Download CHAPTER 11 HW SOLUTIONS

CHAPTER 11 HW SOLUTIONS
5. By Eq. 10-52, the work required to stop the hoop is the negative of the initial kinetic
energy of the hoop. The initial kinetic energy is K  21 I 2  21 mv 2 (Eq. 11-5), where I =
mR2 is its rotational inertia about the center of mass, m = 140 kg, and v = 0.150 m/s is the
speed of its center of mass. Eq. 11-2 relates the angular speed to the speed of the center of
mass:  = v/R. Thus,
 v2  1
1
2
K  mR 2  2   mv 2  mv 2  140 kg  0.150 m/s 
2
R  2
which implies that the work required is – 3.15 J.
11. To find where the ball lands, we need to know its speed as it leaves the track (using
conservation of energy). Its initial kinetic energy is Ki = 0 and its initial potential energy
is Ui = M gH. Its final kinetic energy (as it leaves the track) is K f  21 Mv 2  21 I 2 (Eq.
11-5) and its final potential energy is M gh. Here we use v to denote the speed of its
center of mass and  is its angular speed — at the moment it leaves the track. Since (up
to that moment) the ball rolls without sliding we can set  = v/R. Using I  25 MR 2
(Table 10-2(f)), conservation of energy leads to
MgH 
1
1
1
2
7
Mv 2  I  2  Mgh  Mv 2  Mv 2  Mgh  Mv 2  Mgh.
2
2
2
10
10
The mass M cancels from the equation, and we obtain
v
b g
ib
g
10
10
2
g H h 
9.8 m s 6.0 m  2.0 m  7.48 m s .
7
7
d
Now this becomes a projectile motion of the type examined in Chapter 4. We put the
origin at the position of the center of mass when the ball leaves the track (the “initial”
position for this part of the problem) and take +x rightward and +y downward. Then
(since the initial velocity is purely horizontal) the projectile motion equations become
1
x  vt and y   gt 2 .
2
Solving for x at the time when y = h, the second equation gives t  2 h g . Then,
substituting this into the first equation, we find
xv
2  2.0 m 
2h
  7.48 m/s 
 4.8 m.
g
9.8 m/s 2
17. (a) The derivation of the acceleration is found in §11-4; Eq. 11-13 gives
acom  
g
1  I com MR02
where the positive direction is upward. We use I com  950 g  cm2 , M =120g, R0 = 0.320
cm and g = 980 cm/s2 and obtain
| acom | 
980 cm/s 2
1   950 g  cm
2
 120 g  0.32 cm 
 12.5 cm/s 2  13 cm/s 2 .
2
(b) Taking the coordinate origin at the initial position, Eq. 2-15 leads to ycom  21 acomt 2 .
Thus, we set ycom = – 120 cm, and find
t
2  120cm 
2 ycom

 4.38 s  4.4 s.
acom
12.5 cm s 2
(c) As it reaches the end of the string, its center of mass velocity is given by Eq. 2-11:
vcom  acomt   12.5 cm s 2   4.38s   54.8 cm s ,
so its linear speed then is approximately | vcom |  55 cm/s.
(d) The translational kinetic energy is
1
2
2
mvcom

1
2
.
kgg
b0120
b0.548 m sg 18.  10
2
2
J.
(e) The angular velocity is given by  = – vcom/R0 and the rotational kinetic energy is
c
hb
h
g
5
2
2
1
1
vcom
1 9.50  10 kg  m 0.548 m s
2
I com  I com 2 
2
2
2
R0
2
3.2  103 m
c
which yields Krot = 1.4 J.
(f) The angular speed is
2

vcom
R0

0.548 m/s
 1.7 102 rad/s  27 rev s .
3
3.2 10 m

23. If we write r  xi  yj  zk, then (using Eq. 3-30) we find r  F
is equal to
dyF  zF ii  bzF  xF gj  dxF  yF ik.
z
y
x
z
y
x
ˆ
(a) Plugging in, we find    3.0m  6.0N    4.0m  8.0N   kˆ  (50 N  m) k.


 
(b) We use Eq. 3-27, | r  F |  rF sin  , where  is the angle between r and F . Now
r  x 2  y 2  5.0 m and F  Fx2  Fy2  10 N. Thus,
b gb g
rF  5.0 m 10 N  50 N  m,
the same as the magnitude of the vector product calculated in part (a). This implies sin 
= 1 and = 90°.

 


  mr  v, where r is the position vector of the object, v is its velocity
vector, and m is its mass. Only the x and z components of the position and velocity
 
vectors are nonzero, so Eq. 3-30 leads to r  v   xvz  zvz j. Therefore,
29. (a) We use
b
g
r
l  m   xvz  zvx  ˆj   0.25 kg     2.0 m  5.0 m s    2.0 m  5.0 m s   ˆj  0.


 then (using Eq. 3-30) we find r  F is equal to
(b) If we write r  xi  yj  zk,
dyF  zF ii  bzF  xF gj  dxF  yF ik .
z
y
x
z
y
x
With x = 2.0, z = –2.0, Fy = 4.0 and all other components zero (and SI units understood)
the expression above yields



e
j
  r  F  8.0 i  8.0 k N  m.
39. (a) A particle contributes mr2 to the rotational inertia. Here r is the distance from the
origin O to the particle. The total rotational inertia is
I  m  3d   m  2d   m  d   14md 2  14(2.3102 kg)(0.12 m) 2
2
2
2
 4.6 103 kg  m2 .
(b) The angular momentum of the middle particle is given by Lm = Im, where Im = 4md 2
is its rotational inertia. Thus
Lm  4md 2  4(2.3102 kg)(0.12 m) 2 (0.85 rad/s)  1.110 3 kg  m 2 /s.
(c) The total angular momentum is
I  14md 2  14(2.3102 kg)(0.12 m)2 (0.85 rad/s)  3.9 103 kg  m2 /s.
45. (a) No external torques act on the system consisting of the two wheels, so its total
angular momentum is conserved. Let I1 be the rotational inertia of the wheel that is
originally spinning at  i and I2 be the rotational inertia of the wheel that is initially at
rest. Then I1ωi= (I1+I2) ωf and ωf
I1
f 
i
I1  I 2
b g
where  f is the common final angular velocity of the wheels. Substituting I2 = 2I1 and
 i  800 rev min, we obtain  f  267 rev min.
(b) The initial kinetic energy is Ki  21 I1 i2 and the final kinetic energy is
1
K F  ( I 1  I 2 ) F2 . We rewrite this as
2
Kf 
I
1
1
( I I  2 I1 )( 1 i ) 2  I1i2
2
11  I 2
6
Therefore, the fraction lost, ( K i  K f ) / K i  is
I i2 / 6 2
1
 1  2   0.667.
Ki
I i / 2 3
Kf
CHAPTER 13 HW SOLUTIONS
7. At the point where the forces balance GM e m / r12  GM s m / r22 , where Me is the mass of
Earth, Ms is the mass of the Sun, m is the mass of the space probe, r1 is the distance from
the center of Earth to the probe, and r2 is the distance from the center of the Sun to the
probe. We substitute r2 = d  r1, where d is the distance from the center of Earth to the
center of the Sun, to find
Me
Ms
=
.
2
2
r1
 d  r1 
Taking the positive square root of both sides, we solve for r1. A little algebra yields
r1 =
150 10 m 
9
d Me
Ms + Me
=
5.98 1024 kg
1.99 10 kg + 5.98 10 kg
30
24
= 2.60 108 m.
Values for Me, Ms, and d can be found in Appendix C.
12. All the forces are being evaluated at the origin (since particle A is there), and all

forces (except the net force) are along the location-vectors r which point to particles B
and C. We note that the angle for the location-vector pointing to particle B is 180º –
30.0º = 150º (measured ccw from the +x axis). The component along, say, the x axis of

one of the force-vectors F is simply Fx/r in this situation (where F is the magnitude of

F ). Since the force itself (see Eq. 13-1) is inversely proportional to r2 then the
aforementioned x component would have the form GmMx/r3; similarly for the other
components. With mA = 0.0060 kg, mB = 0.0120 kg, and mC = 0.0080 kg, we therefore
have
GmAmB xB
GmAmC xC
Fnet x =
+
= (2.77  1014 N)cos(163.8º)
3
rB
rC 3
and
Fnet y =
GmAmB yB
GmAmC yC
+
= (2.77  1014 N)sin(163.8º)
rB 3
rC 3
where rB = dAB = 0.50 m, and (xB, yB) = (rBcos(150º), rBsin(150º)) (with SI units
understood). A fairly quick way to solve for rC is to consider the vector difference
between the net force and the force exerted by A, and then employ the Pythagorean
theorem. This yields rC = 0.40 m.
(a) By solving the above equations, the x coordinate of particle C is xC = 0.20 m.
(b) Similarly, the y coordinate of particle C is yC = 0.35 m.
24. (a) What contributes to the GmM/r2 force on m is the (spherically distributed) mass
M contained within r (where r is measured from the center of M). At point A we see that
M1 + M2 is at a smaller radius than r = a and thus contributes to the force:
Fon m 
G  M1  M 2  m
a2
.
(b) In the case r = b, only M1 is contained within that radius, so the force on m becomes
GM1m/b2.
(c) If the particle is at C, then no other mass is at smaller radius and the gravitational
force on it is zero.
30. (a) The gravitational potential energy is
U =
 6.67 10
GMm
=
r
11
m3 /s 2  kg   5.2 kg  2.4 kg 
19 m
=  4.4 1011 J.
(b) Since the change in potential energy is
U = 
GMm  GMm 
2
11
11

 =   4.4  10 J  = 2.9 10 J,
3r
r 
3

the work done by the gravitational force is W =  U = 2.9  1011 J.
(c) The work done by you is W´ = U = 2.9  1011 J.
36. (a) From Eq. 13-28, we see that
v0  GM / 2RE in this problem. Using energy
conservation, we have
1
2
2 mvo – GMm/RE = – GMm/r
which yields r = 4RE/3. So the multiple of RE is 4/3 or 1.33.
(b) Using the equation in the textbook immediately preceding Eq. 13-28, we see that in
this problem we have Ki = GMm/2RE, and the above manipulation (using energy
conservation) in this case leads to r = 2RE. So the multiple of RE is 2.00.
(c) Again referring to the equation in the textbook immediately preceding Eq. 13-28, we
see that the mechanical energy = 0 for the “escape condition.”
46. Kepler’s law of periods, expressed as a ratio, is
3
2
 aM 
 TM 
 TM 
3

 
  (1.52)  

 1y 
 aE 
 TE 
2
where we have substituted the mean-distance (from Sun) ratio for the semi-major axis
ratio. This yields TM = 1.87 y. The value in Appendix C (1.88 y) is quite close, and the
small apparent discrepancy is not significant, since a more precise value for the semimajor axis ratio is aM/aE = 1.523 which does lead to TM = 1.88 y using Kepler’s law. A
question can be raised regarding the use of a ratio of mean distances for the ratio of semimajor axes, but this requires a more lengthy discussion of what is meant by a ”mean
distance” than is appropriate here.
56.
The two stars are in circular orbits, not about each other, but about the two-star
system’s center of mass (denoted as O), which lies along the line connecting the centers
of the two stars. The gravitational force between the stars provides the centripetal force
necessary to keep their orbits circular. Thus, for the visible, Newton’s second law gives
F
Gm1m2 m1v 2

r2
r1
where r is the distance between the centers of the stars. To find the relation between r
and r1 , we locate the center of mass relative to m1 . Using Equation 9-1, we obtain
r1 
m1 (0)  m2 r
m2 r
m  m2

 r 1
r1 .
m1  m2
m1  m2
m2
On the other hand, since the orbital speed of m1 is v  2 r1 / T , then r1  vT / 2 and the
expression for r can be rewritten as
m  m2 vT
r 1
.
m2 2
Substituting r and r1 into the force equation, we obtain
4 2Gm1m23
2 m1v
F

2 2 2
(m1  m2 ) v T
T
or
m23
v3T (2.7 105 m/s)3 (1.70 days)(86400 s/day)


 6.90 1030 kg
2
11
3
2
(m1  m2 )
2 G
2 (6.67 10 m /kg  s )
 3.467 M s ,
where M s  1.99 1030 kg is the mass of the sun. With m1  6M s , we write m2   M s
and solve the following cubic equation for  :
3
 3.467  0 .
(6   )2
The equation has one real solution:   9.3 , which implies m2 / M s  9 .