Tài liệu Physics exercises_solution: Chapter 09 pptx

1.50 m
 0.60 rad  34.4.
2.50 m

9.1: a)

b)

(14.0 cm)
 6.27 cm.
(128)(π rad 180)

c)

(1.50 m)(0.70 rad)  1.05 m.

rev   2π rad   1 min 

1900


  199 rad s.
min   rev   60 s 

(35   rad 180) (199 rad s)  3.07  10 3 s.

9.2: a)
b)

dωz
 (12.0 rad s3 ) t , so at t  3.5 s, α  42 rad s 2 . The angular acceleration

dt
is proportional to the time, so the average angular acceleration between any two times is
the arithmetic average of the angular accelerations. b) ωz  (6.0 rad s 3 )t 2 , so at
t  3.5 s, ωz  73.5 rad s. The angular velocity is not linear function of time, so the
average angular velocity is not the arithmetic average or the angular velocity at the
midpoint of the interval.
9.3: a) αz 

9.4:

a)

b)

αz ( t) 

dω z
dt

 2 βt  (1.60 rad s 3 )t.

α z (3.0 s)  (1.60 rad s 3 )(3.0 s)  4.80 rad s 2 .

ω(3.0 s)  ω(0)  2.20 rad s  5.00 rad s

 2.40 rad s 2. ,
3.0 s
3.0 s
which is half as large (in magnitude) as the acceleration at t  3.0 s.
αav  z 

9.5: a) ωz  γ  3 βt 2  (0.400 rad s)  (0.036 rad s 3 )t 2 b) At t  0, ωz  γ 
.50 rad
0.400 rad s. c) At t  5.00 s, ωz  1.3 rad s, θ  3.50 rad, so ωav  z  35.00 s  0.70 rad s.
The acceleration is not constant, but increasing, so the angular velocity is larger than the
average angular velocity.


9.6: ωz  (250 rad s)  (40.0 rad s 2 )t  (4.50 rad s3 )t 2 , αz  (40.0 rad s 2 ) 
(9.00 rad s3 )t. a) Setting ωz  0 results in a quadratic in t ; the only positive time at which
ωz  0 is t  4.23 s. b) At t  4.23 s, α z  78.1 rad s 2 .
c) At t  4.23 s, θ  586 rad  93.3 rev.
d) At t  0, ωz  250 rad s. e) ωav  z  586 rad  138 rad s.
4.23 s
9.7: a) ωz 

dθ
dt

 2bt  3ct 2 and α z 

dw z
dt

 2b  6ct. b) Setting α z  0, t 

b
3c

.

9.8: (a) The angular acceleration is positive, since the angular velocity increases steadily
from a negative value to a positive value.
(b) The angular acceleration is
α

ω  ω0 8.00 rad s  (6.00 rad s)

 2.00 rad s 2
t
7.00 s

Thus it takes 3.00 seconds for the wheel to stop (ω 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 .
(c) We have
θ  0  0t  1 t 2
2

 0  (6.00 rad s) (7.00 s)  1 (2.00 rad s 2 ) (7.00 s) 2
2
 42.0 rad  49.0 rad   7.00 rad.
Alternatively, the average angular velocity is

 6.00 rad s  8.00 rad s
 1.00 rad s
2
Which leads to displacement of 7.00 rad after 7.00 s.
9.9: a) ω  θ0  200 rev, ω0  500 rev min  8.333 rev s, t  30.0 s, ω  ?

ω  ω

θ  θ0   0
 t gives ω  5.00 rev s  300 rpm
 2 
b) Use the information in part (a) to find α :
ω  ω0  t gives α  0.1111 rev s 2
Then ω  0,   0.1111 rev s 2 , ω0  8.333 rev s, t  ?
ω  ω0  αt gives t  75.0 and
 ω  ω
θ  θ0   0
 t gives   0  312 rev
 2 


9.10: a) ωz  ω0 z  α z t  1.50 rad s  (0.300 rad s 2 )(2.50 s)  2.25 rad s.
b) θ  ω0 z t  1 2 α z t 2  (1.50 rad s)(2.50 s)  1 (0.300 rad s 2 )(2.50 s) 2  4.69 rad.
2
min
(200 rev min  500 rev min)  160 s 
rev
 1.25 2 .
(4.00 s)
s
The number of revolutions is the average angular velocity, 350 rev min, times the time
interval of 0.067 min, or 23.33 rev. b) The angular velocity will decrease by another
min
1
200 rev min in a time 200 revmin  1.25 rev s 2  2.67 s.
60 s

9.11: a)

9.12: a) Solving Eq. (9.7) for t gives t 

ω z  ω0 z
αz

.

Rewriting Eq. (9.11) as θ  θ0  t (ω0 z  1 α z t ) and substituting for t gives
2

 ω  ω0 z 
1
 ω0 z  (ωz  ω0 z ) 
θ  θ0   z

 α

2



z
1
 ω  ω0 z 
 (ωz  ω0 z ) z

2




1

(ωz2  ω20 z ),
2
which when rearranged gives Eq. (9.12).





2
b) α z  1 2 1 θ ωz2  ω0 z   1 2 1 (7.00 rad)  16.0 rad s   12.0 rad s  
2

2

8 rad s 2 .

9.13: a) From Eq. 9 . 7 , with ω0 z  0, t 

ωz
z

b ) From Eq. 9.12, with ω0 z  0, θ  θ0 

rad
 136.0rad ss2  24.0 s.
.50
( 36.0 rad s) 2

2 (1.50 rad) s 2

 432 rad  68.8 rev.

9.14: a) The average angular velocity is 162 rad  40.5 rad s, and so the initial angular
4.00 s
velocity is 2ωav  z  ω2 z  ω0 z , ω0z   27 rad s.
ωz 108 rad s  (27 rad s)
αz 
b)

 33.8 rad s 2 .
t
4.00 s


9.15: From Eq. (9.11),
2
θ  θ0 α z t 60.0 rad (2.25 rad s )(4.00 s)
ω0 z 



 10.5 rad s.
2
4.00 s
2
t

9.16: From Eq. (9.7), with ω0 z  0, α z 

ωz
t

2

rad
 140.00 s s  23.33 rad s  The angle is most
6

easily found from θ  ωav  z t  (70 rad s)(6.00 s)  420 rad.
9.17: From Eq. (9.12), with ω 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 9.0 rev.


9.18: The following table gives the revolutions and the angle θ through which the wheel
has rotated for each instant in time and each of the three situations:

t

(a )
rev's θ

( b)
rev's

0.05

0.50 180

0.10
0.15

θ

(c )
rev's θ

0.03

11.3

0.44 158

1.00 360

0.13

45

0.75 270

1.50 540

0.28 101

0.94 338

0.20 2.00 720

0.50 180
1.00 360
––––––––––––––––––––––––––––––––––––––
The θ and ωz graphs are as follows:
a)

b)

c)


9.19: a) Before the circuit breaker trips, the angle through which the wheel turned was
(24.0 rad s) (2.00 s)  (30.0 rad s 2 ) (2.00 s) 2 2  108 rad, so the total angle is
108 rad  432 rad  540 rad. b) The angular velocity when the circuit breaker trips is
24.0 rad s   30.0 rad s 2 2.00 s   84 rad s, so the average angular velocity while the
432 rad
wheel is slowing is 42.0 rad s, and the time to slow to a stop is 42.0 rad s  10.3 s, so the





time when the wheel stops is 12.3 s . c) Of the many ways to find the angular
acceleration, the most direct is to use the intermediate calculation of part (b) to find that
while slowing down ωz   84 rad s so  z  84 .rad s   8.17 rad s 2 .
10 3 s
α
9.20: a) Equation (9.7) is solved for ω0 z  ωz  αz t , which gives ωz  ave  ωz  2z t , or

θ  θ0  ωz t  1 α z t 2 . b) 2

2



ωz
t



 2θ   0.125 rad s 2 . c) ωz  α z t  5.5 rad s.
t

9.21: The horizontal component of velocity is rω , so the magnitude of the velocity is
a) 47.1 m/s
2

b)


 π rad/s  
2
 (5.0 m)(90 rev / min ) 
   (4.0 m/s)  47.3 m/s.


 30 rev/min  


9.22: a)

1.25 m s
25.0 10 3 m

 50.0 rad s ,

1.25 m
58.0 10 3

 21.55 rad s , or 21.6 rad s to three figures.

b) (1.25 m s ) (74.0 min) (60 s min) = 5.55 km.
c)  z 

50.0 rad s  21.55 rad s
( 74.0 min) (60 s min)

 6.41  10 3 rad s 2 .

9.23: a) ω2 r  (6.00 rad s) 2 (0.500 m)  18 m s 2 .
b) v  ωr  (6.00 rad s) (0.500 m)  3.00 m s , and

v2
r



( 3.00 m s ) 2
( 0.500 m )

 18 m s 2 .

9.24: From arad  ω2 r ,

400,000  9.80 m s 2
a

 1.25  10 4 rad s,
r
2.50  10 2 m
1 rev 2 π rad
which is (1.25  10 4 rad s) 1 min 60 s  1.20  10 5 rev min.

ω





9.25: a) arad  0, atan  αr  0.600 rad s 2  0.300 m   0.180 m s 2 and so a  0.180 m s 2 .
b)θ  π rad, so arad  ω2 r  20.600 rad s 2  π 3 rad 0.300 m   0.377 m s 2 .
3

The tangential acceleration is still 0.180 m s 2 , and so on
a

0.180 m s   0.377 m s 
2 2

2 2

 0.418 m s 2 .

c) For an angle of 120, a rad  0.754 m s 2 , and a  0.775 m s 2 , since a tan is
still 0.180 m/s 2
9.26: a) ωz  ω0z  α z t  0.250 rev s  0.900 rev s 2  0.200 s   0.430 rev s
(note that since ω0 z and α z are given in terms of revolutions, it’s not necessary to
convert to radians). b) ωav  z t  (0.340 rev s) (0.2s)  0.068 rev . c) Here, the conversion
to radians must be made to use Eq. (9.13), and
 0.750 m 
v  rω  
 0.430 rev/s  2π rad rev  1.01 m s.
2 

d) Combining equations (9.14) and (9.15),

a  a 2 rad  a 2 tan  (ω2 r ) 2  (αr ) 2
 [((0.430 rev s  2π rad rev)2 (0.375 m)) 2  ((0.900 rev s 2  2π rad rev)(0.375 m))2
 3.46 m s 2 .
9.27:

r

(3000)(9.80 m s 2 )
arad

 rad s
ω2
(5000 rev min) 30 rev min







2

 10.7 cm,

so the diameter is more than 12.7 cm, contrary to the claim.



1
2


9.28: a) Combining Equations (9.13) and (9.15),

v
arad  ω2 r  ω2    ωv.
ω
a rad
0.500 m s
b) From the result of part (a), ω  v  2.00 m s  0.250 rad s .

9.29: a) ωr  (1250 rev min)
b)

v2
r



( 0.831 m s )

2

a tan
r
50.0 m s
0.200

rad s
π
30 rev min



12.710 3 m
2

  0.831 m s .

 109 m s 2 .

(12.7 10 3 m ) 2

9.30: a) α 



2

.0 m
 10.200 ms   50.0 rad s 2 b) At t  3.00 s, v  50.0 m s and
0

ω v 
 250 rad s and at t  0, v  50.0 m s  (10.0 m s 2 )
r
(0  3.00 s)  80.0 m s , so ω  400 rad s . c) ωavet  (325 rad s)(3.00 s)

 975 rad  155 rev. d) v  arad r  (9.80 m s 2 )(0.200 m)  1.40 m s . This speed will
be reached at time

50.0 m s 1.40 m s
10.0 m s

 4.86 s after t  3.00 s, or at t  7.86 s . (There are many

equivalent ways to do this calculation.)

9.31: (a) For a given radius and mass, the force is proportional to the square of the
angular velocity;



  2.29 (note that conversion to rad s is not necessary for this

640 rev min 2
423 rev min

part). b) For a given radius, the tangential speed is proportional to the angular velocity;
640
 1.51 (again conversion of the units of angular speed is not necessary).
423
c) (640 rev min)

arad 

v2
r

9.32: (a)



2

(15.75 m s)
( 0.470 m 2 )



π rad s
30 rev min



0.470 m
2

  15.75 m s, or 15.7 m s to three figures, and

 1.06  103 m s 2  108 g.

vT  Rω
 7.5 rev   1 min   2π rad 
2.00 cm s  R 



 min   60 s   1 rev 
R  2.55 cm

b)

D  2 R  5.09 cm
aT  Rα

α

aT 0.400 m s 2

 15.7 rad s 2
R
0.0255 m

vr 5.00 m s

 15.15 rad s.
r 0.330 m
The angular velocity of the front wheel is ωf  0.600 rev s  3.77 rad s
Points on the chain all move at the same speed, so rr ωr  rf ωf
rr  rr ωf ωr   2.99 cm

9.33: The angular velocity of the rear wheel is ωr 

9.34: The distances of the masses from the axis are
the moment of inertia is
2

2

L
4

, L and 34L , and so from Eq. 9.16,
4
2

L
L
 3L  11
I  m   m   m   mL2 .
4
4

 4  16

9.35: The moment of inertia of the cylinder is M
moment of inertia of the combination is

L2
12
2

M
12  m L .
2

2

and that of each cap is m L4 , so the

9.36: Since the rod is 500 times as long as it is wide, it can be considered slender.
a) From Table 9.2a ,
I

1
1
2
ML2  0.042 kg 1.50 m   7.88  10  3 kg  m 2 .
12
12

b) From Table 9.2b ,

1
1
2
I  ML2  0.042 kg 1.50 m   3.15  10 2 kg  m 2 .
3
3
c) For this slender rod, the moment of inertia about the axis is obtained by considering it
as a solid cylinder, and from Table 9.2f ,
I

1
1
MR 2  (0.042 kg) (1.5  10 3 m)2  4.73  10 8 kg  m 2 .
2
2


9.37: a) For each mass, the square of the distance from the axis is
2(0.200 m)2  8.00  102 m 2 , and the moment of inertia is
4(0.200 kg) (0.800  102 m 2 )  6.40  102 kg  m 2 . b) Each sphere is 0.200 m from the
axis, so the moment of inertia is 40.200 kg 0.200 m   3.20  102 kg  m 2 .
a) The two masses through which the axis passes do not contribute to the moment of
2





2

I  2(0.2 kg) 0.2 2 m  0.032 kg  m 2 .

inertia.

9.38: (a) I  I bar  I balls 


1
 L
M bar L2  2mballs  
12
2

2

1
4.00 kg 2.00 m 2  20.500 kg 1.00 m 2  2.33 kg  m 2
12

1
(b) I  mbar L2  mball L2
3


1
4.00 kg 2.00 m 2  0.500 kg 2.00 m 2  7.33 kg  m 2
3

c) I  0 because all masses are on the axis
(d) I  mbar d 2  2mballd 2  M Total d 2

 (5.00 kg)(0.500 m) 2  1.25 kg  m 2

9.39:

I  I d  I r (d  disk, r  ring)
3

disk : md  (3.00 g cm )πrd2  23.56 kg
1
md rd2  2.945 kg  m 2
2
ring : mr  (2.00 g cm3 ) π (r22  r12 )  15.08 kg
Id 

1
mr (r12  r22 )  5.580 kg  m 2
2
I  I d  I r  8.52 kg  m 2

Ir 

(r1  50.0 cm, r2  70.0 cm )


9.40: a) In the expression of Eq. (9.16), each term will have the mass multiplied by
f 3 and the distance multiplied by f , and so the moment of inertia is multiplied by
f 3 ( f ) 2  f 5 . b) (2.5)(48)5  6.37  108.

9.41: Each of the eight spokes may be treated as a slender rod about an axis through an
end, so the moment of inertia of the combination is

m

I  mrim R 2  8  spoke  R 2
 3 


8


 (1.40 kg )  (0.20 kg) (0.300 m)2
3


2
 0.193 kg  m

9.42: a) From Eq. (9.17), with I from Table (9.2(a)),
1 1
1
rev 2π rad rev 2
(117 kg )(2.08 m) 2 (2400
K
mL2ω2 

)  1.3  10 J.
2 12
24
min 60 s min
b) From mgy  K ,





1.3  10 6 J
K
y

 1.16  10 3 m  1.16 km.
2
mg (117 kg)(9.80 m s )

9.43: a) The units of moment of inertia are [kg] [m 2 ] and the units of ω are equivalent
to [s 1 ] and so the product 1 Iω 2 has units equivalent to [kg  m  s 2 ]  [kg  (m s) 2 ] ,
2
which are the units of Joules. A radian is a ratio of distances and is therefore unitless.
b) K  π 2 Iω2 1800 , when is in rev min.

9.44: Solving Eq. (9.17) for I,
2K
2(0.025J)
I 2 
 2.25  103 kg  m 2 .
2 π rad s 2
ω
(45 rev min  60 rev min )


2
9.45: From Eq. (9.17), K 2  K  1 I (ω2  ω12 ), and solving for I,

2

I 2
2

K 2  K1 
2
(ω2  ω12 )

(500 J)
((520 rev min) 2  (650 rev min ) 2 )





2
 rad s
30 rev min

 0.600 kg  m 2 .

9.46: The work done on the cylinder is PL, where L is the length of the rope. Combining
Equations (9.17), (9.13) and the expression for I from Table (9.2(g)),
PL 

1ω 2
1 ω v2
(40.0 N)(6.00 m s) 2
v , or P 


 14.7 N.
2g
2 g L 2(9.80 m s 2 )(5.00 m)

9.47: Expressing ω in terms of αrad , ω2 

a rad
R

. Combining with I 

1
MR 2 , Eq. (9.17)
2

11
(70.0 kg)(1.20 m)(3500 m s) 2
MRarad 
 7.35  104 J.
becomes K 
22
4
9.48: a) With I  MR 2 , with expression for v is
v

2 gh
.
1 M m

b) This expression is smaller than that for the solid cylinder; more of the cylinder’s
mass is concentrated at its edge, so for a given speed, the kinetic energy of the cylinder is
larger. A larger fraction of the potential energy is converted to the kinetic energy of the
cylinder, and so less is available for the falling mass.


9.49: a) ω 

2
T

, so Eq. (9.17) becomes K  22 I T 2 .

b) Differentiating the expression found in part (a) with respect to T,
dK
 (4 2 I T 3 ) dT .
dt
dt

c) 2 2 (8.0 kg  m 2 ) (1.5 s) 2  70.2 J, or 70 to two figures.
d) (4 2 (8.0 kg  m 2 ) (1.5 s) 3 )(0.0060)  0.56 W.
9.50: The center of mass has fallen half of the length of the rope, so the change in
gravitational potential energy is
1
1
 mgL   (3.00 kg )(9.80 m s 2 )(10.0 m)  147 J.
2
2

9.51: (120 kg)(9.80 m s 2 )(0.700 m)  823 J.

9.52: In Eq; (9.19), I cm  MR 2 and d  R 2 , so I P  2 MR 2 .
2
2
4
MR 2  MR 2  Md 2 , so d 2  R 2 , and the axis comes nearest to the center of
3
5
15
the sphere at a distance d  (2 15 ) R  (0.516) R.
9.53:

9.54: Using the parallel-axis theorem to find the moment of inertia of a thin rod about an
axis through its end and perpendicular to the rod,
2

I p  I cm

M 2
M 2
 L
 Md 
L  M  
L.
12
3
2
2

1
9.55: Ιp  Ι cm  md 2 ,so Ι  12 Μ a 2  b 2   Μ ( a ) 2  ( b ) 2  , which gives

2
2
1
1
1
Ι  Μ a 2  b 2   a 2  b 2 , or Ι  Μ a 2  b 2 .
12
4
3














1
9.56: a) Ι  12 Μa 2

9.57:

1
b) Ι  12 Μb 2

In Eq. 9.19 , Ι cm 

M
12

L2 and d  L 2  h , so
2
1 2 L
 
Ιp  Μ  L    h  
2
 
12



1
1

 Μ  L2  L2  Lh  h 2 
4
12

1

 Μ  L2  Lh  h 2  ,
3

which is the same as found in Example 9.12.

9.58: The analysis is identical to that of Example 9.13, with the lower limit in the integral
being zero and the upper limit being R, and the mass Μ  πLρR 2 . The result is
Ι  1 ΜR 2 , as given in Table 9.2(f) .
2

9.59: With dm 

M
L

dx
L

M
M x3
dx 
Ι  x
L
L 3
0

L



2

0

M 2

L.
3


9.60: For this case, dm   dx.
L

a)

x2
Μ   dm   γx dx  γ
2
0
L

L

0

yL2

2

4 L

4
x
 γL  Μ L2 .
4
2

4 0
0
This is larger than the moment of inertia of a uniform rod of the same mass and length,
since the mass density is greater further away from the axis than nearer the axis.

b)

Ι   x 2 (γx)dx  γ

L

c)

Ι   ( L  x) 2 γxdx
0

L

   ( L2 x  2 Lx 2  x 3 )dx
0

L

 x2
x3 x 4 
  L2  2 L  
 2
3 4 0



L4

12
M 2

L.
6

This is a third of the result of part (b), reflecting the fact that more of the mass is
concentrated at the right end.

9.61: a) For a clockwise rotation, ω will be out of the page. b ) The upward direction


crossed into the radial direction is, by the right-hand rule, counterclockwise. ω and r are
 

perpendicular, so the magnitude of ω  r is ωr  v. c) Geometrically, ω is



perpendicular to v , and so ω  v has magnitude ωv  arad, and from the right-hand
rule, the upward direction crossed into the counterclockwise direction is inward, the

direction of arad . Algebraically,







a rad  ω  v  ω  ω  r 
  
  
 ω ω  r   r ω  ω 

   2r ,


where the fact that ω and r are perpendicular has been used to eliminate their dot
product.


9.62:

For planetary alignment, earth must go through 60 more than Mars:
θE  θM  60
wE t  ωM t  60
60
t
ωE  ωM

wE 
t

360
360
and wM 
1yr
1.9 yr

 365d 
60
 0.352 yr 
 1yr   128 d

360
 1.9 yr



360
1yr

9.63: a) v  60 mph  26.82 m s
r  12 in.  0.3048 m
v
ω   88.0 rad s  14.0 rev s  840 rpm
r
b) same ω as in part (a) since speedometer reads same
r  15 in.  0.381 m
v  rω  (0.381m)(88.0 rad s)  33.5 m s  75 mph
c) v  50 mph  22.35 m s
r  10 in.  0.254 m
v
ω   88.0 rad s; . this is the same as for 60 mph with correct tires, so
r
speedometer read 60 mph.
9.64: a) For constant angular acceleration θ  ωα , and so arad  ω2 r  2αθr.
2
b) Denoting the angle that the acceleration vector makes with the radial direction as β ,

and using Equations (9.14) and (9.15),
1
a
αr
αr
tan β  tan  2 

,
arad ω r 2 αθ r 2θ
1
so   2 tan β  2 tan136.9  0.666 rad.
2


d
 2γt  3 βt 2  (6.40 rad s 2 )t  (1.50 rad s3 )t 2 .
dt
dwz
αz 
 2γ  6 β t  (6.40 rad/s 2 )  (3.00 rad/s3 )t.
dt

9.65: a) ωz 
b)

c) An extreme of angular velocity occurs when  z  0, which occurs at

t

γ

3β

3 20 rad/s
 1..50 rad/s 3  2.13 s, and at this time
2

ωz  (2γ)(γ / 3 β )  (3 β )(γ / 3 β ) 2  γ 2 / 3 β 

(3.20 rad/s 2 ) 2
 6.83rad/s.
3(0.500 rad/s3 )

9.66: a) By successively integrating Equations (9.5) and (9.3),
β
z  γt  t 2  (1.80 rad/s 2 )t  (0.125 rad/s3 )t 2 ,
2
γ
β
θ  t 2  t 3  (0.90 rad/s 2 )t 2  (0.042 rad/s3 )t 3 .
2
6

γ
b) The maximum positive angular velocity occurs when αz  0, t  , the angular

velocity at this time is
2

 γ β γ
1 γ 2 1 (1.80 rad/s 2 ) 2

 6.48 rad/s.
ωz  γ     

 β 2 β
2 β 2 (0.25 rad/s3 )
 
 
2γ
The maximum angular displacement occurs when ωz  0, at time t 
(t = 0 is an
β
inflection point, and θ (0) is not a maximum) and the angular displacement at this time is
2

3

  2 
β  2  2  3 2 (1.80 rad/s 2 )3
 62.2 rad.
=
θ      
2  
6  β  3 β 2 3 (0.25 rad/s3 ) 2
 
 
9.67: a) The scale factor is 20.0, so the actual speed of the car would be 35 km h  9.72 m

b) (1 2)mv 2  8.51 J. c)  

2K

I

 652 rad s.


9.68: a) α 

a tan
r



3.00 m s 2
60.0 m

 0.050 rad s 2 . b) αt  (0.05 rad s 2 )(6.00s)  0.300 rad s.

c) arad  2 r  (0.300 rad s) 2 (60.0 m)  5.40 m s 2 .
d)

2

e) a  a 2 rad  a 2 tan  (5.40 m s ) 2  (3.00 m s 2 ) 2  6.18 m s 2 ,
and the magnitude of the force is F  ma  (1240 kg)(6.18 m s 2 )  7.66 kN.
f) arctan

   arctan   60.9.
a rad
a tan

5.40
3.00

9.69: a) Expressing angular frequencies in units of revolutions per minute may be
accomodated by changing the units of the dynamic quantities; specifically,
2W
ω2  ω12 
I
 2(4000 J ) 
 (300 rev min)  
 16.0 kg  m 2 



 211 rev min.
2

  rad s 

 30 rev min 




2

b) At the initial speed, the 4000 J will be recovered; if this is to be done is 5.00 s, the
power must be 4000sJ  800 W.
5.00

9.70: a) The angular acceleration will be zero when the speed is a maximum, which is at
the bottom of the circle. The speed, from energy considerations, is
v  2 gh  2 gR (1  cos ), where  is the angle from the vertical at release, and

2g
2(9.80 m s 2 )
v
(1  cos ) 

(1  cos 36.9)  1.25 rad s.
(2.50 m)
R
R
b)  will again be 0 when the meatball again passes through the lowest point.
c) arad is directed toward the center, and arad  ω2 R, arad  (1.25 rad s 2 ) (2.50 m)  3.93 m



d) arad  2 R  (2 g R )(1  cos β ) R  (2 g )(1  cos β ), independent of R.
9.71: a) (60.0 rev s)(2π rad rev)(0.45  102 m)  1.696 m s.
b) ω  v 
r

1.696 m s
2.00  10  2 m

 84.8 rad s.

9.72: The second pulley, with half the diameter of the first, must have twice the angular

velocity, and this is the angular velocity of the saw blade.
a)

  rad s   0.208 m 
(2(3450 rev min)) 
  75.1 m s.
 30 rev min  

2



2


  rad s    0.208 m 
4
2
b) arad   r   2(3450 rev min) 
  5.43  10 m s ,
 30 rev min   


2 




so the force holding sawdust on the blade would have to be about 5500 times as strong as
gravity.

2


2
2
arad  2 r  0 r  (2  0 )r

9.73: a)

   0   0  r
   0 

(  0 )t  r
 t 

 [2 (  0 )r.
b) From the above,
αr 

arad (85.0 m s 2  25.0 m s 2 )

 2.00 m s 2 .
2θ
2(15.0 rad)

c) Similar to the derivation of part (a),

1
1 2
1

K  ω2 I  ω0 I  [α ][2θ ]I  I.
2
2
2
d) Using the result of part (c),

(45.0 J  20.0 J)
K

 0.208 kg  m 2 .
2
 ((2.00 m s ) /(0.250 m))(15.0 rad)

I

9.74: I  I wood  I lead

2
2
mw R 2  mL R 2
5
3
4
mw  ρwVw  ρw πR 3
3


mL  σ L AL  σ L 4π R 2
2
4

2

I   ρ w π R 3  R 2  ( L 4 π R 2 ) R 2
5
3
3

8
ρ R

 π R4  w  σL 
3
 5

 (800 kg m3 )(0.20 m)

8
 20 kg m 2 
(0.20 m)4 
3
5


2
 0.70 kgm



9.75: I approximate my body as a vertical cylinder with mass 80 kg, length 1.7 m, and
diameter 0.30 m (radius 0.15 m)

I

1
1
mR 2  (80 kg) (0.15 m)2  0.9 kg  m 2
2
2

9.76: Treat the V like two thin 0.160 kg bars, each 25 cm long.
1

1
  2 mL2   2  (0.160 kg)(0.250 m)2
3

 3
3
 6.67  10 kg  m 2

9.77: a)   90.0 rpm  9.425 rad s
K

1 2
2 K 2( 10.0  106 J)
Ιω so Ι  2 
 2.252  105 kg  m 2
2
( 9.425 rad s) 2
ω

m  ρV  ρπR 2t ( ρ  7800 kg m 3 is the density of iron and t=0.100 m is the
thickness of the flywheel)
1
1
Ι  mR 2  ρπtR 4
2
2
R  (2 I ρπt )1 4  3.68 m; diameter  7.36 m
b) ac  Rω2  327 m s 2

9.78: Quantitatively, from Table (9.2), I A  1 mR 2 , Ι B  mR 2 and Ι C  2 mR 2  a) Object A
2
3
has the smallest moment of inertia because, of the three objects, its mass is the most
concentrated near its axis. b) Conversely, object B’s mass is concentrated and farthest
from its axis. c) Because Ι sphere  2 5 mR 2 , the sphere would replace the disk as having the
smallest moment of inertia.


9.79: a) See Exercise 9.50.

2π 2 Ι 2π 2 (0.3308)(5.97  1024 kg)(6.38  106 m)2
K 2 
 2.14  1029 J.
2
(86, 164 s)
T
2

b)

1  2πR 
2π 2 (5.97  1024 kg)(1.50  1011 m)2
M
 2.66  1033 J.
 
2  T 
(3.156  107 s) 2

c) Since the Earth’s moment on inertia is less than that of a uniform sphere, more of
the Earth’s mass must be concentrated near its center.
9.80: Using energy considerations, the system gains as kinetic energy the lost potential
energy, mgR. The kinetic energy is
1
1
1
1
1
K  Ιω2  mv 2  Ιω2  m(ωR) 2  ( Ι  mR 2 )  2 
2
2
2
2
2
2
1
Using Ι  2 mR and solving for ω,

ω2 

4g
4g
, and  
.
3R
3R


9.81: a)

Consider a small strip of width dy and a distance y below the top of the triangle. The
length of the strip is x   y h b.
1
The strip has area x dy and the area of the sign is bh, so the mass of the strip is
2
 xdy 
 yb  2 dy   2M 
dm  M  1   M  
   2  y dy
 bh 
 h  bh   h 
2 

dI 

1
3

dm x 2  2Mb
4

3h

2

y 3 dy

2Mb 2 h 3
2Mb 2
y dy 
0
3h 4 0
3h 4
b) I  1 Mb 2  2.304 kg  m 2
6
ω  2.00 rev s  4.00π rad s
h

I   dI 

K  1 Iω2  182 J
2

1 4 h 1
2
 y |0   Mb
4
6



9.82: (a) The kinetic energy of the falling mass after 2.00 m is
2
KE  1 mv 2  1 8.00 kg 5.00 m/s   100 J. The change in its potential energy while
2
2





falling is mgh  8.00 kg  9.8 m/s2 2.00 m   156.8 J
The wheel must have the “missing” 56.8 J in the form of rotational KE. Since its
outer rim is moving at the same speed as the falling mass, 5.00 m s :

v  r


KE 

v 5.00m/s

 13.51 rad/s
r 0.370m

1 2
I ; therefore
2
2 KE
256.8 J 

I 2 
 0.6224 kg  m 2 or 0.622 kg  m 2
2

13.51 rad s 

(b) The wheel’s mass is 280 N 9.8 m s 2 = 28.6 kg. The wheel with the largest possible
moment of inertia would have all this mass concentrated in its rim. Its moment of inertia
would be
I  MR 2  28.6kg 0.370m   3.92 kg  m 2
The boss’s wheel is physically impossible.
2





9.83: a) 0.160 kg  0.500 m  9.80 m s 2  0.784 J. b) The kinetic energy of the stick
is 0.784 J, and so the angular velocity is



2k
2k


I
ML2 3

2(0.784 J)

 5.42 rad s .
0.160 kg 1.00 m 2 3

This result may also be found by using the algebraic form for the kinetic energy,
K  MgL 2, from which   3 g L , giving the same result. Note that ω is independent
of the mass.
c) v  L  5.42 rad s 1.00 m   5.42 m s
d) 2gL  4.43 m s ; This is 2 3 of the result of part (c).


9.84: Taking the zero of gravitational potential energy to be at the axle, the initial
potential energy is zero (the rope is wrapped in a circle with center on the axle).When
the rope has unwound, its center of mass is a distance R below the axle, since the length
of the rope is 2R and half this distance is the position of the center of the mass. Initially,
every part of the rope is moving with speed  0 R, and when the rope has unwound, and
the cylinder has angular speed  , the speed of the rope is R (the upper end of the rope
has the same tangential speed at the edge of the cylinder). From conservation of energy,
using I  (1 2) MR 2 for a uniform cylinder,
 M m 2 2  M m 2 2
   R 0     R   mgπ R.
 4 2
 4 2
Solving for  gives
4πmg R  ,
2
ω  ω0 
M  2 m 
and the speed of any part of the rope is v  R.
9.85: In descending a distance d, gravity has done work mB gd and friction has done
work   K m A gd , and so the total kinetic energy of the system is gd mB  μK mA . In

terms of the speed v of the blocks, the kinetic energy is
1
1
1
K  m A  m B v 2  I 2  m A  m B  I R 2 v 2 ,
2
2
2
where ω  v R, and condition that the rope not slip, have been used. Setting the kinetic
energy equal to the work done and solving for the speed v,
2 gd mB   k m A 
v
mA  mB  I R 2 .
9.86: The gravitational potential energy which has become kinetic energy is
K  4.00 kg  2.00 kg  9.80 m s 2 5.00 m   98.0 J. In terms of the common speed v of
the blocks, the kinetic energy of the system is





1
1 v
K  (m1  m2 )v 2  I  
2
2  R
 v2

2

(0.480 kg  m 2 ) 
1
 4.00 kg  2.00 kg 
  v 2 (12.4 kg).
2
(0.160 m) 2 



Solving for v gives v 

98.0 J
12.4 kg

 2.81 m s.