INTRODUCTION to BIOMECHANICS for HUMAN MOTION ANALYSIS, SECOND EDITION

SOLUTIONS to ODD-NUMBERED PROBLEMS

by D. Gordon E. Robertson, PhD, FCSB

School of Human Kinetics, University of Ottawa © Copyright 2004 (revised 12 February 2010)

INTRODUCTION (p. 12)

Conversion factors are taken from Table 1.3 on page 8. 1. (a) 350 × 0.4536 × 9.81 = 1557 N (b) 6.50 / 0.4536 = 14.33 lbs. (c) 168.5 × 2.54 = 428 cm (d) (10 × 100) /2.54 = 394 inches

(e) m/s 3.31km1

m 1000s3600

hour 1mile 1

km 609.1hourmiles0.70 =×××

(f) m/s 7.128mile 1

km 609.1hourmiles0.80 =×

(g) 8.35 × 12 × 2.54 = 255 cm (h) 440 × 0.9144 = 402 m (i) (800 / 0.9144) × 3 = 2620 feet (j) 50.0 × 1.609 = 80.5 km (k) 25.0 / 1.609 = 15.54 miles (l) 3.00 × 9.81 = 29.4 newtons 3.

N 5.111281.94.113

kg 4.113lbs 1

kg 4536.01

lbs250

=×==

=×

mgW

Thus, the 1200 N person weighs more than the 250 lbs. person. 5. m 716.13cm 6.137154.2540in. 5401245ft. 45 ==×==×= Thus, 13.75 m is longer than 45 feet.

2FUNDAMENTAL CONCEPTS (p. 26)

⎟⎠⎞

⎜⎝⎛==

+==

−

xyry

yxrrx

1

22

tanθ θsin

θcos

1.

(a) o5.54)250/350(tanθ

cm 4303502501

22

==

=+=−

r (b) o6.26)00.2/000.1(tanθ

kN 24.2000.100.21

22

==

=+=−

r

(c) o6.116)00.10/0.20(tanθ

m/s 22.40.2000.101

22

−=−−=

=−+−=−

r (d) o4.63)1000/2000(tanθ

N 2240200010001

22

==

=+=−

r

(e) o5.54)250/350(tanθ

m/s 0.43)0.35(0.251

222

−==

=−+=−

r (f)

o4.108)00.5/00.15(tanθ

kN 81.1500.1500.51

22

=−=

=+−=−

r

3.

(a) m/s 57.100.25sin0.25

m/s 7.220.25cos0.25

==

==o

o

y

x (b)

N 00.50.30sin00.10

N 66.80.30cos00.10

==

==o

o

y

x

(c) 2

2

m/s 7.700.45sin0.100

m/s 7.700.45cos0.100

−=−=

=−=o

o

y

x (d)

m 00.5)2/πsin(00.5m 0.0)2/πcos(00.5

====

yx

(e) N 55.80.160sin0.25

N 5.230.160cos0.25

==

−==o

o

y

x (f)

m/s 68.17)4/πsin(0.25m/s 17.68)4/πcos(0.25

====

yx

5. Sum of sides is always greater than length of hypotenuse. 7. (a) cm )00.10 ,00.10()0.400.50 ,0.200.30( =−− (b) N )300 ,0.100()400700 ,500400( −=−− 9. R = (750, 1000) N = 1250 N at 53.1 deg 11. km 93.525.575.2 22 =+=r

3RESOLUTION of FORCES into COMPONENTS (p. 33)

θsin θcos

weight 221

FFFFr

mGmFmgW

yx

G

==

===

1. N 49581.95.50 =×== mgW

3.

N 6699.8168.2mgW

kg 2.68lbs 2.2045

kg 1lbs. 150

=×==

=×

5.

kg 7.6381.9

625===

gWm

7.

)6.21 ,03.2()10)5(256.1 ,0)25.1(253.4( )()00.7 ,25.11()51210 ,25.1100( )(

)68.9 ,34.12()556.13 ,25.153.43( )()4.40 ,5.39()56.11230 ,53.41025( )(

−=−−+−−−+=−−=−−−−=

−=−−×−×==−+++=

RdRcRbRa

)00.18 ,00.15()]3012( ),2510[( )()00.7 ,25.11()10125 ,01025.1( )(

)50.10 ,75.8()1230(41 ),1053.2(

41 )(

)78.2 ,87.7()256.1(21 ,6.5)53.4(

21 )(

=−−−=−−=+−−+−−=

=⎥⎦⎤

⎢⎣⎡ ++=

−=⎥⎦⎤

⎢⎣⎡ −−+=

RhRg

Rf

Re

9.

N 00.0

kg 0.60

N 1.980.60)81.9(61

=

=

==

space

space

moon

F

m

F

11.

2

26

2311

2

m/s 71.3

)10396.3(10419.610673.6

=

××

××=

×=

−

mars

mars

rmGg

4MOMENT of FORCE (p. 42-3)

θsinrFMFdM

==

1.

N 303650.1

500===

dMF

3. (a) N 00.00.500.1250.250.50 =−+−−=Σ= FR (b)

N.m00.550.1275.1825.10

)25.0(0.50)15.0(0.125)05.0(0.25)0.0(0.50=−+−=

−+−=Σ= iiR dFM

(c)

0.1100.012.50)/125(1.25d50.0(0.25)25.0(0.05)00)125.0(d

050.0(0.25)25.0(0.05))00(050)125.0(d0

C

C

C

A

=+=++=

=−−−∴=Σ

...

M

The force at C should be moved 11.00 cm to the right of A. 5. (a) ΣM = 0.50)20.2(250)50.2(200 −=− Therefore, Cathy will go up. (b) Jill must move 20.0 cm towards Cathy so the two moments are equal. (c)

33300150050d

0150.0(d)250(2.20)200(2.50)0

...

M

=−−

=

=−−∴=Σ

Therefore, Ian must sit 33.3 cm from fulcrum on Cathy’s side. 7. (a) N.m 0.75)25.0(300 −=−== FdM A (b) N.m 635)(0.20)20(200θ(0.20) O .coscosFM CA −=−== Note, vertical component at C cancels vertical component at D and horizontal component at D has no moment about axis at A. 9. N.m 7.36)650.0(5.56 === FdM

5LAWS of STATICS (p. 58-60)

0)( 0

0 0

)(F 0

=×Σ=Σ=Σ

=Σ=Σ

−=×==Σ

FrMM

FF

kFrFrrMF

A

y

xyyxx

1. (a) N.m 00.18)0.5020.00.8035.0( kM =×−×= (b) N.m 00.13)0.3020.00.2035.0( kM =−×−×= (c) N.m 00.16)]0.300.50(30.0)0.200.80(10.0[ kM −=−×−+×−= (d) N.m 00.7)0.2030.00.2010.0( =×−×−=M (e) N.m 275)0.8550.20.50250.1( −=×−−×=M (f) N.m 5.20)0.8530.00.5010.0( kM −=×−−×−= 3.

N.m 20.5)51.46205.073.311366.0(

)73.31 ,51.46()3.34sin3.56 ,3.34cos3.56()θsin,θcos(kM

FFF−=×−×=

=== oo

5. (a) N.m 600.0N.cm 0.60)0.2500.80.3500.4( −=−=×−×=M (b) N.m 40.3N.cm 340)0.2500.80.3500.4( −=−=×−×−=M (c) N.m 40.3N.cm 340)0.25)00.8(0.3500.4( −==×−−×=M (d) N.m 600.0N.cm 0.60)0.25)00.8(0.3500.4( ==×−−×−=M 7.

cm 101.9m 019.1)81.9(0.60

30000.2==

×=

×=

mgFLC A

9.

0)()(

0

0

=×+×+=Σ

=−+=Σ

=+=Σ

ggkneekneekneecg

ykneeygy

xkneexgx

FrFrMM

mgFFF

FFF

6 15.

cableper N 392

2)81.9(0.80

02 :0

==

=−=Σ

cable

cabley

F

mgFF

17. (a) jacking up a car, prying with a bottle opener, shoveling (b) throwing a dart, kicking, jumping 21.

N 30040030sin200

0 :0

N 2.17330cos200

0 :0

21

21

2

21

21

=+−=+=

=−+=Σ

==

=

=+−=Σ

o

o

mgFF

WFFF

F

FF

FFF

xx

yyy

x

xx

xxx

23.

N 1650.4030sin250

0 :0

N 21730cos250

0 :0

=+=+=

=−−=Σ

==

=

=−=Σ

o

o

mgFF

WFFF

F

FF

FFF

xloadxknee

yloadykneey

xknee

xloadxknee

xloadxkneex

7DRY FRICTION (pp. 71-2)

normalkinetickinetic

normalstaticstatic

FFFF

µµ

==

1.

N 4415.49090.0µN 4665.49095.0µ

5.49081.9500 :0

=×===×==

=×===−=Σ

normalkinetickinetic

normalstaticstatic

normal

normaln

FFFF

mgFmgFF

3.

direction negativein N 8.1884.34390.0µ:motionin isobject Since

4.34381.9350 :0

=×==

=×===−=Σ

normalkinetickinetic

normal

normaln

FF

mgFmgFF

5.

incline theup N 0.12725882.081.95015sin: thatsee we triangle thengconstructi be

0 :0:rule trianglethe

applycan you staticsfor zero toaddmust they and forces only three are thereSince

=××==

=++=Σ

WF

WFFF

friction

normalfriction

7.

2.1174.14680.0µ4.146sin45 20012cos81.930

0sin4512cos

:0

=×===−××=

=+−

=Σ

normalstaticstatic

normal

appliednormal

n

FFF

FmgF

F

oo

oo

incline. down the N 2.80 isfriction an smaller th is valueabsulute thissince

2.80cos45 20012sin81.930

0cos4512sin

:0 Assume

static

equilbrium

appliedequilbrium

t

F

F

FmgF

F

−=−××=

=+−

=Σ

oo

oo

9.

).(friction moving isbody thean greater th is valueabsolute since

2.487cos13 500

0cos13 :0 Assume

N 4.48260380.0µ0.603sin13 50081.950

0sin13 :0

kineticstatic

equilbrium

appliedequilbriumt

normalstaticstatic

normal

appliednormaln

FF

F

FFF

FFF

FmgFF

=

−=−=

=+=Σ

=×===+×=

=−−=Σ

o

o

o

o

811.

N 5.16641.4310.12310 sin250

010 cos :0

. equalmust friction themovingbox get the To

N 10.1232.246500.0µ2.246cos10 250

010 cos :0

=+=+=

=+−−=Σ

=×====

=−=Σ

o

o

o

o

staticapplied

appliedstatict

static

normalstaticstatic

normal

normaln

FF

FWFF

F

FFF

WFF

13.

N 1324)81.90.150(900.0

N 1398)81.90.150(950.0=×=µ=

=×=µ=

normalkinetickinetic

normalstaticstatic

FFFF

15.

204.01.980.20

255.01.980.25

N 1.9881.900.10

===µ

===µ

=×=

normal

kinetickinetic

normal

staticstatic

normal

FF

FF

F

17.

731.03.294

215

765.03.294

225N 3.29481.90.30

===µ

===µ

=×=

normal

kinetickinetic

normal

staticstatic

normal

FF

FF

F

9LINEAR KINEMATICS (pp. 87-8)

tvvss

ssavv

attvss

atvv

fiif

ifif

iif

if

)(

(2

21

)22

221

++=

−+=

++=

+=

1. (a)

0 to 40 m: )(2

22

if

if

ssvv

a−

=−

2

22

m/s 0125.18081

)40(29

)040(209

==

=−

−=a

40 to 70 m: a = 0 m/s2

70 to 100 m: 2½atvtss if ++=

2

2

m/s 750.016

)3630(2

)4(21)4(970100

−=−

=

++=

a

a

(b)

s 22.16433.389.8s 00.4

s 33.39

4070

s 89.80125.1

09

10070

7040

400

=++==

=−

=

=−

=−

=

−

−

−

total

if

tt

t

avv

t

(c)

m 66.1258081

2100 2 =⎟

⎠⎞

⎜⎝⎛++=fs

(d)

m/s 00.64439100 =⎟

⎠⎞

⎜⎝⎛−=v

3. (a)

s 11.7005.0

50.3536.3

m/s 536.3

)025)(005.0(25.3 22

=−+

=−

=

±=

−+±=

avv

t

v

if

f

(b)

m 3.3525.035

10)005.0(21105.30

21

22

2

=+=

+×+=

++= attvss iif

5. (a)

s 33.7300.0

20.20=

−−

=−

=a

vvt if

(b)

m 0780.3)(7.33)(21

(7.333)2020

at21(7.333)vss

2

2iif

.

.

=−

++=

++=

7.

ms 24.0 s 2400.01736

67.410

1736

)0500.0(267.410

)(2

:catcher 2

ms 14.40 s 40014.02894

67.410

2894

)0300.0(267.410

)(2

:catcher 1

m/s 67.416.3km/h 150

222

nd

222

st

==−−

=−

=

−=

−−

=−

−=

==−−

=−

=

−=

−−

=−

−=

=÷=

avv

t

ssvv

a

avv

t

ssvv

a

v

if

if

if

if

if

if

The difference in the accelerations is 1158 m/s2. The difference in the times is 9.60 ms.

109.

m/s36.8500.1/546.12/

m 546.12232.700.12

22

===

=⎟⎠⎞

⎜⎝⎛+=

tsv

s

Ball speed must be greater than 8.36 m/s. 11. (a)

m/s 417.0s 60

min 1min 4

m 100=×==

tsv

(b)

m 2850.1007.266

m 7.266min 4.00km 1

m 1000min 60h 1

hkm00.4

22

=+=

=×××==

total

currentriverdown

s

tvs

(c) m/s 187.1)0.6000.4/(8.284/ =×== tsv total 13.

90.40.24

)000.6)(00.2(2)(2

)(202

2

±=±=

−−−=−−=

−+=

i

ifi

ifi

v

ssav

ssav

The initial velocity must be 4.90 m/s.

11PROJECTILE MOTION (pp. 94-5)

tvvss

ssgvv

gtss

gtvv

fyiyiyfy

iyfyiyfy

iyfy

iyfy

)(21

)(222

++=

−−=

−=

−=

1. (a)

m/s 179.25sin25θsinm/s 90.245cos25θcos

m/s 0.25s 3600

h 1km 1

m 100090

=°===°==

=××

vvvv

hkm

y

x

(b)

st

tvss xixfx

402.09.24010

=−

=

−=

(c)

cm 2.24m 242.00)81.9(2)179.2(0

2

)(2

2

22

22

==+−−

=

+−

−=

−= −

iyiyfy

fy

iyfyiyfy

sgvv

s

ssgvv

(d)

m/s 6.28350.0

010=

−=xv

3. (a)

2

22

m/s 81.19

m/s 81.194.392

)020)(81.9(20

)(2

−=

±=±=

−−−±=

−= −

f

fy

iyfyiyfy

v

v

ssgvv

(b)

s 02.281.9

081.19=

−−−

=−

=−

gvv

t if

(c) m 04.4)02.2(20 =+=−= tvss xixtx

5. (a)

m 835.1)81.9(2

00.62

00

2

22

22

==+

++=

−

−= +

gv

gvv

ss

iy

iyfyiyfy

(b)

s 16.281.9

624.15 velocitynegative select the

m/s 24.152.232

)100)(81.9(26

)(22

22

=−

−−=

±=±=

−−=

−−=

−

−=

t

v

ssgvv

gvv

t

f

ifif

if

(c) m 082.1)16.2(500.0 === tvs xx

7. θ vx vy ymax xmax time 30 ̊ 8.66 5.00 1.270 8.83 1.019 45 ̊ 7.07 7.07 2.55 10.19 1.442 60 ̊ 5.00 8.66 3.82 8.83 1.765

9.

m/s 3.31319.010

s 319.081.9

013.3

13.381.9

81.9)5.10.1(202

==

=−

−−=

−=

−=±=

=−−=

−

x

if

fy

fy

v

gvv

t

v

gv

11.

fps 8.294206.012

4026.0

m/s 94.3

)795.0)(81.9(202

==

==

−=

−−=

fps

f

f

f

v

gv

t

v

v

12

ANGULAR KINEMATICS (p. 99)

t

tt

t

fiif

ifif

iif

if

)(θθ

θθ(2

θθ

21

)22

221

ω+ω+=

−α+ω=ω

α+ω+=

αω=ω +

1. (a)

22

2if

rad/s 429rad/s 3

r/s 50012

25tωωα

.

.

=π=

=−

=−

=

(b)

srevolution 00.72)5.1(21)2(20

α21ωθθ

2

2

=++=

++= ttiif

3.

srevolution 227

36)35.0(2100θ 2

=

+=

5.

rad/s 75.9)5(750.06

αωω

=+=

+= tif

7.

rad/s 2.2189.0π6

89.0 3ω ===r

9.

rad/s 00.500.3)200.0(60.5αωω =−+=+= tinitialfinal 11.

s 33.13500.1

0.200αω-ω

=−

−== initialfinalt

13RELATIONSHIP between LINEAR and ANGULAR MEASURES (pp. 104-5)

22

22

tr

tr

t

t

aaa

rvra

rarv

+=

=ω=

α=ω=

1. (a) m/s 25.11)15(75.0ω === rvt (b)

2

2222

22

m/s 203

75.1685.117

75.168)15(75.0ω

5.117)150(75.0α

=

+=+=

===

===

rt

r

t

aaa

ra

ra

3. (a)

m/s 00.9

deg180rad πdeg573900.0ω

=

×⎟⎠⎞

⎜⎝⎛==

srvt

(b)

2rad/s 67.65.1

010=

−=

ω−ω=α

tif

(c)

2

2222

22

m/s 4.90

0.9000.9

0.90)10(9.0ω

00.9)10(9.0α

=

+=+=

===

===

rt

r

t

aaa

ra

ra

5.

m/s 50.7

)00.10(75.0ω

==

= rvt

2

2222

2

222

m/s 0.75

5.175

m/s 500.1)00.2(75.0α

m/s 0.75)10(75.0ω

=

+=+=

===

===

rt

t

r

aaa

ra

ra

7. (a)

m/s 00.13)10(30.1ω

m/s 0.23)10(30.2ω

===

===

rv

rv

tcg

tfeet

(b)

rad/s 00.52

010ω =−

==rv

(c)

m 29.616.845.14

m 16.8)3.1(π2π2

m 45.14)30.2(π2π2

=−=

===

===

rs

rs

cg

feet

(d)

2

22

m/s 230

)00.10(30.2ω

=

== ra feetr

9. (a)

m 1714.07.175.8375.16

ω5.170.1

,ω

;ω

m/s 875.14)75.8(70.1ωm 700.170.00.1

=−=

+=+

==

====+=

r

vr

vrrv

rvr

t

t

total

(b)

2

22

m/s 12.13

)75.8(1714.0ω

=

== rar

11.

m/s 681.0

)81.6(100.0m/s 25.2)81.6(330.0

rad/s 81.6s 60

min 1r 1rad 2

minr65

=

=ω===ω=

=

×π

×=ω

rvrv

pushwheel

rim

14

LAW of ACCELERATION (p. 110)

yy

xx

maFmaF

amF

==

=

ΣΣΣ

1.

N 0813501060maF

m/s 35016081

030290

xx2vva 2

2

if

2i

2f

.).(.

.)()(

−=−==

−=−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=

3.

N 472)81.9(40)2(40

00.21

02

=+=+=

=−

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

−==Σ

mgmaFt

vva

mgFmaF

ylifter

yiyfy

lifteryy

5.

m 49.8)886.5(2

10002

Thus,

m/s 886.570

N 4126.07.6867.686)81.9(70

222

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

+=⎟⎟⎠

⎞⎜⎜⎝

⎛ −+=

−=−

=

−==Σ=×==

===

x

ixfxif

x

kineticxx

kineticnormalkinetic

normal

avv

xx

Fkinetica

FmaFFFmgF

µ

7.

s 70.2111.1

03m/s 111.1900/1000

90010002

=−

=−

=

==

===Σ

avv

t

a

amaF

if

9.

N 00.12)67.66(180.0

m/s 67.663.0200 2

−=−==

−=−

=−

=

maFt

vva if

11.

2

2

m/s 41.820/2.168

m/s 05.220/41

N 2.1682.19628232025

N 0.41525835

−=−=

==

−=−=−+−==Σ

=−+==Σ

y

x

yy

xx

a

a

mgmaF

maF

15MOMENT of FORCE (p. 115)

d F MIM

=α=Σ

1.

2rad/s 50.172/35α ===I

M

3.

2rad/s 10.2/α

N.m 5.52)35.0(150

==

===

IM

FdM

5.

N.m 44.13)42.0(0.32α

420.000.5

45.435.2ωωα

−=−==

−=−

=−

=

IMt

if

7.

N.m 0.14035.0400 =×== FdM 9.

2kg.m 00.1020/200α/

N.m 20040.0500

===

=×==

MI

FdM

16MOMENT of INERTIA (p. 122)

2cgaxis

2cgcg

cg

mrII

mkI

/LkK

+=

=

= cg

1.

2

22

kg.m 000.1

)25.0(85.0

m 250.085.0

=

+=+=

===

mrIImI

k

cgaxis

cgcg

3.

22

2

222

kg.m 90.6)50.0(00.1290.3

kg.m 90.3)57.0(00.12

m 57.0)95.0(60.0

=+=

+=

===

===

mrII

mkI

KLk

cghip

cg

5. (a)

2

22

kg.m 0.1842.1688.15

)450.1(0.8080.15

=+=

+=+= mrII cgbar

(b)

m 444.01975.00.80

80.15===k

7.

m 1581.00.90/25.2/

kg/m 25.220/45

rad/s 0.201/20/)ωω(α

N.m 0.45)15.0(300

2

2

===

===

==−=

===

mIk

M/aI

t

FdM

if

9.

cm 2.41m 241.005.7/453.0/

kg.m 410.0

)1925.0(05.71489.02

22

====

=

+=+=

mIk

mrII

proximalproximal

cgproximal

17LAW of REACTION (p. 129)

1.

2

2

m/s 19.158.53/)81.98.531345(

m/s 22.48.53/227/

=×−=

=−=Σ

===

==Σ

y

ygyy

gxx

xgxx

a

mamgFF

mFa

maFF

3.

deg 2.21)3875.0(tan

81.983.3111tanθ

83.31π/100θ/θ

1

221

==

×=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

====

−

−

rgv

srrs

5.

km 37.6m 637181.9

25022

2

====

=

gvr

rvmmg

7.

N 686

)81.975.0(0.65

=

+=+=

−==Σ

mgmaF

mgFmaFy

ygy

gyy

9.

N 0.1755)00.2(50.3

ω/2

22tr

−=−=

−=−= mrrmvma

11.

N 720 14)81.920(75 =×== maF

18LINEAR IMPULSE and MOMENTUM (p. 140)

∫∫

−=

==

==

if mvmvFdt

tFFdt

mvp

impulse

momentum

1. N 655)909.10(60100.1

00.1200.60 −=−=⎟⎠⎞

⎜⎝⎛ −

== amF

3. N.s 5.1370)50.2(0.55Impulse =−=−== if mvmvFt

5. 7.

N 772)87.12(0.60

m/s 87.12400.0

)127.5(0

m/s 1247.5476.26

487.26)0350.1)(81.9(20

)(2

2

22

===

=−−

=−

=

−==

=−−−=

−−=

landinglanding

iflanding

fy

ifiyfy

maFt

vva

v

yygvv

9.

N 0.40050.0

)400(005.0−=

−=

−=

−=

tmvF

mvmvtF

i

if

11.

m/s 00.64/)1(0.240 =+=+=m

tFvv if

m/s 37.50.70/86.37586.3752.1)81.9(7012000

m/s 86.20.70/200

2000

===−+=

−+=

==

+=+=

∫

∫

fy

yiyfy

fx

xixfx

v

WtdtFmvmv

v

dtFmvmv

19ANGULAR IMPULSE and MOMENTUM (p. 145)

∫∫

−=

==

==

if IIMdt

tMMdt

IL

ωω

impulseangular

ωmomentumangular

1. 3. 5.

N.m.s 8.168

500.0)350.1(250)(=

== tdFtM

7.

N.m.s 8.108

5)320.0(0.68)(=

== tdFtM

9.

/skg.m 75.4)55.2(8641.1ω

kg.m 8641.1)4026.0(50.11

m4026.0)235.1(326.0

2

222

===

===

===

IL

mkI

Klk

/skg.m 6.27

)52.5(00.5ω2=

== IL

/skg.m 9.79)137.14(65.5

1rad π225.265.5ω

2==

⎟⎠⎞

⎜⎝⎛ ×==

rsrIL

20CONSERVATION of MOMENTUM (p. 153)

constantωILLconstantmvpp

if

if

======

1. (a) /skg.m 0.40)20(2ω 2=== IL (b)

2

2

2

kg.m 504.16.26/

kg.m 333.10.30/

/kg.m 0.40ω

==

==

====

LI

LI

sLLLI

land

top

landtopstarttoptop

3. (a)

rad/s 19.1943.0/25.8ωω430.025.8

ω

===

= IL

(b)

N.m 50.165.0

019.1943.0α =⎟⎠⎞

⎜⎝⎛ −

== IM

5.

s 42.115504.02

ω0θ

θθω

5504.026.16/95.8/ω

==−

=

−=

===

pt

t

IL

f

if

7. 2kg.m 95.1025.3/6.35ω/ === LI

21WORK-ENERGY THEOREM (p. 162)

22 ω½½energy

work

ImvmgyE

EEEW if

++==

−=∆==

1.

kJ 5.83joules 583050407.789

00.12)70(½15070(9.81)1.½ 22

==+=+=+= mvmgyE

3. J 120.0 00.2)0.60(½½ 22 === mvE 5.

J 987416.31)00.2(½60min1

12

min300)00.2(½½Iw

2

22

==

⎟⎠⎞

⎜⎝⎛ ××==

srrE π

7.

kJ 4.60)1.123(5.49001.123)81.9(50

m 1.123)s 0.60(20sin00.620sin

==−=

−=−====

if mgymgyEiEfWtvy oo

9. J 313½(25.0)5½00 22 −=−=−=−=−= mvEEEW iif 11. (a)

m 00.6

50.10.9

)75.0(230

2

750.04

30

222

=−−

=−−

=−

+=

−=−

=−

=

avv

ss

tvivfa

ifif

(b) N 00.15)75.0(0.20 −=−== amF (c) J 0.90)00.6(00.15 −=−== FsW

22WORK of a FORCE or MOMENT of FORCE (p. 168-9)

θφ==

+=⋅=

φ=

)(θ sinFrMW

sFsFsFWcosFsW

moment

yyxxforce

force

1. (a)

J 1.185517.0358 =×== FsW (b)

J 218

92.22.63325.13.25=

×+×=

+= yyxx sFsFW

3.

J 69730 cos (23.0) 35.0

cos

==

φ=o

FsWforce

5.

J 491

00.102.025.245J 1962

00.102.196N 2.196800.025.245

µN 25.245

)81.9(0.25

2.0

8.0

=

××==

×===×=

==

==

=

=

µ

µ

W

sFW

FF

mgF

kinetic

kineticnormalkinetic

normal

7.

kJ 77.11J 11772)0.60(2.196

)00.6(10)81.9(20)(

===

=== sLgFsW

9.

kJ 3.137J 340 137

4000)81.9(50.3)(

===

== sLgFsW

11.

J 206

350.0)81.9(60=

== mgyW

13. (a)

J441

1500.0)81.9(300=

== mgyWtotal

(b)

J 0.54)rad 1)(600.00.90(

θ)(θ=×=

== FdMWmoment

(c)

cycles 8.17 1/6 and 8

54441

/

===

= momenttotal WWn

15. (a)

J 09020.0 (0.450) ½

½02

2

.IEEW if

−=−=

ω−=−=

(b)

s 00.3667.6

0.200αωωrad/s 667.6

450.0300.000.10α

N 00.10)800.0(50.12

µ

2

=−

−=

−=

−=

×===

−=−=

=

if

kineticnormalfriction

t

IFd

IM

FF

23POWER (p. 173)

MωP

vFvFvFP

FvPtEtWP

moment

yyxxforce

force

=

+=⋅=

φ===cos

/∆/

1.

W687)602/()840(1.98/

joules 1.9881.900.10metres 84000.60.702

=×==

=×==××=

tsWPWs

L

L

3.

W382)606/(4000)81.95.3(/)( =×××== tsgWP L 5. 0=P An isometric contraction does no mechanical work. 7.

W5.17439.30.50ω

49.3deg360rad π2deg200ω

=×==

=×=

MPs

9.

W103785.180.55ω

rad/s 85.181rad π23ω

=×==

=×=

MPrs

r

11. W9.124555.0225 =×== FvP J 43750.39.124 =×== PtW

24CONSERVATION of MECHANICAL ENERGY (p. 178)

constant== if EE

1.

m/s 67.73)81.9(22

½

J 221000.381.90.752

===

=

=××==

gyv

mgymv

mgyWG

3.

m/s 61.144.2130.65

69362

J 69365.63767.559

00.10)81.9(0.655½(65.0)4.10½

½½22

22

==×

=

=+=

+=+

+=+

=

f

f

iiff

if

v

mv

mgymvmgymv

EE

5.

m 542.1

62.1925.30

)81.9(250.5

2

½22

2

====

=

gvy

mvmgy

7.

m/s 98.2868.80.60

2662

J 266 ½

J 266452.0)81.9(0.60

½

2

2

==×

=

=

==

==

takeoff

takeoff

takeofftop

v

mv

W

mvmgyW

9.

m/s 2.329.1036100.0

85.512

J 85.51 ½

J 85.51)35.0)(81.9)(100.000.15( ½

2

2

==×

=

=

=+==

v

mv

mgymvmgy