38
Jump to first page Chapter 6 Plane Kinetics of Rigid Body
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Chapter 6
Plane Kinetics of Rigid Body
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Plane Kinetics of Rigid Body
Rotation about a fixed axis :
GamF
QQ MH and
if Q :(1) has zero acceleration, (2) is the centre of mass G or (3)has acceleration parallel to
iiiiQ
vrmH
MH
Q is an arbitrary point on z-axis (zero acceleration).
For a system of particles :
GQr /
z
iQH
iv
Q
i
i
ir im
iR
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For the angular momentum of the whole rigid body about a fixed axis z :
)( iiiiiiiQ RrmvrmH
2
)sin(
sin
sin
ii
iiii
iiii
iiQiQ
Rm
Rrm
Rrm
HHz
zz
iiizQz
IH
RmHH
)( 2
QzQz MH The z-component of : QH
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Where i
iiz RmI 2
zz MH
(moment of inertia of a rigid body about z axis)
zz
zz
MI
MI
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If Mz is zero or the time for integration is small enough, we have
Kinetic energy :
2
112
t
t zzzz dtMHHH
222
222
2
1
2
1
2
1
2
1
zi
ii
iii
iii
IRm
RmvmT
zzz HHorH 120 (conservation of angular momentum)
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Rectilinear motion Rotation about a fixed axis
ntdisplacemeangularsntDisplaceme
velocityangularvvelocity
onacceleratiangularaonaccelerati
zIinertiaofmomentmmass
zMtorqueFforce
zHmomentumangularGmomentum
zz IMmaF
sva
sv
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dMVVFdsVV
ITmvT
IHmvG
MPFvP
dMUFdsU
OO
zO
S
S
O
z
zz
z
z
22
2
1
2
1
Rectilinear motion Rotation about a fixed axis
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e.g. cylinder : -density of rigid body
dr
r
L
a
adrrL
rdrLrdmrI
0
3
22
2
2
2
4
0
4
2
1
2
1
4
12
maI
LarLa
Lam 2
Calculation of moment of inertia :dmRRmI i
ii 22
)2( rdrLdm
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Moment of Inertia :
44
2
1
22
2
aam
maI
43
22 bamI
dmrI 2
a2
axis
axis
b2
aa
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b2
a
aaxis
33
22 bamI
sphere
axis
a 222
5
2
55ma
aamI
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k
b
n
amIG
22
spherical
circular
straight
kn
5
4
3
,
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Parallel axis theorem :
2mhII G
x
G
axis
h
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'xx
'yy
im
G
iy'
ix'
iR'iR
hO
Gx
Gy
ix
iy
)( 222ii
ii
iiiO yxmRmI
iGi
iGi
yyy
xxx
'
'
22
2222
22
00'
'2'2''
''
mhRm
yxmymyxmxyxm
yyxxmI
ii
i
GGi
ii
iiGii
iGiii
i
iGiGi
iO
2mhII GO
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2
2
1
2
2
rR
mR
mrI
g
g
G
e.g. Circular disc
2gG mRI
Rg - radius of gyration about G G
shapeAny
Gr
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General Plane motion(rigid body)
y
x
G
G
System of particlesGG
G
HM
amF
For general plane motion :
GGG
G
IHM
amF
} 3 different equations
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Kinetic energy :
22
222
22
22
2
1
2
1
)(2
1
2
1
)(2
1
2
1
2
1
2
1
GG
iii
G
iii
G
iGii
G
Imv
Rmmv
Rmmv
vmmvT
y
x
im
ir
Gr G
Gir /
iGi Rr /
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Find : T1 , T2 and aG .
The motion of ABC is purely translational. AG aa
0.3m 0.3m
0.3m
n̂Cut
Av
o700.1m
o70
t̂
1T 2T
A B
C kg10
DE
G F
Example
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free body diagram
0GG
G
IM
amF
070sin3.070cos1.0
70sin3.070cos1.0
108.91070sin)(
1070cos
22
11
21
21
oo
ooG
Gyo
y
Gxo
x
TT
TTM
aTTF
aTTF
kg10
o70 0.1mo70
1T 2T
A B
C
Gx
y
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G
oA
oA
AA
A
AAA
a
jviv
tva
v
nAE
vtva
70cos70sin
ˆ
0
ˆˆ2
oAGy
oAGx
va
va
70cos
70sin
(while BF is cut)
oA
o
oA
o
vTT
vTT
70cos108.91070sin
70sin1070cos
21
21
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Hence
070sin3.070cos1.070sin3.070cos1.0 2211 oooo TTTT
#2#1 5.407.51 NTNT
#2ˆ36.3 smtaG
#236.3 msvA
Problems on Plane Kinetics of Rigid Bodies :
(3), (7), (8), (11), (12), (15), (17), (18), (22).
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222
22
2
2
1)5.0(3)5.03(3
1
)(3
1
5.060cos/8.93
5.060cos
mkg
mrmr
mhII
msmkg
mmgM
GA
o
oA
Example
A
B
f
NR
G
o60mg
omg 60cos hr
x
yFind and S .
Take moment about A : MA=IA
mrkgmsrad 5.032
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Ns
A
A
R
f
sradI
M
#2/358.7
ooAG jirkk 60sin60cos5.0;2;358.7 /
ji
jikk
jik
rraa
oo
oo
AGAGAG
572.3186.2
60sin60cos5.022
60sin60cos5.0358.70
//
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Nf
maF Gxx
558.6)186.2(3
#35.0714.18
558.6
714.18
572.338.93
Ns
N
N
Gyy
R
f
NR
R
maF
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Example
A B
CDP Q
h
FG
m3.1
m2.0Em4.2
m4.2
m6.0
m65.0
x
yFind aEx and aEy.
E-accelerometer
0
PP IM
pEpEpE rraa //
)0( Pa
65.01065.0
8.9/10106
66
mgM
mNmg
P
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26
22222
1047.0
65.08.12.12.13
1
mkg
mmmhII Gp
#
//
2
55.210.1
8.085.138.1
8.085.1;
38.1
/38.1
ji
jik
jirra
k
srad
PEPEE
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Example
GT
Ga
f NR
mm450
mm150
x
y
Find aG.
Radius of gyration about G = Rg= 0.25m
NT
kgm
k
S
15520.0
25.070
2gG mRI
08.970;0
70155;
Ny
GGxx
RF
afmaF
2)25.0(7015515.045.0 fIM GG
,,,: NG RfaunknownsFour
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45.0.: GaslippingwithoutrollingisItAssume
We can then solve :
NRNfsmasrad NG 68776/13.1/51.2 22
For static friction :
NNNRNS 7617268725.0
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Example
NS
NG
RNf
NRsmasrad
NTIf
186
687/77.2/15.6
38022
So there must be slipping in the motion, then
NkRf (k is the kinetic frictional coefficient)
4 equations and 4 unknowns
NfNR
sradsma
N
G
4.137687
/10.1/47.3 22
45.0 Ga
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ExampleT
TGaA B
o30
Rolling without slipping
mB=4.5kg
Find aG and T.
Ba B
T
gmB
B 4.5 9.8 x 4.5 - T a
raG
GGB araa 2
)()2(5.48.95.4 AaT G
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x
Ga
gmA
fNR
G
T
rmrkgmA 76.023
)(232
1CaTf G
r
aG 276.0232
1
)(23
30sin8.923
Ba
fTF
G
ox
GG ITrfrM
Solve the above three equations
##2 3.48/467.0 NTsmaG
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Example
o10
N200
m5)(
15.040
gyrationofradius
mRkgm g
Find vG.
N200
mgf NR
Gm1.0
m25.0Gv
5
0200 dsf
dsdFW G
2GG mRI
The workdone on the spool :
5
0 25.0)2001.025.0(ds
f
)25.0/( dsd
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)(1400
5805200
forcesveconservatinonfromJ
fdsfds
#
222
22
/24.6
140010sin58.940
25.015.040
2
140
2
1
10sin581.9402
140
2
1
smv
J
vv
Iv
VTW
G
o
GG
oGG
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Examplem22.1 m92.0
Gv
NRCG
omrkgm 603.08.6
Find vG, RN when = 0o.
0
60cos192.08.98.6
2
1
2
1 22
VT
V
ImvT
o
GG
#2 /54.230.08.6
5
2smvI GG
Rolling without slipping 0Cv
rvrvv GCGCG /
By conservation of energy :
#2 11492.0/8.68.68.9 NRvR NGN
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Example
Unstrained length = 95mm
Find : of the block,
K.E. of the block
after impact.
Let v be the velocity of the block before impact.
smv
v
/378.0
1.08.3108010951042
198.0
2
1 23332
Regard to point O, there is no torque acting to the block.Therefore the angular momentum about O will be conserved.
mm mm
A B
80 50 100mm
mm25vFO
m
Nf 8.3h
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Before the impact, the angular momentum about 0 = mv x r = 0.98 x 0.378 x 25 x 10-3
#
23
232323
2
/68.5
1063.1
2102598.01025102598.03
1
srad
mkg
mhII Go
After the impact, the angular momentum = Io
K.E. after impact :
#232 026.068.51063.1
2
1
2
1JIo
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Example
A
B
yR
xR
h
m1
m5.0G
AF
srada /3
x
y
C
kgm 20
Find : (i) h with min reaction at B. (ii) FA.
Let b be the angular velocity of the rod just before impact :
5.08.9202
1 2 bBI
srad
I
b
B
/425.5
)5.0(205.0203
1 22
(Lost of P.E.)GI
2mh
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Refer to the centre of mass G :
4213
425.55.035.02002.0
]425.55.035.0[02.0
0
Ax
Ax
GG
Ax
FR
FR
mGdtFRba
hFM
I
IHdtM
AB
B
BBB
425.53
02.0
0
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42132808
2808)425.8(02.0
hR
hFIhF
X
ABA
If RX = 0, h = 0.67 m #
(So the point at h is called the centre of percussion.)
FA = 4213 N #
