Tài liệu Physics exercises_solution: Chapter 09 pptx

rad2500, dt dw z dt dθ 9.8: a The angular acceleration is positive, since the angular velocity increases steadily from a negative value to a positive value.. 9.12, with ω z 0, the numb

9.1: a) 0.60rad 34.4

m502

m50

.

)180rad)(

(128

cm)0.14

min1rev

rad2min

is proportional to the time, so the average angular acceleration between any two times is the arithmetic average of the angular accelerations b) ω z (6.0rad s3)t2, so at

s

rad73.5s,

t 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

9.4: a) α (t) dω dt 2βt ( 1.60rad s3)t

b) α z(3.0s)(1.60rad s3)(3.0s)4.80rad s2

,srad40.2s

0.3

srad00.5srad20.2s

0.3

)0(s)0

rad, 3.50s,

rad3.1s,5.00At

9.6: (250rad s)(40.0rad s ) (4.50rad s ) , (40.0rad s2)

z 2 3

at which time

positiveonly

the

;

in quadratica

in results0Settinga)

93.3rad586s,

4.23

At b)s

α t

t

s

rad138e)

s

rad2500,

dt

dw z dt

dθ

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

2

s00.7

s)rad00.6(srad00

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 0rad s to8.00rad s (c) We have

rad

7.00rad

49.0rad0.42

s)(7.00)srad00.2(s)(7.00s)rad00.6(

2 1

2 2

1 0 0

srad00.8srad00.6





Which leads to displacement of 7.00 rad after 7.00 s

9.9: a)ωθ0 200rev,ω0 500rev min8.333revs, t 30.0s,ω?

2

and 75.0gives

?s,rev 333.8,srev1111.00,

Then

srev1111.0gives

0

0 0

0

0 2

2 0

θ

t αt ω

ω

t ω

ω

α t

ω

ω

9.10: a) 1.50 rad s (0.300rad s2)(2.50s) 2.25 rad s.

s00.4(

)minrev500minrev200(

2 s

60min1

9.12: a) Solving Eq (9.7) for t gives 0

z z z

α ω ω

t Rewriting Eq (9.11) as ( 21 )

21

2)(

1

)(

21

0 2

0 0

0 0

0 0

z z

z z z z

z z z

z

z z

ω ω

ω ω ω ω

ω ω ω

α

ω ω θ θ

z z

)srad27(srad

9.15: From Eq (9.11),

s

rad5.102

s)00.4)(

srad(2.25s

4.00

rad0.602

2 0

t

θ θ

z

s 00 6 s rad 140

z

easily found from θ ωavzt(70rad s)(6.00s)420rad

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:

θ θ

θ

)c( srev'

)b(

srev'

)a( 0.05 0.50 180 0.03 11.3 0.44 1580.10 1.00 360 0.13 45 0.75 2700.15 1.50 540 0.28 101 0.94 338

0.20 2.00 720 0.50 180 1.00 360––––––––––––––––––––––––––––––––––––––

The θand ω graphs are as follows: z

a)

b)

c)

9.19: a) Before the circuit breaker trips, the angle through which the wheel turned was

1082s)(2.00)srad(30.0s)

rad

108   b) The angular velocity when the circuit breaker trips is

24.0rad s30.0rad s2 2.00s84rad s,so the average angular velocity while the wheel is slowing is 42.0rad s, and the time to slow to a stop is 42.0432radrads 10.3s, so the time when the wheel stops is 12.3s 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 ω z  84rad s so 8.17rad s2

s 3 10 s rad

)min/rev90m)(

.

74

(

s rad 55 21 s rad

.sm

2

2

1025

1  4 rad s,which is (1.25104 rad s)11revmin2π60rads  201 105 rev min

(note that since ω0z and α are given in terms of revolutions, it’s not necessary to z

convert to radians) b)ωavzt(0.340rev s)(0.2s)0.068rev c) Here, the conversion

to radians must be made to use Eq (9.13), and

0.430rev/s 2 rad rev 1.01m s

2

m750

2srev900.0((

))m375.0(rev)rad2srev

2 2

2 2

2 2

2 tan

2 rad

αr r

ω a

3000(

2 min rev s rad 30

2 2

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

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

.2

2

ω

v ω r ω

ω

9.29: a) (1250rev min)  12 7 102 m 0.831m s

min rev s rad 30

.

12

(

) s m 831

.

0

(

3 2 2

be reached at time 50 0 m10.s0m1 40s m s4.86safter 3.00s,or at 7.86s

t

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; 640423revrevminmin22.29 (note that conversion to rad is not necessary for this spart) b) For a given radius, the tangential speed is proportional to the angular velocity; 51

.108s

m1006

) 2 m 470

.

0

(

s) m 75

g

cm09.52

cm55.2

rev1

2s60

min1min

5.7s

cm00.2

rad π rev

R

m0255.0

sm400





R a α

9.33: The angular velocity of the rear wheel is 15.15rad s.

m330.0

sm00.5r

r

v ω

The angular velocity of the front wheel is ωf 0.600rev s3.77rad s

Points on the chain all move at the same speed, so rrωr rfωf

114

34

4

2 2

2 2

mL

L m

L m

L m

112

13

I

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 ,

.mkg1073.4m)105.1(kg)042.0(2

12

I

9.37: a) For each mass, the square of the distance from the axis is

,m1000.8m)

2 bar balls

1

L m L m

mkg33.7m00.2kg500.0m00.2kg00.43

d

2 2

2

2 1 r r

2 1

2 1

2 2

3 r

2 2

d d d

2 d

3 d

mkg52.8

mkg580.5)(

2

1

)cm 70.0cm,

50.0(

kg15.08)

()cmg00.2

1

kg56.23)

cmg00.3

I

r r m I

r r

r r π m

r m I

πr m

9.40: a) In the expression of Eq (9.16), each term will have the mass multiplied by

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

2

2

2 spoke 2

rim

mkg193.0

m)(0.300kg)

(0.203

8)kg40.1(

38

revrad2min

rev2400()m08.2)(

kg117(24

112

kg117(

J103

9.43: a) The units of moment of inertia are [kg][m2] and the units of ω are equivalent

to [s 1] and so the product 2

2

1Iω has units equivalent to [kgms 2][kg(m s)2], which are the units of Joules A radian is a ratio of distances and is therefore unitless b) K π2Iω2 1800, when is in rev min

9.44: Solving Eq (9.17) for I,

.mkg1025.2)min

rev(45

J)025.0(2

2 min rev s rad 60 2

9.45: From Eq (9.17), ( 2),

1

2 2 2

))minrev650()minrev520((

)J500(2

)(

2

2

2 min rev s rad 30 2 2

2 1

2 2

1 2

K K I

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)),

N

7.14)m00.5)(

sm80.9(2

)sm00.6)(

N0.40(2

1

or ,2

1

2

2 2



L

v g

ω P v

g

ω PL

9.47: Expressing ω in terms of , 2 a rad

m20.1)(

kg0.70(2

12

2

m M

gh v





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 2T, so Eq (9.17) becomes K 22I T2.

b) Differentiating the expression found in part (a) with respect to T,

.)4

dt dT T I dt

c) 22(8.0kgm2) (1.5s)2 70.2J,or 70 to twofigures

d) (42(8.0kgm2) (1 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

J

147)

m0.10)(

sm80.9)(

kg00.3(2

12

so ,5

23

2MR2  MR2 Md2 d2  R2 and the axis comes nearest to the center of

the sphere at a distance d (2 15)R(0.516)R

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,

.32

12

2

2 2

112

b a Μ Ι b a b

a Μ

, h Lh L Μ

h Lh L L Μ

h

L L Μ Ιp

2 2

2

2 2

31

4

1121

212

1

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 πLρR2 The result is

2 L

0

3 2

L

M dx L

M x Ι

9.60: For this case, dm dx.

a)

22

2 0 0

x γ

γx dx

dm Μ

L L

0

2 2 4 0

4

4)

.612

43

22

)2

2 0

3 2 2

L M L

x x L

x L

dx x Lx x L

L L

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  c) Geometrically, ω v  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 arad

ω

ω

r ω ω v ω

M E

t ω t w

θ θ

yr9.1

360and

yr136060

M E

M E

ω ω t

128d

yr1

d365352

.060

yr 9

1360yr 1

m3048.0.in12

m381.0(

m0.381in

15

samereads

r speedometesince

(a)part

in assame

10

sm22.35mph

50c)

rad0.88

rad ω r αθr

b) Denoting the angle that the acceleration vector makes with the radial direction as β ,

and using Equations (9.14) and (9.15),

,2

12

rr

rad

tan

θ r αθ

α r ω

α a

3 2

)rad/s20.3(3/)3/)(

3()3/)(

2

2 2 2

,)rad/s125.0()rad/s80.1(2

3 3 2

2 3

2

2 3 2

2

t t

t

β t

γ θ

t t

t

β γt z

γ β β

γ γ

ω z

2 22

1

)rad/s80.1(2

1

3

2 2

2

3

22

)rad/s80.1(3

2

2 3

3 2



9.67: a) The scale factor is 20.0, so the actual speed of the car would be 35km h9.72 m

s

rad652c)

J

51.8)

2

1

(

9.68: a) 0.050rad s2 b) (0.05rad s2)(6.00 ) 0.300rad s.

m 0 60 s m 00 3

9.69: a) Expressing angular frequencies in units of revolutions per minute may be

accomodated by changing the units of the dynamic quantities; specifically,

min

rev211

minrev

srad30m

kg16.0

)J4000(2min)rev300(

2

2 2

2

2 1 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 40005.00sJ 800 W

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



), wherecos

1(2

)sm80.9(2)cos1(

v

b)  will again be 0 when the meatball again passes through the lowest point

c) arad is directed toward the center, and , (1.25rad s2 (2.50m) 3.93m

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

2

m208.0minrev

srad30min))rev3450

srad30min)rev3450(

2 2

9.73: a) a r r ( 2)r

0 2 2

0

2 rad     

0

0 0

0 0

r

r t t

)sm0.25sm0.85(2

2 2

2][

[2

12

12

)sm00.2((

J)0.20J0.45

2 L

2 w

343

25

2

πR ρ V ρ

m

R m R

23

45

2

4

2 2 L 2

3 w

2 L L L L

R R π R

R π ρ I

R π σ A σ

8

σ R ρ R π

2

2

3 4

kgm70

0

mkg205

)m20.0)(

mkg800(m)20.0(3

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)

2 2

2 (80kg)(0.15m) 0.9kg m2

12

2 2

mkg1067.6

m)250.0kg)(

(0.1603

123

12

2

6 2

s)rad4259

J1001022so2

(

)

( ω

K Ι Ιω K

mkg7800



ρV ρπR t ρ

thickness of the flywheel)

4 2

2

12

1

ρπtR mR

m36.7diameter m;

68.3)2

 I ρπt

R

2 2

b)a  Rω 

3 2 2

2 2

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 2

sphere 2 mR5

Ι  , the sphere would replace the disk as having the smallest moment of inertia

9.79: a) See Exercise 9.50.

J

1014.2s)

164,86(

m)1038.6kg)(

1097.5)(

3308.0(2

2

2 6 24

2 2

)s10156.3(

m)1050.1kg)(

1097.5(22

2

2 7

2 11 24

2 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

)(

2

1)(2

12

12

12

1Ιω2 mv2 Ιω2 m ωR 2 Ι mR2

,for solvingand

4

and ,3

42

R

g R

g

dy h

yb M bh

xdy M

2 0

13

23

2

Mb y

h

Mb dy

y h

Mb dI

9.82: (a) The kinetic energy of the falling mass after 2.00 m is

8.00kg5.00m/s2 100J.2

1 2

m/s00

J8.562

2  28.6kg 0.370m 3.92kgm

 MR

I

The boss’s wheel is physically impossible

9.83: a) 0.160kg0.500m 9.80m s20.784J b) The kinetic energy of the stick

is 0.784 J, and so the angular velocity is

0.160kg1.00m 3 5.42rad s.

J)2(0.7843

22

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  below the axle, since the length R

of the rope is 2Rand half this distance is the position of the center of the mass Initially, every part of the rope is moving with speed 0R, and when the rope has unwound, and the cylinder has angular speed  the speed of the rope is ,  (the upper end of the rope R

has the same tangential speed at the edge of the cylinder) From conservation of energy, using I (1 2)MR2for a uniform cylinder,

242

4

2 2 2

0

R m

R πmg ω

ω







and the speed of any part of the rope is vR

9.85: In descending a distance d, gravity has done work m B gd and friction has done work Km A gd, and so the total kinetic energy of the system is gdm B μKm A In

terms of the speed v of the blocks, the kinetic energy is

2

12

12

v R I m m I

v m m

whereω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 k 2

R I m m

m m

gd v

B A

A B

the blocks, the kinetic energy of the system is

2 2

2

1)(

)m160.0(

)mkg480.0(kg002kg00.42

9.87: The moment of inertia of the hoop about the nail is 2MR2(see Exercise 9.52), and the initial potential energy with respect to the center of the loop when its center is directly

below the nail is gR (1cosβ) From the work-energy theorem,

minrev

srad60

2minrev3000m)

kg)(0.901000

(2

12

1

7

2 2

J1000.2

4 7 ave

which is about 18 min

9.89: a) ((0.80kg)(2.50 10 m) (1.60kg)(500 10 m) )

2

12

12

2 2

2 1 1

.1()mkg1025.2((

1(

m)00.2)(

sm80.9(2

)(

12

2 2

3

2 2

gh v

c) The same calculation, with R instead of 2 R gives 1 v4.95m s This does make sense, because for a given total energy, the disk combination will have a larger fraction of the kinetic energy with the string of the larger radius, and with this larger fraction, the disk combination must be moving faster

9.90: a) In the case that no energy is lost, the rebound height h is related to the speed

9.91: We can use Κ(cylinder)250J to find  for the cylinder and v for the mass.

2 1 2 2

v

h

9.92: Energy conservation: Loss of PE of box equals gain in KE of system.

sm68.3kg)

00.7(kg50.1

m))(1.50s

mkg)(9.8000

.3(

4

14

12

1

2

12

1

2

12

12

1and2

1

2

12

1

4 1 2

C 4

1 p 4

1 B 2 1

B B

2 B C

2 B P

2 B B B

2

C

B 2 C C

2 2

P

2 B B B

cylinder

Box cylinder

p

Box pulley

2 cylinder cylinder

2 pulley pulley

2 box box box

gh m v

v m v

m v

m gh m

r

v r m

r

v r m v

m gh m

r

v ω

r

v ω

ω I

ω I v

m gh m

p

B p

9.93: a) The initial moment of inertia is 2

92

42

116

2 2

2

MR R

247512

92

MR MR

MR

512 383 2 2

1 2 2

2

9.94: a) From the parallel-axis theorem, the moment of inertia is

I P (2 5)MR2ML2,and

.5

21

i i

i i i

9.96: Each side has length a and mass M4 , and the moment of inertia of each side about

an axis perpendicular to the side and through its center is 2 48

perpendicular axis theorem,  2 12

2 4 48

2

Ma   The total moment of inertia is the sum of the contributions from the four sides, or 4Ma122  Ma32