Tài liệu Physics exercises_solution: Chapter 10 pdf

10.1: Equation (10.2) or Eq. (10.3) is used for all parts.
a)
m,N00.4090sin N)m)(10.000.4( 
out of the page.
b)
m,N6.34120sin N)m)(10.000.4( 
out of the page.
c)
m,N0.2030sin N)m)(10.000.4( 
out of the page.
d)
m,N3.1760sin N)m)(10.0000.2( 
into the page.
e) The force is applied at the origin, so
0.τ
f)
.0180sin N)m)(10.000.4( 
10.2:
m,N40.0m)N)(5.0000.8(
1
τ

m,N0.1230sin m)N)(2.00(12.0
2
τ
where positive torques are taken counterclockwise, so the net torque is
m,N0.28 
with
the minus sign indicating a clockwise torque, or a torque into the page.
10.3: Taking positive torques to be counterclockwise (out of the page),
m,N2.34N)m)(26.0(0.09m,N1.62N)(180.0m)090.0(

21
 ττ
 
m,N78.1N)(14.0m)090.0(2
3


so the net torque is
m,N50.2 
with the
direction counterclockwise (out of the page). Note that for
3

the applied force is
perpendicular to the lever arm.
10.4:
RFFRFRFττ )(
122121


m.N726.0m)N)(0.3307.50N30.5( 
10.5: a)
b) Into the plane of the page.
c)
]
ˆ
N)00.4(
ˆ
n)00.5[(]
ˆ

m)150.0(
ˆ
m)450.0[( iijiFr 





ˆ
m)N1.05(
ˆ
N)00.5m)(150.0(N)00.4(m)450.0(
k
k


10.6: (a)
CCWm,N8.7m)2.0)(60N)(sin 50(
A
τ

CWm,N10m)2.0(N)50(
CWm,N5m)2.0)(30N)(sin 50(
0
D
C
B




τ
τ
τ
(b)
mN10mN5mN8.7 τ

CWm,N3.6 
10.7:
kg00.2kg,40.8 where,2
22
3
2
 mMmRMRI

2
mkg600.0 I

mN0524.0 ,
;srad0.08726gives
?s,0.30;srad5.236rpm0.50;srad7.854rpm75.0
2
0
0



IατIατ
ααtωω
α
tωω

f
10.8:
 
 
 
m.N1.13
s00.8
minrev400
mkg50.2 a)
minrev
srad
60
2
2







t
II
α
b)
.J102.19
minrev
srad
60
2

minrev400)mkg50.2(
2
1
2
1
3
2
22










π
I
10.9:
 
 
in found that assame the,sm2.1m0.2sm36.022
2
 asv
Example 9-8.
10.10:
  
 

.srad00.2
mkg0.5
m250.0N0.40
2
2



I
FR
I
τ
α
10.11: a)


















Mm
mM
g
Mm
m
MgTMgn
21
3
21

b) This is less than the total weight; the suspended mass is accelerating down, so the
tension is less than mg. c) As long as the cable remains taut, the velocity of the mass
does not affect the acceleration, and the tension and normal force are unchanged.
10.12: a) The cylinder does not move, so the net force must be zero. The cable exerts
a horizontal force to the right, and gravity exerts a downward force, so the normal force
must exert a force up and to the left, as shown in Fig. (10.9).
b)
 
  
N,490sm80.9kg)50(N0.9
2
2
2
n
at an angle of arctan
 
 1.1
490
0.9

from the
vertical (the weight is much larger than the applied force F ).
10.13:
 
n
tMR
Rn
I
n
R
n
f
2
0
k







   
 
  
.482.0
N160s50.72
minrev850m260.0kg0.50
minrev
srad

30


10.14: (a) Falling stone:
2
2
1
atg 

 
 
)2(
::Pulley
)1(::Stone
sm80.2
s00.3m6.12
2
1
R
2
2
1
2
2
1
2
2
2
1
MaT

MR
αMRTRIατ
maTmgmaF
a
a
a






Solve (1) and (2):
kg00.2
m/s80.2m/s80.9
m/s80.2
2
kg0.10
2
22
2




























M
ag
aM
M
(b)From (2):
 
 
N0.14
m/s80.2kg0.10
2
1
2

1
2


T
MaT
10.15:
  
2
2
2
1
2
2
1
mkg02320.0m0750.0kg8.25  mRI

 
N47.7so

rad/s046.8gives2
?rad,33.0rev25.5;0rad/s;23.04rpm220
k
k
kk
2
0
2
0
2

00





Rμ
Iα
nIαnRμ
nRμRfττ
Iατ
αθθαωω
αθθωω
f
10.16: This is the same situtation as in Example 10.3. a)
N.0.42)21(  MmmgT
b)
s.m11.82(12  mMghv
c) There are many ways to find the time of fall.
Rather than make the intermediate calculation of the acceleration, the time is the distance
divided by the average speed, or
 
s.69.12 vh
d) The normal force in Fig.
(10.10(b)) is the sum of the tension found in part (a) and the weight of the windlass, a
total 159.6 N (keeping extra figures in part ( a)).
10.17: See Example 10.4. In this case, the moment of inertia I is unknown, so
 
  
.

2
2121
RImmgma 
a)
   
,m/s75.3s80.0m20.12
2
2
1
a
 
N.2.18andN50.7so
122111
 agmTamT
b) The torque on the pulley is
 
m,N0.803
12
 RTT
and the angular acceleration is
.mkg0.016so,rad/s50
22
1
 IRaα
10.18:
.
3
2
3
1

Ml
F
Ml
Fl
I



10.19: The acceleration of the mass is related to the tension by
,
cm
TMgMa 
and the
angular acceleration is related to the torque by
2
cmcm
and/ where,/or , MRIRStaαMTaTRτIα 
have been used.
a) Solving these for
N.882.02/gives  MgTT
b) Substituting the expression for T
into either of the above relations gives
which from,2/
cm
ga 
s.553.042
cm
 ghaht
rad/s.9.33c)
cmcm

 RtaRvω
10.20: See Example 10.6 and Exercise 10.21. In this case,
.srad33.9,and
cmcm
2
cm2
 RvωghvMvK
10.21: From Eq. (10.11), the fraction of the total kinetic energy that is rotational is
 
     
,
1
1
/1
1
2121
21
cm
2
22
cmcm
2
cm
2
cm
2
cm
I
MR
vIMIMv

I







where
 Rv
cm
for an object that is rolling without slipping has been used.
a)
,5)MR2(Ib).31isratioabove theso,)21(
22
cm
 MRI
so the above ratio is
,32c) .72
2
MRI 
so the ratio is
,MR85d) .52
2
I
so the ratio is
.135
10.22: a) The acceleration down the slope is
,sin
M

f
θga 
the torque about the
center of the shell is

,
3
2
3
2
2
MRa
R
a
MR
R
a
II
αRfτ 
.so
3
2
a
M
f

Solving these relations
a
for
f

and simultaneously gives
or,sin
3
5
θga 
.N83.4)smkg)(3.6200.2(
3
2
3
2
,sm62.30.38sin)sm80.9(
5
3
sin
5
3
2
22


Maf
θga
The normal force is Mg cos
θ
, and since
,
s
nμf 
.313.0tan
5

2
cos
sin
3
2
cos3
2
cos
5
3
3
2
s
 θ
θ
g
θg
θg
a
θMg
Ma
n
f
μ
b)
2
sm62.3a
since it does not depend on the mass. The frictional force, however,
is twice as large, 9.65 N, since it does depend on the mass. The minimum value of
s

μ
also
does not change.
10.23:

)1eq.()cos(sin
cossin
cos
s
s
aθμθg
ma
θmgμθmg
mgn



n
and
mg
act at the center of the ball and provide no torque.
αmRθmgμIατ
mRIθRmgμττ
f
2
5
2
s
2
5

2
s
cosgives
;cos


No slipping means
(eq.2) cos so ,
5
2
s
agμRaα 
We have two equations in the two unknowns a and
.
s

Solving gives
613.00.65 tan tanandsin
7
2
7
2
7
5
 θμθga
s
b) Repeat the calculation of part (a), but now
.
2
3

2
mRI 
858.00.65 tan tanandsin
5
2
5
2
s
5
3
 θμθga
The value of
s

calculated in part (a) is not large enough to prevent slipping for the
hollow ball.
c) There is no slipping at the point of contact.
10.24:
slippingnofor
cm

Rv 
a) Get v at bottom:
ghv
R
v
mRmvmgh
I
ωmvmgh
7

10
5
2
2
1
2
1
2
1
2
1
2
22
22















Now use energy conservation. Rotational KE does not change

h
g
gh
g
v
h
KEhmgKEmv
7
5
22
2
1
7
10
2
RotRot
2






hhhmgmgh 



(b)
With friction on both halves, all the PE gets converted
back to PE. With one smooth side, some of the PE remains as rotational KE.

10.25:
cm
2
2
1
0
2
cm1
)2/1( mvwIKWwh
f

Solving for
h
with
Rwv 
cm
 
m.7.11
N392
J3500

]s)rad0.25(m)600.0()rad0.25(m)600.0)(800.0([
2222
sm80.9
2
1
2




w
s
h
w
10.26: a)
The angular speed of the ball must decrease, and so the torque is provided by a friction
force that acts up the hill.
b) The friction force results in an angular acceleration, related by
.fRI 

The
equation of motion is
cm,
sin mafβmg 
and the acceleration and angular acceleration
are related by
Rαa 
cm
(note that positive acceleration is taken to be down the incline,
and relation between
cm
a
and

is correct for a friction force directed uphill).
Combining,
 
,571sin
2
ma

mR
I
ma
βmg 







 
.sin75 which from
cm
βga 
c) From either of the above relations between if f and
,
cm
a
,cossin
7
2
5
2
sscm
βmgμnμβmgmaf 
from which
 
. tan72
s


μ
10.27: a)
    
  
 
rad/s,3086.0s0.15mkg2100m40.2N0.18
2
 tIFRtαω
or 0.309 rad/s to three figures.
b)
   
2121
2
2
 IωKW
 
 
J.100srad3086.0mkg00.2
2
2

c)
W.67.6,or either From
ave
 PtWPτωP
10.28: a)
  
 
m.N519

rev/min
rad/s
30
rev/min2400
hp/ W746hp175








π
ω
P
τ
b)
  
J.32612mN519  πθτW
10.29: a)
t
ω
IIατ



   
 
 

m.N377.0
s5.2
minrev
srad
30
minrev1200m100.0kg50.121
2










π
b)
  
rad.157rev0.25
s/min60
s5.2rev/min600
ave
t

c)
J.2.59θτ
d)
2

2
1
IωK 

 
J, 2.59
rev/min
rad/s
30
rev/min)1200(m)kg)(0.1005.1)(2/1(
2
1
2
2

















π
the same as in part (c).
10.30: From Eq. (10.26), the power output is
W,2161
rev/min
rad/s
60
2
rev/min 4800m)N30.4(









τωP
which is 2.9 hp.
10.31: a) With no load, the only torque to be overcome is friction in the bearings
(neglecting air friction), and the bearing radius is small compared to the blade radius, so
any frictional torque could be neglected.
b)
N.6.65
m)086.0(
rev/min
rad/s
30
rev/min)(2400

W/hp)hp)(7469.1(/








π
R
ωP
R
τ
F
10.32:
22
2
1
2
2
1
mkg2.42m)kg)(2.08117(  mLI
a)
.rad/s2.46
mkg42.2
mN1950
2
2





I
τ
α
b)
rad/s.9.53rev)2rev0.5)(rad/s2.46(22
2


c) From either
(10.24),Eq.or
2
1
2

 KW
J. 106.13rad/rev)2revN.m)(5.001950(
4
 πτW
d), e) The time may be found from the angular acceleration and the total angle, but the
instantaneous power is also found from
hp).kW(141105 τωP
The average power is
half of this, or
kW.6.52
10.33: a)
m.N358
rev/min

rad/s
30
rev/min)(400 W)10150(/
3
















π
ω
Pτ
b) If the tension in the rope is
N. 1079.1/soand ,
3
 RτwwFF
c) Assuming ideal efficiency, the rate at which the weight gains potential energy is the
power output of the motor, or
m/s.8.83so,  wPvPwv

Equivalently,
.Rv


10.34: As a point, the woman’s moment of inertia with respect to the disk axis is
2
mR
,
and so the total angular momentum is
s./mkg1028.5
rad/rev)2rev/s500.0(m)00.4(kg50.0kg110
2
1

2
1
)(
23
2
2
womandiskwomandisk

















π
ω
RmMωIILLL
10.35: a)
,s/mkg115sin
2
φmvr
with a direction from the right hand rule of into the
page.
b)
      
,smkg125mN12536.990sinm8kgN8.9kg2
2
2
 τdtdL
out of the page.
10.36: For both parts,
.IωL 
Also,
,rv

so
).( rvIL 

a)
mvrrvmrL  ))((
2

smkg1067.2)m1050.1(s)m1098.2(kg)1097.5(
24011424
L
b)
))(52(
2
ωmrL 

smkg1007.7
hr))s3600hr(24.0rad2()m1038.6)(kg1097.5)(52(
233
2624


L
10.37: The period of a second hand is one minute, so the angular momentum is
s.mkg1071.4
s60
2
)m100.15(
3
kg100.6
2
3
2622
3

2

















T
l
M
I
ωL
10.38: The moment of inertia is proportional to the square of the radius, and so the
angular velocity will be proportional to the inverse of the square of the radius, and the
final angular velocity is
s.rad106.4
km16
km100.7
ds400,86(d)(30

rad2
3
2
5
2
2
1
12































R
R
10.39: a) The net force is due to the tension in the rope, which always acts in the radial
direction, so the angular momentum with respect to the hole is constant.
b)
,,
2
222
1
2
11
rmLrmL


and with
.rad00.7)(,
2
211221
srrωωLL 
c)
J.1003.1))()(()21(
22

11
2
22

 rωrωmK
d) No other force does work, so
J1003.1
2

of work were done in pulling the cord.
10.40: The skater’s initial moment of inertia is
,mkg56.2)m80.1)(kg00.8(
2
1
)mkg400.0(
222
1
I
and her final moment of inertia is
.mkg9.0)m1025)(kg00.8()mkg400.0(
222
2


I
Then from Eq. (10.33),
s.rev14.1
mkg9.0
mkg56.2
s)rev40.0(

2
2
2
1
12




I
I
ωω
Note that conversion from rev/s to
srad
is not necessary.
10.41: If she had tucked, she would have made
40.0)mkg18)mkg6.3()2(
22

rev in
the last 1.0 s, so she would have made
60.0)0.15.1)(rev40.0( 
rev in the total 1.5 s.
10.42: Let

.mkg1360)m00.2)(kg0.40(mkg1200
,mkg1200
2222
02
2

01


mRII
II
Then, from Eq. (10.33),

s.rad924.0
kg.m1360
kg.m1200
s6.00
rad2
2
2
2
1
12








π
I
I
10.43: a) From conservation of angular momentum,
 

 
 
srad385.1
1207021
srad3.0

21
1
21
21
1
22
2
1
2
0
1
12









Mm
ω
mRMR

MR
ω
mRI
I
ωω

or 1.39
srad
to three figures
b)
      
kJ,80.1srad00.3m00.2kg1202121
22
1
K
and
    
 
J.499m00.2kg7021
2
2
2
02
 ωIK
In changing the parachutist’s horizontal
component of velocity and slowing down the turntable, friction does negative work.
10.44: Let the width of the door be l;
 
   
   

      
s.rad223.0
m500.0kg500.0m00.1kg0.4031
m500.0sm0.12kg0.500

231
2
22
2
2





lmMl
lmv
I
L
ω
Ignoring the mass of the mud in the denominator of the above expression gives
,srad225.0ω
so the mass of the mud in the moment of inertia does affect the third
significant figure.
10.45: Apply conservation of angular momentum
,L

with the axis at the nail. Let object
A be the bug and object B be the bar.
Initially, all objects are at rest and

.0
1
L
Just after the bug jumps, it has angular momentum in one direction of rotation and the
bar is rotating with angular velocity
B
ω
in the opposite direction.
srad120.0
3
gives
andm00.1 where
2
3
1
21
2
3
1
2



rm
vm
ω
rmrvmLL
rmIr
ωIrvmL
B

AA
B
BBAA
BBBBAA
10.46:
(a) Conservation of angular momentum:
srad885
)m00.2(
sm9.80
N0.90
3
1
)m50.1)(sm00.6)(kg00.3()m(1.50s)m0.10)(kg00.3(
2
2
2
2
3
1
101
.ω
ω
ω
Lmvdmdvm












(b) There are no unbalanced torques about the pivot, so angular momentum is
conserved. But the pivot exerts an unbalanced horizontal external force on the system, so
the linear momentum is not conserved.
10.47:
10.48: a) Since the gyroscope is precessing in a horizontal plane, there can be no net
vertical force on the gyroscope, so the force that the pivot exerts must be equal in
magnitude to the weight of the gyroscope,
 
 
N,617.1sm80.9kg165.0
2
 mgωF
1.62 N to three figures.
b) Solving Eq. (10.36) for
,ω

 
 
 
 
,srad7.188
s20.2rad2mkg1020.1
m1000.4N617.1
24
2









πI
ωR
ω
which is
.minrev1080.1
3

Note that in this and similar situations, since

appears in
the denominator of the expression for
,

the conversion from
srev
and back to
minrev
must be made.
c)
10.49: a)
P
MR

P
K
22
))2/1)((2/1(



 
  
s,1021.2
W1046.7
rev/min)(500m)kg)(2.00000,60)(2/1()2/1(
3
4
2
rev/min
rad/s
30
2




or 36.8 min.

m.N1010.1
360
rad2
/s)00.1(
rev/min

rad/s
30
rev/min)500(m)kg)(2.00000,60)(2/1(
b)
5
2

















ππ
Iτ
10.50: Using Eq. (10.36) for all parts, a) halved b) doubled (assuming that the
added weight is distributed in such a way that
r
and I are not changed) c) halved
(assuming that

w
and
r
are not changed) d) doubled e) unchanged.