d
 0, or
 b cos  v02 cos sin 
g
 v02

0  cos b 
sin  
g


The solution cos  0 corresponds to min. Thus
sin   bg v0 2 will give the max. Evaluate ymax for
this value of  .
Page 4 of 16
2/6/2016
ymax
2
 bg 
v
0
1
 b 2  
2
g
 v0 
 b g v0
2
2
  bg  2 
1  
 
 v 2  
  0  
2
2
v
b
g
0
1
1


2
g
2
v
2
0
2
v0 2
b
g
1
1


2
g
2
v
2
0
This is measured from the center of the wheel.
To measure from the bottom of the wheel, add b.
 ymax from ground
2
2
v
b
g
0
1
1
b

2
g
2
v
0
2
Page 5 of 16
2/6/2016
4.8 Textbook, page 166
A gun is located at the bottom of a hill of constant
slope  . Show that the range of the gun measured
up the slope of the hill is
2v0 2 cos sin    
g cos2 
where  is the angle of elevation of the gun, and
that the maximum value of the slope range is
v0 2
g 1  sin  
Firing the bullet at an elevation angle  with initial
velocity v0 gives the initial conditions to be:
x0  0
y0  0
vx 0  v0 cos
v yo  v0 sin 
ax  0
ay  g
Thus
Page 6 of 16
2/6/2016
x  0  v0 cos t
y  0  v0 sin  t  1 gt 2
2
or eliminating time between the two givens:
2
x
y  tan x  1
2 v 2 cos2 
0
g
The equations of the straight line is y  tan x .
Bullet hits the ground at point where curves
intersect.
x2
g
1
y  tan x  tan x 
2 v 2 cos2 
0

g
0  x  tan  tan 
2
2
2
v
cos


0

x

 tan  tan 
2
x
  2v0 cos2 
g


 tan  tan 
y  tanx  tan  2v0 2 cos2  

g


Range  x 2  y 2  x 2  tan 2 x 2  x sec
Range 
tan  tan
1
2v02 cos2 
g
cos
Page 7 of 16
2/6/2016
sin  sin  2v0 2 cos2 



cos cos g cos
sin  cos  sin  cos 2v0 2 cos2 



cos cos
g
cos
sin     cos 2v0 2
R 

g
cos2 
To maximize this range:
dR
cos   cos  sin    sin  2v0 2
0

2
d
g
cos 
or
cos    cos  sin    sin   0
1  tan    tan
or
    
 2 

2

2

Evaluate R for this value of  .
Page 8 of 16
2/6/2016
2v0 2 sin  cos  cos sin   cos
R
g
cos2 
2v0 2 sin  cos cos  cos2  sin 

g
cos2 
1 sin 2 cos  1  cos 2 sin 
2v
2
2
 0
2
g
cos 
2
2v0 2

g cos2 
1
1


1 cos      sin  
sin


cos






 2

2 2
2


 2



2v0 2 1

cos cos  sin   sin 2 
2
g cos  2
v0 2
1  sin  

2
g cos 
v0 2 1  sin 
v0 2
1


g 1  sin 2 
g 1  sin 
 Rmax
v0 2
1


g 1  sin 

Page 9 of 16
2/6/2016
4.11 Textbook page 166
Write down the component form of the differential
equations of motion of a projectile if the air
resistance is proportional to the square of the
speed. Are the equations separated? Show the x
and y components of the velocity are given by
x  x0e s
y  y 0e s
where s is the distance the projectile has traveled
along the path of motion, and   c2 m.
The force equation with a quadratic air resistance
term is:


2 v
ˆ
F  mgk   v  ma
v
mx   vx v
my   v y v
mz  mg   vz v
or investigating the x equation, one has:
x 
d

x   x v or
dt
m
dx
 ds
  m vdt  
dt
x
m dt
Page 10 of 16
2/6/2016
Integrate both sides:
dx
vx
 s


Ln


ds


s
v x 0 x

s

0
vx 0
m
m
vx

 vx  vx 0
 s
e m
Likewise for vy just let x  y.

v y  vgo
 s
e m
Page 11 of 16
2/6/2016
4.18 Textbook, page 167
A particle is placed on a smooth sphere of radius b
at a distance b/2 above the central plane. As the
particle slides down the side of the sphere, at what
point will it leave?
The sum of the forces is the radical directions will
be: N  mg cos90     maradial   m v
2
b
or
mv 2
 N  mg sin  
b
but with N  0, one has the velocity at which N  0
v 2  bg sin 
(When this velocity is achieved N will become zero.)
Look at energy, at the initial point and when N=0.
1 m02  mg b  1 mbg sin    mgb sin 
2
2
2
simplifyin g
mg b  1 mgb sin   mgb sin   3 mgb sin 
2
2
2
Page 12 of 16
 sin   1
2/6/2016
3
It will leave when sin   1
height above the reference plane is b .
3
3
or when
Page 13 of 16
2/6/2016
4.17 Textbook, page 167
An electron moves in a force field due to a uniform
electric field E and a uniform magnetic field B that
is at right angles to E. Let E =jE and B=kB. Take
the initial position of the electron at the origin with
initial velocity v0=iv0 in the x direction. Find the
resulting motion of the particle. Show that the path
of motion is a cycloid:
x  a sin wt  bt
y  a1  cos wt 
z0
Cycloid motion of electrons is used in an electronic
tube called a magnetron to produce the microwaves
in a microwave oven.

E  jE0

B  B0kˆ
v0 x  iv0
v0 y  0 v0 z  0
x0  y0  z0  0


 
F  qE  qv  B
 qE0 j
i
 q vx
0

j
vy
0
k
vz
B0
 qE0 j  q iv y B0  jvx B0  k 0

Page 14 of 16
2/6/2016
Fx  mx  qB0v y
Fy  my  qE0  qvx B0
Fz  mz  0
x 
qB0
vy
m
y  
qB0
qE
vx  0
m
m
Let
B0

m
Let
qE0

m
Integrate each:
d
dy

x
dt


 dt
 dt dt
vx  vx 0    y  y0  : vx  v0  y
v y  vx 0    x  x0    t : v y   x   t
Put these back into D.E.
x    x   t 
x   2 x    t
y   v0  y   
y   2 y  v0  
Page 15 of 16
2/6/2016
Solution for these is:
y
   v0
 A sin  t  B cos t
2
x

t  C sin  t  D cos t

Now apply initial coordinates to find A,B,C, and D.
At t=0, y=0,
   v0
 v0  B

B

0
or
B

2
2
And
At t  0, x  0  0  D or D  0.
At t  0, v y 0  0  A cos t  B sin  t  A  0
At
t  0, x  v0 ,
so x 


 C cos t  D sin  t  v0  C  v0 


Or
v0  

 sin  t
x t


y
   v0
   v0


1

cos

t
 a and
or
let
2
2
Page 16 of 16
2/6/2016

E
b 0

B0
Then our solution can be written as
x  bt  a sin wt
y  a (1  cos  t )
E
with b  0
b0
and
m  E0  v0 B0 
a 
q  B0 2 