PLANER KINETIC F AND PLANER KINETIC FPLANER KINETIC F AND ACC.pdf AND ACC.pdfACC.pdf.pdf

PLANER KINETIC F AND ACC.pdf

1. 17-53. The 80-kg disk is supported by a pin at A. If it is
released from rest from the position shown, determine
the initial horizontal and vertical components of reaction

at the pin.

EEE = m(@),: A, =0 Ans
+TER = mi@),; A, ~ 80(9.81) = —80(1.5)(a)
T+EM, = ha; 80(9.81)(1.5) = [osa

a = 4.36 rad/s*

A, =262N Ans

fe Ne

sodann

2. 17-55. ‘The fan blade has a mass of 2 kg and a moment
of inertia L = 0.18 kg-m? about an axis passing through
its center O. If it is subjected to a moment of M =
3(1 = e”02') N-m, where 1 is in seconds, determine its
angular velocity when 1 = 4 s starting from rest.

(GEM, =ba 3(1-e°") =

180
a=16.67(1-e%*)

dw = ade

Fu fs 6711-22) de

w= 16.67[ t+

&= 20.8 rad/s Ans

3. 17-62, Cable is unwound from a spool supported on
small rollers at A and B by exerting a force of T = 300 N
on the cable in the direction shown. Compute the time
needed to unravel 5 m of cable from the spool if the spool
and cable have a total mass of 600 kg and a centroidal
radius of gyration of ko = 12 m. For the calculation,
neglect the mass of the cable being unwound and the mass
of rollers at A and B. The rollers turn with no friction.

Equation of Motion : The mass moment of inertia of the spool about point O is
given by /, = mk}, = 600( 1.27) = 864 kg-m°. Applying Eq. 17- 16. we have

= 300(0.8) =-864a 0=0.2778rad/s*

5

__ Kinematic : Here, the angular displacement 6 = à

r 08

6.25 rad. Applying
1
tquation $= 8) + @y r+ jar, we have

1 a
(+ 625=0+0+ (0.2778)

1=671s Ans

4. *17-91. Two men exert constant vertical forces of 40 Ib
and 30 Ib at the ends A and B of a uniform plank which
has a weight of 50 Ib. If the plank is originally at rest,
determine the acceleration of its center and its angular
acceleration. Assume the plank to be a slender rod.

Equation of Motion : The mass moment of inertia of the plank about xs mass
ear 1750 2

ener is given by la = ami? = 2 (2015
se pe ello le

29.115 slug- f°, Applying
‘Eq. 17-14, we have

+ TER = mag), : s0+30-50= (2)
&), + E ac

2.9 fus? Ans

a

(do = a: 30(7.5) -40(7.5) =-29.115a
= 2.58 rad/s

30 Ib

5. 17-94. The spool has a mass of 500 kg and a radius
of gyration kg = 1.30 m. It rests on the surface of a
conveyor belt for which the coefficient of static friction is
4s = 0.5. Determine the greatest acceleration ac of the
conveyor so that the spool will not slip. Also, what are
the initial tension in the wire and the angular acceleration
of the spool? The spool is originally at rest.

ae DA = mag): T —0.5N, = 50046
+t OF = mac): N, = 5000.81 = 0

A+ Y] Mo = lou. 0.5, (1.6 — T(0.8) = 50001.30)°a

An = ac + UG

(ap)yi = agi — 0.8ei
ac = 0.8a

Solving;

N, = 4905 N
T=313KN Ans
a = 1.684 rad/s Ans
ac = 1.347 m/s?

Since no slipping

ac = ag + acc

ac = 1.34

(1.684)(1.6)i

500(9.81)

16m

(Gp)

ac = 1.35 m/s? Ans

Also,

A+ Mic = ticas

= 4905 N

0.5N, (0.8) = [500(1.30)? + 500(0.8) le

Since N,

a = 1.684 rad/s

6. 17400. À uniform rod having a weight of 10 1b is pin-
supported at A from a roller which rides on a horizontal
track. Ifthe rod is originally at rest, and a horizontal force
of F = 15 Ib is applied to the roller, determine the
acceleration of the roller. Neglect the mass of the roller
and its size d in the computations.

Equation of Motion: Te mass moment lnea rod bouts mass
10
centers given by fg = Im = 2 (20,923) 20.1038 sf. Atte instant
sense ml = Ie) 0.1035 slug =f, Au
force apd, the angular velocity of be od à 0. Thus, he nomal
component of aceras of he mas ene for he od (a), = 0. APP
Eg, 17-16, webave

LE = ma),

Bie erase
10
Gem zz: 0= (Jen 010350
ar 1649 nats?

Kinematic: Since © = 0, (Aa), =. The aceleran of rer can be
obtain by analyzing he motion of pois A and G. Applying Eq 16= 17. we have

ia

(5) 483=0,- 1449
PEUT Ans

13 tb A

10% (= 19

fe (an),z0
9 144:9 dls
Gum) TT 20.0

7. 17-102. The lawn roller has a mass of 80 kg and a ris
Sr gyration kg = 0.175 m. If it is pushed forward with a
foes 200 N when the handle is at 45°, determine its
angular acceleration. The coefficients of static and kinetic
ne ion between the ground and the roller are Hs = 0.12
‘and ys = 0.1, respectively.

DN

‘Assume no slipping.
200N
la + md? = 80075)" + 8000200) = 5.65 ke a

Ma =200c0545*(0200) = 28284 N-m $00081)N

My = Ina

= 274 _ 5.01 rads? Ans

3.65
Check no slippage assumption. a

AHER=0 —200sinas’ 8008) +N=0

N=9622N
(926.22) = 111.15 N

Ems
lo = 800.175) = 245 kg. m’

45(5.01) = 1227 N-m

Mo = loa

12.27 gan <iisN OK.

8. 17-107. ‘The 16-1b bowling ball is cast horizontally onto
“lane such that initially «o = 0 and its mass center has a
velocity v= 8 ft/s. If the coefficient of kinetic friction
between the lane and the ball is x = 0.12, determine the
distance the ball travels before it rolls without slipping.
For the calculation, neglect the finger holes in the ball
and assume the ball has a uniform density.

à 16
SF, machi OM = 57586

¿TER = much: M716

questa oro a6 Jon le

sonne
D216: ao = 3.864 fur‘; = 25.76 rad/s
en al os without siping ¥= 00375).
(Go ars

042560
(x: ES

v=9.6606

164
= (cot
á 30
e Sore
um
TH

¿=0+8(0.592)-0.860/05927

s=8068 Ans

9. around each of the two
from rest, determine the
the mass of the cord.

cord € is wrapped
If they are released
fixed cord D. Neglect

70m

x p
SP

num [moon] ©

100.81) - 7 = 1000 o

®

Ans

Cm
q

Da

"

won

ono

(A

10. 182. The wheel is made from a 5-kg thin ring and two
kg slender rods. If the torsional spring attached to the
wheel's center has a stiffness k = 2 N-m/rad, so that the
torque on the center of the wheel is M = (20) N-m,
where @ is in radians, determine the maximum angular
velocity of the wheel if it is rotated two revolutions and

‘then released from rest.

alo = 1588

He. =% CE

+ [2040 = 3058)

(49? = 079170"

= 14) rd/s Ans

11. 18-4, The double pulley consists of two parts that are
attached to one another. It has a weight of 50 Ib and a
centroidal radius of gyration of ko = 0.6 ft and is turning
with an angular velocity of 20 rad/s clockwise. Determine
the kinetic energy of the system, Assume that neither
cable slips on the pulley.

1 1
T= Hood + ¿mv +¿m%

e] + (e syn Ans

12. 18-9. A force of P = 20N is applied to the cable, which
causes the 175-kg reel to turn since it is resting on the
two rollers A and B of the dispenser. Determine the
angular velocity of the reel after it has made two
tevolutions starting from rest. Neglect the mass of the
rollers and the mass of the cable. The radius of gyration
of the reel about its center axis is ke = 0.42 m.

T +EU 2 =%
0+ 20(2)(2)(0.250) = [1751 (0.42) Ja?
@= 2.02 rad/s Ans
? 175 (9.30N
9.250m

Na N