Physics 103
9.1.
9.2.
Assignment 9
IDENTIFY: s  r , with  in radians.
SET UP:  rad  180.
s 150 m
EXECUTE: (a)   
 0600 rad  344
r 250 m
s
140 cm
 627 cm
(b) r  
 (128)( rad/180)
(c) s  r  (150 m)(0700 rad)  105 m
EVALUATE: An angle is the ratio of two lengths and is dimensionless. But, when s  r is used,  must be in
radians. Or, if   s /r is used to calculate  , the calculation gives  in radians.
IDENTIFY:   0  t , since the angular velocity is constant.
SET UP: 1 rpm  (2 /60) rad/s.
EXECUTE: (a)   (1900)(2 rad/60 s)  199 rad/s
(b) 35  (35)( /180)  0611 rad. t 
EVALUATE: In t 
  0 0611 rad

 31103 s

199 rad/s
  0
we must use the same angular measure (radians, degrees or revolutions) for both

  0 and  .
9.3.
d z
. Writing Eq. (2.16) in terms of angular quantities gives   0   tt2  z dt.
1
dt
1 n 1
d n
t
t  nt n  1 and  t n dt 
n 1
dt
IDENTIFY  z (t ) 
SET UP:
EXECUTE: (a) A must have units of rad/s and B must have units of rad/s3.
(b)  z (t )  2Bt  (300 rad/s3 )t. (i) For t  0,  z  0. (ii) For t  500 s,  z  150 rad/s2.
(c)  2  1   tt2 ( A  Bt 2 )dt  A(t2  t1)  13 B(t23  t13 ). For t1  0 and t2  200 s,
1
 2  1  (275 rad/s)(200 s)  13 (150 rad/s3 )(200 s)3  950 rad.
EVALUATE: Both  z and z are positive and the angular speed is increasing.
9.8.
IDENTIFY:  z 
 av- z 
 z1   z 2
t2  t1
d z
.   0  av- zt. When z is linear in t,  av-z for the time interval t1 to t2 is
dt
.
SET UP: From the information given, z (t )  600 rad/s  (200 rad/s2 )t.
EXECUTE: (a) The angular acceleration is positive, since the angular velocity increases steadily from a
negative value to a positive value.
(b) It takes 3.00 seconds for the wheel to stop (z  0). During this time its speed is decreasing. For the next
4.00 s its speed is increasing from 0rad/s to  800 rad/s.
 600rad/s  800rad/s
 100rad/s.   0  av-zt then leads to
2
displacement of 7.00 rad after 7.00 s.
EVALUATE: When  z and z have the same sign, the angular speed is increasing; this is the case for
(c) The average angular velocity is
9.9.
t  300 s to t  700 s. When  z and z have opposite signs, the angular speed is decreasing; this is the case
between t  0 and t  300 s.
IDENTIFY: Apply the constant angular acceleration equations.
SET UP: Let the direction the wheel is rotating be positive.
1
EXECUTE: (a) z  0 z   zt  150rad/s  (0300rad/s2 )(250 s)  225 rad/s.
(b)   0  0 z t  12  z t 2  (150 rad/s)(250 s)  12 (0300 rad/s 2 )(250 s) 2  469 rad.
9.11.
   z   150 rad/s  225 rad/s 
EVALUATE:   0   0 z
t  
 (250 s)  469 rad, the same as calculated with
2
2

 

another equation in part (b).
IDENTIFY: Apply the constant angular acceleration equations to the motion. The target variables are t
and   0 .
SET UP: (a)  z  150 rad/s2 ; 0 z  0 (starts from rest); z  360 rad/s; t  ?
z  0 z   zt
EXECUTE: t 
 z  0 z 360 rad/s  0

 240 s
z
150 rad/s2
(b)   0  ?
  0  0 zt  12  zt 2  0  12 (150 rad/s 2 )(240 s) 2  432 rad
  0  432 rad(1 rev/2 rad)  688 rev
EVALUATE: We could use   0  12 (z  0 z )t to calculate   0  12 (0  360 rad/s)(240 s)  432 rad,
9.17.
9.21.
which checks.
IDENTIFY: Apply Eq. (9.12) to relate z to   0 .
SET UP: Establish a proportionality.
EXECUTE: From Eq. (9.12), with 0 z  0, the number of revolutions is proportional to the square of the initial
angular velocity, so tripling the initial angular velocity increases the number of revolutions by 9, to 9.00 rev.
EVALUATE: We don’t have enough information to calculate  z ; all we need to know is that it is constant.
IDENTIFY: Use constant acceleration equations to calculate the angular velocity at the end of two revolutions.
v  r.
SET UP: 2 rev  4 rad. r  0200 m.
EXECUTE: (a) z2  02z  2 z (  0 ). z  2 z (  0 )  2(300 rad/s2 )(4 rad)  868 rad/s.
arad  r 2  (0200 m)(868 rad/s)2  151 m/s2.
(b) v  r  (0200 m)(868 rad/s)  174 m/s. arad 
v 2 (174 m/s) 2

 151 m/s 2 .
r
0200 m
EVALUATE: r 2 and v2/r are completely equivalent expressions for arad .
9.22.
IDENTIFY: v  r and atan  r .
SET UP: The linear acceleration of the bucket equals atan for a point on the rim of the axle.
 75 rev  1 min  2 rad 
EXECUTE: (a) v  R. 200cm/s  R 


 gives R  255 cm.
 min  60 s  1 rev 
D  2R  509 cm.
atan 0400m/s 2

 157rad/s 2 .
R
00255 m
EVALUATE: In v  R and atan  R ,  and  must be in radians.
(b) atan  R .  
9.25.
IDENTIFY: Use Eq. (9.15) and solve for r.
SET UP: arad  r 2 so r  arad / 2 , where  must be in rad/s
EXECUTE: arad  3000g  3000(980 m/s2 )  29,400 m/s2
 1 min  2 rad 

  5236 rad/s
 60 s  1 rev 
  (5000 rev/min) 
Then r 
arad
2

29,400 m/s 2
(5236 rad/s) 2
 0107 m.
2
9.31.
EVALUATE: The diameter is then 0.214 m, which is larger than 0.127 m, so the claim is not realistic.
IDENTIFY: Use Table 9.2. The correct expression to use in each case depends on the shape of the object and the
location of the axis.
SET UP: In each case express the mass in kg and the length in m, so the moment of inertia will be in kg  m2.
EXECUTE: (a) (i) I  13 ML2  13 (250 kg)(0750 m) 2  0469 kg  m 2 .
1 ML2  1 (0469 kg  m 2 )  0117 kg  m 2 .
(ii) I  12
(iii) For a very thin rod, all of the mass is at the axis and
4
I  0.
(b) (i) I  52 MR 2  52 (300 kg)(0190 m) 2  00433 kg  m 2 .
(ii) I  23 MR 2  53 (00433 kg  m 2 )  00722 kg  m 2 .
(c) (i) I  MR2  (800 kg)(00600 m)2  00288 kg  m2.
(ii) I  12 MR 2  12 (800 kg)(00600 m)2  00144 kg  m 2 .
9.33.
EVALUATE: I depends on how the mass of the object is distributed relative to the axis.
IDENTIFY: I for the object is the sum of the values of I for each part.
SET UP: For the bar, for an axis perpendicular to the bar, use the appropriate expression from Table 9.2. For a
point mass, I  mr 2 , where r is the distance of the mass from the axis.
2
EXECUTE: (a) I  I bar  I balls 
1
L
M bar L2  2mballs   .
12
2
1
(400 kg)(200 m)2  2(0500 kg)(100 m) 2  233 kg  m 2
12
1
1
(b) I  mbar L2  mball L2  (400 kg)(200 m)2  (0500 kg)(200 m)2  733 kg  m2
3
3
(c) I  0 because all masses are on the axis.
(d) All the mass is a distance d  0500 m from the axis and
I
9.41.
I  mbar d 2  2mballd 2  M Totald 2  (500 kg)(0500 m)2  125 kg  m2.
EVALUATE: I for an object depends on the location and direction of the axis.
IDENTIFY: K  12 I  2 . Use Table 9.2 to calculate I.
SET UP: I  52 MR 2 . For the moon, M  735 1022 kg and R  174 106 m. The moon moves through
1 rev  2 rad in 27.3 d. 1 d  864 104 s.
EXECUTE: (a) I  52 (735  1022 kg)(174  106 m) 2  890  1034 kg  m 2 .

2 rad
(273 d)(864 104 s/d)
 266 106 rad/s.
K  12 I  2  12 (890  1034 kg  m 2 )(266  106 rad/s) 2  315 10 23 J.
9.43.
315  1023 J
 158 years. Considering the expense involved in tapping the moon’s rotational energy, this
5(40  1020 J)
does not seem like a worthwhile scheme for only 158 years worth of energy.
EVALUATE: The moon has a very large amount of kinetic energy due to its motion. The earth has even more,
but changing the rotation rate of the earth would change the length of a day.
IDENTIFY: K  12 I  2 , with  in rad/s. Solve for I.
(b)
SET UP: 1 rev/min  (2 /60) rad/s. K  500 J
EXECUTE: i  650 rev/min  681 rad/s. f  520 rev/min  545 rad/s. K  K f  Ki  12 I (f2  i2 ) and
I
2(K )
f2
 i2

2(500 J)
(545 rad/s)2  (681 rad/s)2
 0600 kg  m2.
3
EVALUATE: In K  12 I  2 ,  must be in rad/s.
9.47.
IDENTIFY: Apply conservation of energy to the system of stone plus pulley. v  r relates the motion of the
stone to the rotation of the pulley.
SET UP: For a uniform solid disk, I  12 MR 2 . Let point 1 be when the stone is at its initial position and point
2 be when it has descended the desired distance. Let  y be upward and take y  0 at the initial position of the
stone, so y1  0 and y2  h, where h is the distance the stone descends.
EXECUTE: (a) K p  12 I p 2 . I p  12 M p R 2  12 (250 kg)(0200 m)2  00500 kg  m2 .

2Kp
Ip

2(450 J)
00500 kg  m 2
 134 rad/s. The stone has speed v  R  (0200 m)(134 rad/s)  268 m/s. The
stone has kinetic energy Ks  12 mv 2  12 (150 kg)(268 m/s)2  539 J. K1  U1  K2  U 2 gives 0  K2  U 2 .
0  450 J  539 J  mg (h). h 
(b) K tot  K p  Ks  989 J.
9.55.
Kp
K tot
989 J
(150 kg)(980 m/s 2 )

 0673 m.
450 J
 455%.
989 J
EVALUATE: The gravitational potential energy of the pulley doesn’t change as it rotates. The tension in the
wire does positive work on the pulley and negative work of the same magnitude on the stone, so no net work on
the system.
IDENTIFY and SET UP: Use Eq. (9.19). The cm of the sheet is at its geometrical center. The object is sketched
in Figure 9.55.
EXECUTE: I P  Icm  Md 2.
From part (c) of Table 9.2,
1 M ( a 2  b 2 ).
I cm  12
The distance d of P from the cm is
d  (a /2) 2  (b /2) 2 .
Figure 9.55
1 M (a 2  b 2 )  M ( 1 a 2  1 b 2 )  ( 1  1 ) M (a 2  b 2 )  1 M (a 2  b 2 )
Thus I P  I cm  Md 2  12
4
4
12 4
3
9.72.
EVALUATE: I P  4Icm . For an axis through P mass is farther from the axis.
IDENTIFY: Use the constant angular acceleration equations, applied to the first revolution and to the first two
revolutions.
SET UP: Let the direction the disk is rotating be positive. 1 rev  2 rad. Let t be the time for the first
revolution. The time for the first two revolutions is t  0.750 s.
EXECUTE: (a)    0  0 zt  12  zt 2 applied to the first revolution and then to the first two revolutions gives
2 rad  12  z t 2 and 4 rad  12  z (t  0.750 s) 2 . Eliminating  z between these equations gives
2 rad
(t  0.750 s) 2 . 2t 2  (t  0.750 s) 2 . 2t  (t  0.750 s). The positive root is
t2
0.750 s
t
 1.81 s.
2 1
(b) 2 rad  12  z t 2 and t  1.81 s gives  z  3.84 rad/s 2
4 rad 
EVALUATE: At the start of the second revolution, 0 z  (3.84 rad/s 2 )(1.81 s)  6.95 rad/s. The distance the disk
rotates in the next 0.750 s is    0   0 zt  12  zt 2  (6.95 rad/s)(0.750 s)  12 (3.84 rad/s 2 )(0.750 s)2  6.29 rad,
which is two revolutions.
4
5