PHYSICS 171
AQ 2009
Solutions to Homework
#10
1. Giancoli Chapter 11, Problem 8
(a) L  I   12 MR 2 
1
2
rev  2 rad 
2
2
 48 kg  0.15 m   2.8  
  9.50 kg m

s  9.5 kg m 2 s
s   1rev 
(b) If the rotational inertia does not change, then the change in angular momentum is strictly
due to
a change in angular velocity.
L I final  I 0 0  9.50 kg m 2 s



 1.9 m N
t
t
5.0 s
The negative sign indicates that the torque is in the opposite direction as the initial
angular momentum.
2. Giancoli Chapter 11, Problem 24
We use the determinant rule, Eq. 11-3b, to evaluate the torque.
ˆi
ˆj
kˆ
τ  r  F  4.0
0
3.5
6.0 m N
9.0
4.0

  68ˆi  16ˆj  36kˆ  m N

 ˆi  3.5 4.0    6  9    ˆj  6  0    4  4    kˆ  4  9    3 5  m N
3. Giancoli Chapter 11, Problem 26
(a) We use the distributive property, Eq. 11-4c, to obtain 9 single-term cross products.

 

 A B  ˆi  ˆi   A B  ˆi  ˆj  A B  ˆi  kˆ   A B  ˆj  ˆi   A B  ˆj  ˆj  A B  ˆj  kˆ 
 A B  kˆ  ˆi   A B  kˆ  ˆj  A B  kˆ  kˆ 
A  B  Ax ˆi  Ay ˆj  Az kˆ  Bx ˆi  B y ˆj  Bz kˆ
x
x
z
x
x
y
z
x
y
z
z
y
x
y
y
y
z
z
Each of these cross products of unit vectors is evaluated using the results of Problem 21
and Eq.
11-4b.
A  B  Ax Bx  0   Ax B y kˆ  Ax Bz  ˆj  Ay Bx  kˆ   Ay B y  0   Ay Bz ˆi
 
 Az Bx ˆj  Az By ˆi  Az Bz  0 
 Ax B y kˆ  Ax Bz ˆj  Ay Bx kˆ  Ay Bz ˆi  Az Bx ˆj  Az B y ˆi
  Ay Bz  Az By  ˆi   Az Bx  Ax Bz  ˆj   Ax By  Ay Bx  kˆ
(b) The rules for evaluating a literal determinant of a 3 x 3 matrix are as follows. The indices
on
the matrix elements identify the row and column of the element, respectively.
a11 a12 a13
a21
a23  a11  a22 a33  a23a32   a12  a23a31  a21a33   a13  a21a32  a22 a31 
a22
a31 a32 a33
Apply this as a pattern for finding the cross product of two vectors.
ˆi
ˆj
kˆ
A  B  Ax
Ay
Az  ˆi  Ay Bz  Az B y   ˆj  Az Bx  Ax Bz   kˆ  Ax B y  Ay Bx 
Bx
By
Bz
This is the same expression as found in part (a).
4. Giancoli Chapter 11, Problem 29
(a) We use the determinant rule, Eq. 11-3b, to evaluate the cross product.
ˆi
ˆj
kˆ
A  B  5.4
8.5
0  7.0ˆi  10.8ˆj  0.49 kˆ  7.0ˆi  11ˆj  0.5kˆ
3.5
5.6
2.0
(b) Now use Eq. 11-3a to find the angle between the two vectors.
AB 
 7.02   10.82   0.49 2
 5.4 2   3.52
A
 6.435 ; B
A  B  AB sin     sin 1
 12.88
 8.52   5.6 2   2.0 2
AB
 10.37
12.88
 sin 1
 11.1 or 168.9
AB
 6.43510.37 
Use the dot product to resolve the ambiguity.
A B   5.4  8.5   3.5 5.6   0  2.0   26.3
Since the dot product is negative, the angle between the vectors must be obtuse, and so
  168.9  170 .
5. Giancoli Chapter 11, Problem 38
(a)
From Example 11-8, a 
a

 m B  mA  g
m
 mB  I R

1.2 kg   9.80 m
s2
A
2
0

 mB  mA  g
m
A
 mB  I R02
 m B  mA  g
 mA  mB   12 mR02
  0.7538 m s
2

.
2
0
R

 mB  mA  g
mA  mB  12 m
 0.75 m s 2
15.6 kg
(b) If the mass of the pulley is ignored, then we have the following.
a
 mB  mA  g 1.2 kg   9.80 m

15.2 kg
 mA  m B 
s2
  0.7737 m s
2
 0.7737 m s2  0.7538 m s 2 
% error  
  100  2.6%
0.7538 m s 2


6. Giancoli Chapter 11, Problem 42
Take the origin of coordinates to be at the rod’s center, and the axis of
rotation to be in the z direction. Consider a differential element
M
dm 
dr of the rod, a distance r from the center. That element rotates in a
l
circle of radius r sin  , at a height of r cos . The position and velocity of
this point are given by the following.
r  r sin  cos t ˆi  r sin  sin t ˆj  r cos  kˆ
dL

dm

r
 r sin  cos t ˆi  sin  sin t ˆj  cos  kˆ 
v
dr
dt
  r sin  sin t ˆi  r sin  cos t ˆj

 r   sin  sin t ˆi  sin  cos t ˆj
Calculate the angular momentum of this element.
ˆi
dL  dm  r  v   r 
2

M
l
M
l
ˆj
kˆ
dr sin  cos t
sin  sin t
cos 
 sin  sin t
sin  cos t
0


dr   sin  cos  cos t  ˆi    sin  cos  sin t  ˆj  sin 2  cos 2 t  sin 2  sin 2 t kˆ 
Mr 2 sin 
dr   cos  cos t  ˆi    cos  sin t  ˆj  sin  kˆ 
l
Note that the directional portion has no r dependence. Thus dL for every piece of mass has
the same direction. What is that direction? Consider the dot product r dL.

r dL  r sin  cos t ˆi  sin  sin t ˆj  cos  kˆ 
 Mr 2 sin  dr

  cos  cos t  ˆi    cos  sin t  ˆj  sin  kˆ  

l


3
Mr  sin  dr

sin  cos t   cos  cos t   sin  sin t   cos  sin t   cos  sin    0
l
Thus dL  r for every point on the rod. Also, if  is an acute angle, the z component of dL
is positive. The direction of dL is illustrated in the diagram.
Integrate over the length of the rod to find the total angular momentum. And since the
direction of dL is not dependent on r, the direction of L is the same as the direction of dL.
L   dL 
M  sin 
l
  cos  cos t  ˆi    cos  sin t  ˆj  sin  kˆ 
l /2

r 2 dr
l / 2
M l sin 
2
  cos  cos t  ˆi    cos  sin t  ˆj  sin  kˆ 
12
Find the magnitude using the Pythagorean theorem.

L
M l 2 sin 
12
  cos  cos t  2    cos  sin t  2  sin 2  


1/ 2

M l 2 sin 
12
L is inclined upwards an angle of  from the x-y plane, perpendicular to the rod.
7. Giancoli Chapter 11, Problem 48
Angular momentum about the pivot is conserved during this collision. Note that both objects
have angular momentum after the collision.
 Lbullet  Lstick  Lbullet  mbullet v0  14 l
Lbefore  Lafter
collision

collision
initial
mbullet  v0  vf  14 l
I stick

final

mbullet  v0  vf  14 l
1
12
  I stick  mbullet vf  14 l 

final
M stick l

2
stick

3mbullet  v0  vf 
M stick l stick

3  0.0030 kg 110 m s 
 0.27 kg 1.0 m 
 3.7 rad s
8. Giancoli Chapter 11, Problem 50
(a) Linear momentum of the center of mass is conserved in the totally inelastic collision.
pinitial  pfinal  mbeam v0   mbeam  mman  vfinal 
vfinal 
 230 kg 18 m s 

 mbeam  mman 
 295 kg 
mbeam v0

14 m s
(b) Angular momentum about the center of mass of the system is conserved. First
we find the center of mass, relative to the center of mass of the rod, taking down
as the positive direction. See the diagram.
m  0   mman  12 l   65 kg 1.35 m 
yCM  beam

 mbeam  mman 
 295 kg 
 0.2975 m below center of rod
We need the moment of inertia of the beam about the center of mass of the
entire
system. Use the parallel axis theorem.
2
I beam  121 mbeaml 2  mbeam rbeam
; I man  mman  12 l  rbeam 
2
CM rod
0.2975 m
CM rod +
man
Linitial  Lfinal  mbeam v0 rbeam   I beam  I man  final 
final 

mbeam v0 rbeam
 I beam  I man 

mbeam  l
mbeam v0 rbeam
2
2
 mman  12 l  rbeam 
  mbeam rbeam
 230 kg 18 m s  0.2975 m 
2
2
2
1
230 kg  2.7 m    230 kg  0.2975 m    65 kg 1.0525 m 
12 
1
12
 5.307 rad s  5.3rad s
1
2
2