Notes on

Engineering Mechanics "Dynamics"

Chapter

(18)

By

Ahmed Elsamman

Eng.ahmedelsamman@live.com

0111 011 93 97002

Work and Energy

Problem 5.1

(1) The double pulley consists of two parts that are

attached to one another.It has a weight of 50 lb and a radiusof gyration about its center of h:0.6 ft. If it rotates withan angular velocity of 20 radfs clockwise, determine thekinetic energy of the system. Assume that neither cable slips on the pulley.

r .l-6lrlo^rr doJ ct s -

:. tqpr^ilr4 )

k. = o'6

es = i{l r;itlfi

*e7

fl^t , 2-a r\5

Yy\xP.,tlt4

=$o-\L(f,

?o

\l..) = 2-b

o

\,<. rn*.{ t q -e,'.'f'61 CT) --t

rryt h:

*!^do' * | vilR Lt^, . "^f

*

*nni esVt

t\=_z

\-Z

* **C**tftul}-

p,('. ()' c*)'* + +h Qo.'f- t+ht'"* o'$l

3;- L

= L83' \r' It

St*b3?.L

\tf-.I14

(2) Ttre handthe workgear at I

G,trl"-",

rA

oA

Yh=

{B r

fz

Wovt( (tt) = ,ryh

seB? f,

F

fo $tr

ht Y'

fo "0-t

t'A @n --

'u CPt)

.', @g,

winch is used to lift the 50-kg load. Determinerequired to rotate the handle five revolutions. Thehas a radius of 20 mm.

J^f".:, -

--5t ry= S (-$ = 2'T L6)

= toT rJ

! (3 v*v'21

l3s s^v.l

lclct i/Aw\

(r.)

@ts

(elu)

Ye os

? ljo c) g

+'833 e r*al

L..0 (+'g 3ra)

o . +t332- r/!A

v

(3) The pendulum of the Charpy impact machine has a

mass of 50 kg and a radius of gyration of kd:I.75 m . If itis released from rest when 0:0o , determine its angularvelocity just before it strikes the specimen S,g:901

t.7 s

-rt

"ioe

=O

=- 1o

-Lz

l, +

*Jta

2u,\--AL

+ ,.3hTA

€ sz- =2-'83

r+-6l rue,,... d,gl a o- -

YVI-: $o ryt-A

W, io

\lJ ='L/l,

,4.

o .t 5o*"t'l\

\ T w_z .tr i^t -"^ kl= _z _A z

- \, ( 7'ofurt)t *fx\.2-5 j n \-

".J k

(a) The spool has a mass of 60 kg and a radius ofgyration kgc:0.3 m . If it is released from rest, determinehow far its center descends down the smooth plane before itattains an angular velocity of a:6raqs . Neglect frictionand the mass of the cord which is wound around the central core.

G l*+-

rA=6"&f"ry

3

It

W, -:Q

!"6Vat*)*t')!r d,'

o * (6. rol.arx$ 6in3o) ?

$ F o'6(\

3r**{E PLe*4

dL = 6 f'Al^$=2.?

T+ Lv\l-+ L

.{- ,"r g h 'c y:.,L

rv\ ha a 4aa! ( qG.rfq6t)"Z \ ,/

T=o

^K;wfz 6lfc

=T-

C^*

(5) The 30-lb ladder is placed against the wall atangle'of. 0:45o as shown. If it is releaseddetermine its angular velocity at the instant0:0? Neglect friction and assume the ladder is

anfrom rest,just beforea uniform slender rod.

G;u*"" &tt cr B -r4^fl -, 3. t bt/, --o ^t' O= 45'

t*)o=,?t J O=-

Wt^' l,r^ {1"^,')-/ n' J

ZJ"\\4L

mg h =!z

3" (+ srn 4:)

g+,85 =

\?-

. z-\ -z+y.^d ) w

=T.

I t^)ttc

= + ( r-

(+Gl;t'f*Gtd\-)'F*, i z-'\zrJh 7

(5) The tub of the mixer has a weight of 70 lb and a radius ofgyration t<c:I.3 ft about its center of gravity. If a constant

torque M-60lb.ft is applied to the dumping wheel,

determine the angular velocity of the tub when it has rotatedg:90". Orieinally the tub is at rest when g:0o.

J^rt

€-^,t14z

G:**n "t'f a L-

N--

wt t o

l"r,:. M

+'I b

t-B frG tb.f t

C --ao --\"

s*-

\."

KG

M

=T.

Me-^3h ?\I'.^,-?-o

6o C*) - 7"G's) =+ (**r'f *t.,( 6'*)*'

itr- v :i{l

.Eng.ehmed ebamman

0't11 011 93 97

(Report D-10) The uniform door has a mass of.20 kg and can betreated as a thin plate having the dimensions shown. If it isconnected to a torsional spring at { which has a stiffness ofk:80 N.my'ad, determine the required initial twist of thespring in radians so that the door has an angular velocity ofI?radls when it closes at 0:0" after being opened at0:90o and released from rest. Hinl:For a torsional springM:k0 whenk is the stiffness and 0 is the ansle of twist.

G;t'^ A^f'n t -tA -- 2p v1

a go fv.nn lnJ

='U)*+

.:Q

=L,A3so

1T"oo

!I .^lzzFz

+ (+ s%"' lo"*B

l*"o'

'z-

e3'oe\'zJ

A

e"L^J r

Lrr"'l

t l'Z ""dlS

M ' KA = 8oe

Tn 2q =T.

MdO =

oAo

$o

Q.o Cg" "€\o

!o 7'z

\l.vrn

c'{

tz\nA t-".* I

.: e\-- \-/- -3 -a-\.66

= 3o 7'z

(1) The 30 kg pdirdulum has its mass center at G and aradius of gyration about pointG of lq;:300 mm . If it is released from rest when g:0o ,

determine its angular velocity at the instant 6):90" . Spring 4fi has a stiffness of

k:300 N/m and is unstretched when g:0o

{}.{r n"r

,(* X'T.la 5o

o =(+ T 4r")*

o = .+[ a"^o'f) * (so'"'th]*f o'.1

++ (j"l (L

r o'\S")

I

,-^-l =

- T ' 'Z -t L='*4:-| = J-+t^^d o'J' %- -'q

'c, Ct

.n$ = L- Lo

,Ull --.,

V,--o I

*- Va

I.1f,aa4 tl + a kst

- (3o &9'b\ x

3nA, t-

k

#l

.{n'u

{}.}5t-

k = -i1)t) \,/ln

o.6L+o.4Sa

h, --o

(2) A spring having a stiffness of k:300 N7tn isattached to the end of the 15-kg rod, and it is unstretchedwhen 0:0o . If the rod is released from rest when 0:0o ,

determine its angular velocity at the instant 6:30o . Themotion is in the vertical plane.

6],'*+'. J"f" : ^

r^ jLS F:K= \o" */^

L,,= ^f @=o

wt--a lf €)=o

UJ-s?? ^,t O --\o

va,,/ti cA Pts'^t

S =".6 Stnjo

r c'\.8\ G 3 af' El

, 3.. ot ,41x

* = i{X} N$n

@ =+ I".-' + lu*t - '$hi (+ (,*)k'.)'+Qr) G.)') '{':* ! 3os Q r ai,rg")'

T".= T * *tt J"L

-r rn (err.

Vg

fBtX. :.''3

_tlz-

r9

Aavry\ L

N

(e li.'r'.J =e"3

d1 d,',t2\--a'- ulli-rt

') o'd si''njo

= o"3

K* y -- T- *V.Jo 5/o

o?

e t5

t^).

."3.0r-"iF*..011't 011 93 97

Ic

(3) me motion of the uniform 80-lb garage door isguided at its ends by the track. Determine the requiredinitial stretch in the spring when the door is open, b:0n " sothat wheh it falls freely it comes to rest when it just reachesthe fully closed position, g:90" . Assume the do,or can betreated as a thin plate, and there is a spring and pulleysystem on each of the two sides of the door.

,\ fl

9

@

-(* Vr ?

5oXo{'€o

F d t*'rx&/VF $)r'^^tl'5o

1 f," =

T S =2't+ 4 [t #1(+*$f- s" L4)

L'$n+ZS"o :ASA a e-NS,

or B +

AS" '- -! 9t

ZAS5

G.r[r,^ d.af,u :

^fl :- $o lb€ --'za

l,/\/Z S O

O tolrrJ, ='

* $ trhli'r

$+As3-+z+

(4) When the slender 10-kg bar AB is horizontal it is atrest and the spring is unstretched. Determine the stiffness kof the spring so that the motion of the bar is momentarilystopped when it has rotated clockwise 90'.

-rlGl',1*^,n dolJcl 9 . orA*-1" t-Jl"t), io

Sl sc

W-s o

K ='u)

.XX=y+v.jo Jo tso

O = -r"^th t-s'r f

--

a * F to x\.8\ G."s) * + \^ ({ 3-- *r.\r

ITz

-\,1

(5) The arm and seat of the amusement-park ride havea mass of 1.5 Mg, with the center of mass located at point e .

The passenger seated at Ahas a mass of 125 kg, with thecenter of mass located at q.If the arm is raised-io a positionwhere g:150o and released from rest, determine the speedof the passenger at'the instant 0:0o .The arm has a radius ofgyration of kot:72 m about its center of mass e. Neglect

i + t^J'6

1Z,^t /R: + rn c)

9r

.X5o

+ V, = T-+V. 6 cal-- ^ K;' ;q Ke, = Zo

(ryr t h) r(t'^e3 h)u"4r

(lsoo *ot'E\r 4rtrn6o) * ( ?,3 xot'b\ x zrsr-6o)| (

zaa \so" L+\a)'0"= i \tsoorlr

I e8 r q ,Bl (?t)- ta 2 _ \Soorr.a.b\r4 o\e-5 Utl) *

=e r**1*:Q

-r l.

the size of the passenger.

u\) , \'oj6l f"r"Ib

(Report D-11)The uniform garage door has a mass of 150 kg ancl isguided along smooth tracks a't its ends. Lifting is done usingthe two springs, each of which is attached to the anchorbracketatAand to the counterbalance shaft at B andCAs the door is raised, the springsbegin to unwind from the shaft, thereby assisting the lift. If each springprovidesatorsionalmoment of M:e.7il)Nm , where 0 is in radians,determine the angle ft at which both the left-wound andright-wound spring should be attached so that the cloor iscompletely balanced by the springs, i.e., when the door is inthe vertical position and is given a slight force upwards, thesprings will lift the door along the side tracks to the horizontalplane with no final angular velocity. l''/ote: The elastic potentialenergy of a torsional sprin gis yu:l|,- r1-' L

l= too'. where and

in this case /d:0.7 N.m,tadr'}.

GtUu 6fuiAu ! '

rt^ -- l5o k4J

\I = e|+ €

> 7/7 rnr

INr'-o

4{xd)* =z( r ^th

T €)o =5(.\srJ

.X" v\Jo

L(

o,7 €

lt/' ,^^

@ --o

brt* ic

= F*r[-9o

; l5or(q.\\ Ct.sl

(Report E-l1) T'he 30-kg rod is released from rest when g:0o.Determine the angular velocity of the rocl when0:90o.The spring is unstretched when g:0o.

lnr

oz +

x- {. r.,.vz_

z!d'

o

-L x3ota

rB o i(

ks'hJ.] - Bo rot 'b\ * "' tS

- ".r)<)a !rz\

.Lz

< \r..J uJ2

rl

z.jLt t* /U

-

* = S{i irl'rn