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RIGID BODY DYNAMICS - 1
TOPIC : RIGID BODY DYNAMICS
PART - I
1. Let m1= mass of the square plate of side 'a'
and m2= mass of the square of side 'a/2'
Then m1=
2
2
a!"
#$%
&' ; m2= ( )
2a' ; ('being the areal density)
and m2m1= M.
* + =-.
-/0
-1
-23
!"
#$%
&45
2
1
21
22
4
am
6
)2/a(m
6
am=
-.
-/0
-1
-23
!"
#$%
&!
"
#$%
&'4
'5
'2244
4
a.
2
a
6
)2/a(
6
a
=./0
123
65
65'
164
1
616
1
6
1a4 =
./0
123
6556
'1216
32)162(a4
+ =./0
123
6'
1612
27a4
Also ; M =2a
4
11 !
"
#$%
&5' * '= 2a
M
3
4* + =
./0
123
6!
"
#$%
&1612
27a.
a
M
3
4 42 * + =
16
aM3 2
2. M.I. about Ois2
MR2
By parallel-axis theorem2
MR2= +cm+
2
2.3
R4M !
"
#$%
&7
*+cm=22
3
R4.2M
2
MR!"
#$%
&7
5
3. From FBD
Equation in horizontal direction
T = Nx ...............(1)For Rotational equation about P
T. 2 = 1.5 300T = 225 N
Nx= 225 N
Ny= 300 NAnd Ng = mg = 300 N
PHYSICS SOLUTIONSADVANCE LEVEL PROBLEMS
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RIGID BODY DYNAMICS - 2
4.
There is no slipping between pulley and thread.
So, (a = 8r) ..........(i)For point mass :
mg T = ma ...........(ii)Equation of torque for disc
Tr = +.8
Tr = 8.2
mr2
T =2
mr8= !
"
#$%
&2
mg...........(iii)
mg 2
mg
= ma
mg =2
mg3
a =3
g2.
5.
R
R
V
2V2V
2R
x
x =g
R4v2
g
2R2v2 9
:=
g
Rv16 2
6.F sin 8
F
)8 Fcos 8
mg
f
Rr
(F sin 8+ N = mg)F cos 8 increased linear speed for pure rolling friction force acting leftward direction and thread winds.
7. As ;
* 150 >0= 180 > * >= 5/6 >0 Ans.
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RIGID BODY DYNAMICS - 3
8. Impulse = change in momentum
= P.2
!=
12
m 2!.> (about centre of AB)
* >=!m
P6
For ?= 27
; 2
7= >t * t = >
72 = p62
m
67 !
* t = p12
m!7
9.
?
G
g sin?a
Circular pipe is rest
so g sin ?= a
10. Cons. of ang. momentum about P gives
MV2
L=
12
)L2()M2( 2w *
2
V=
3
wL2
w =L4
V3Ans. (C)
11. Rotation energy =2
2
1>+ += mK2
linear energy =2mv
2
1K = gyration radius
Total energy =2
2
1>+ + 2mv
2
1v = >R
Frictional of its total energy associated with rotation.
=22
2
mv2
1
mv2
12
1
4
>+=
2222
22
RmmK
mK
>4>
>= !
!"
#$$%
&
4 22
2
KR
K
12.* (A)
2= (12 >
2)
Man covers 27angle relative to disc
t
27= >
r= 12 >
2
(27= 12 >2t)
>2t = !"
#
$%
&7
6
Some time t angle taken by disc
Ans.60 East of South 60 East of South
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RIGID BODY DYNAMICS - 6
PART - II
1.CM
! x
>
m
12
m 2
CM
!9+ += +
CM+ mx2
x =m
CM+5+=
m12
m 2!5+
x = m34.060
79
2. (i) x
dx
A
B
+AB
= B 2dmx
+AB
= B!
0
3dxax = !!"
#$$%
&4
a 4!.
(ii)x
dm
dx
xcm
=
B
BQ
0
0
2
dxax
dxax
!
= !"
#$%
&!
3
2
2 /3!
+cm
dx
+ABA
B
+AB
= +cm
+ m
2
3
2!"
#$%
&!
+cm
= +AB
9
m4 2!
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RIGID BODY DYNAMICS - 7
m = B!
0
dxax =2
a 2!
+cm
=9
a2
4
a 44 !!5
CCD
E
FFG
H9+ 36
a
4
cm
!
Ans.
3. (a)
For 2kg mass,
T12g sin 45 = 2a ......(i)For 4kg mass
4g sin45T2= 4a ............(ii)
For pulley,
r(T2T
1) = +8= +(a/r) ............(iii) (+= !
!"
#$$%
&
2
mr2
)
From eq. (i),(ii) and (iii)
a =!
"
#$
%
& +44
?5
2r24
sing)24(
a =
!"
#$%
&44
665
01.0
5.024
52/110)24(.
a = 0.248 = (0.25 m/s2).
(b) m1= 4kg m
2= 2kg
I= 0.2 (between inclined plane and 2kg block)+= 0.5 kg-m2 r = 0.1 mm
1gsin?T
2= m
1a .........(i)
T1(m
2gsin?+ Im
2gcos?) = m
2a .........(ii)
r(T1T
2) = +.8= !
"#$
%&+
r
a .........(iii)
From eq. (i),(ii) and (iii)
m1g sin?(m
2g sin?+ Im
2gsin?) + 2r
a+= m
1a + m
2a
Put values :
4g sin45 (2g sin45 + 0.2 2g sin45) +01.0
5.0a = 6a
* 27.80(13.69 + 6.95) = 56a
= a =567 = (0.125 m/s2).
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RIGID BODY DYNAMICS - 8
4.
N2sin 8= f (i)
N1+ N
2cos 8= mg (ii)
Torque about point A
( N2cos 8) !
"
#$%
&8tan
b+ N
2sin 8b = mg 2
a cos 8 )
N2=
( )b2
sincosmga 88
N2cos 8= ( )b2sincosmga 2 88
From equn. ....(ii)
N1= mg N
2cos 8= mg
b2
sincosmga 2 88
N1= mg
b2
)sincosab2( 2 88
N2sin 8= IN
1
I=1
2
N
sinN 8
I =
b2
)sincosab2(mg
b2
sincosamg
2
2
88
88
* I=CCD
E
FFG
H
88
88
sincosab2
sincosa2
2
5.
Ny
mg
C
bb
oNx C
bb
oNx C
bb
oNx C
bb
oNx C
bb
oNx C
bb
oNx C
bb
oNx
mg 2/b = +8, + =6
mb2+ m
2
2
b!!"
#$$%
&
! I=6
mb2+
2
mb2=
2
mb2 !
"
#$%
&4
3
11
+= 3mb2
2
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RIGID BODY DYNAMICS - 9
Hence2
mgb=
3
mb2 28!8=
b22
g3
Accnof corner C = 22 bb 4 8= 2g3
Acceleration of O in horizontal direction is zero So Nx= 0
mg Ny= m 2
b
8
= m2
b !
!"
#$$%
&
b22
g3 =
4
mg3
= Ny=
4
mg
6. (a)
L
21kxsin30 kxsin30
30 30B
A
mg
(i)
Before cutting 2k xsin30 = mg
kx = mg (T = kx = mg)After cutting(ii) Torque about COM
(Tsin30) x 2
!
= +8
4
mg!= 8.
12
m 2!
8= !"
#$%
&!
g3(clockwise)
(b)acceleration of point Ama
x= T cos30
ax=
m2
3mg=
2
g3= a
AC@
mg T sin30 = may
mg 2
mg= ma
y
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RIGID BODY DYNAMICS - 10
aAy
= !"
#$%
&5
2
g+
2
!8=
2
g5 +
2
g3= (g) (J)
aA= gji2
3!!"
#$$%
&4
(c) aBx= @g23
aBy
=22
g !84!
"
#$%
&= 2g ( K )
aB= gj2i2
3!!"
#$$%
&5
(ii)
kxsin30
30
kxsin30
30
12
L/3 L/3 L/3
mg
Before cuttingmg = 2kx sin30 = kx = TT = mg.
After cutting
(a) Torque about COM
(T sin30) !"
#$%
&6
!= +.8
(mg) !"
#$%
&2
1!
"
#$%
&6
!= 8.
12
m 2!
8= !"
#$%
&!
g(cw).
(b) (T cos30) = max
mg 2
3= max
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RIGID BODY DYNAMICS - 11
ax= !
!"
#$$%
&
2
g3
aAx
= )i(2
g35
mg 2
ma
= may
ay=
2
g( j5 )
aAy
= (ay8
2
!) = !
"
#$%
&2
g !
"
#$%
&2
g !
!= 0
!!"
#$$%
&59 ig
2
3aA"
(c) ig2
3a
cx
59"
j22
gacy !
"
#$%
&8459
!"
= gj
ac= a
cx i + acyj = !!"
#$$%
&4 jgig
2
3.
7.
Angular momentum about point A
Li= m
1v
s! (u
s: Final velocity of ball after collision)
L= 3
m 22!
>+ m1us!
Li= L
(m1v
s!=
3
.m 22 >!+ m
1u
s!)
2 5 =3
2.18 >66b + (2 u
s)
10 =10
32>+ 2u
s............ (i)
Coefficient of restitution
e =s
s
v
u5>!
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RIGID BODY DYNAMICS - 12
0.8 =s
s
v
u5>!
5
4=
( )5
u2.1 s5>
4 =5
6>u
s
us= !
"
#$%
& >5
206............ (ii)
Put equation (ii) in equation (i)
10 =10
32>+ 2 !
"
#$%
& >5
206
10 =10
32>+
5
4012>
100 = 32>+ 24>80
>=14
45
Put >in equation (ii)
us= !
"
#$%
& 5>5
206
us=
5
2014
456 5!
"
#$%
&
us=
514
280270
6=
514
10
6= !
"#$
%&5
7
1
So direction is (L) us !
"
#$%
&7
1
8. (a)
Let coordinates of instantaneous axis of rotation be P(x,y).
then velocity of P w.r.t. C is zero.
* 0ivCP 946>"
* bt ( k ) 0iv]jyi)tvx[( 9445
* x = vt
and Myt = Vfrom these eliminating t
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RIGID BODY DYNAMICS - 13
1x
.v
y9
NM
or xy =M
2v
* locus of P is a Hyperbola.
(b) Here coordinate at point C = !"
#$%
&0,Nt
2
1 2
= 0ivCP 946>"
* 0itwjyi)tw2
1x(k 2 94C
D
EFG
H456> * x = 2tw
2
1
>y = wt
from these eliminating t,
* x =2
2
yw
w2
1!"
#$%
&>* x =
w2
2>y2
Eqn. of parabola.
9. a = 8Rmg sin 300T = ma .........(1)
or2
mgT = ma .........(2)
8=I
!
= 2
2
1
TR
MR
8 =MR
2T.........(3)
Solving Equations (1), (2) and (3) for T, we get
T =2
1
mM 2
mgM
4Substituting the value, we get
T = !"
#$%
&2
1
./0
123
4 (0.5)(2)2.8)(2)(0.5)(9
= 1.63 N
T = 1.63 N
(ii) From Eq. (3) , angular retardation of drum
8=
MR
2T=
)2.0)(2(
(2)(1.63)= 8.15 rad/s2
or linear retardation of blocka = R8= (0.2) (8.15) = 1.63 m/s2
At the moment when angular velocity of drum is
>0= 10 rad/s
The linear velocity of block will bev
0= >
0R = (10) (0.2) = 2 m/s
Now, the distance (s) travelled by the block until it comes to rest will be given by
s =2a
v 02
[ Using v2= v0
2 2as with v = 0 ]
= )63.1(2
(2)2
m or s = 1.22 m
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RIGID BODY DYNAMICS - 14
10.
Angular momentum conservation about A
xO
3R10
mvR + mv !"
#$%
&5
10
R3R = 2mR2>
v ' = 2
v7.1
*./0
123
!"
#$%
&9>
R2
v7.1'
By Energy conservation
2mv2
1+
2
1+>2= mg 0.3 R +
2
1mv'2 +
2
1+>'2
mv2= mg 0.3 R + mv'2
v2= g 0.3 R +2
2
v2
7.1!"
#$%
&
vmin
= gR3.07.1
2
11. Minimum velocity required by block mto complete the motion in gR5
conserving mech. energy
2
1
+>2= Mg . 2R
* >= +
MgR
Cons. angular momentum wrt P before & after collision.
+>= m.R gR5
++
MgR= mR gR5
MgR+= m2R25gR
putting +=3
ML2
m
M= 15
Ans. : 15m
M9
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RIGID BODY DYNAMICS - 15
12. (i) L = Natural length of spring
(ii)
(a) By energy conservation from (i) to (ii)
CD
EFG
H!"
#$%
&54>94> x
2mgI
2
1kx
2
1I
2
1 21
22 !............. (i)
I = Icm
+ ,
2
x2
!"
#$%
&5
!
I =12
m 2!+ m
2
x2
!"
#$%
&5
!............. (ii)
!"#$
%& 549 Lyxx 22 ............. (iii)
Put equation (ii) and (iii) in equation (i)
!!
"
#
$$
%
&!
"
#$%
&54
22
x2
m12
m
2
1 !!>2+
2/122 LyxK
2
1!"#$
%& 54
= !!
"
#
$$
%
&!
"
#$%
&54
22
x2
m12
m
2
1 !!>
12+ mg !
"
#$%
&5
8x
!
x = 150 mm, y = 20 mm, != 450 mm, K = 300 N/m
m = 3 kg, >= 4 rad/sec
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RIGID BODY DYNAMICS - 16
Put all the data
>1= 86
3
2rad/sec
(b) rotating to 180 condition is
This is like a initial condition so >2= >
>2= 4 rad / sec
13. Force moment relative to point O
N"
=dt
Md"
= tb2"
Let the angle between M" and N"
8= 45 at t = t0
Then2
1=
NM
N.M""
=0
40
22
20
bt2tba
)bta(
4
4"
=
0422
30
2
bt2.tba
tb2
4=
40
22
20
tba
bt
4
Solving, t0=
b
a(as t
0cannot be nagative)
Thereforeb
ab2tb2N 0"""
99
14. 8= angular acceleration
8= angular accelerationFor the plank
F = m1>
1....... (i)
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RIGID BODY DYNAMICS - 17
For sphere torque about point C
fr = Ic8=
5
2m
2r28 ....... (ii)
assuming >2is the acceleration of COM of sphere at point A
(>1= >
2+ 8r) ....... (iii)
From equation (i), (ii) and (iii)
>1=
!"
#$%
&4 21 m
7
2m
F and >2= !
"#$
%& >1
72
15.
For cylinder
Mg + T12T = Ma ......(i)
Torque about axis of rotation
2TR + T1(2R) = +O8= !
"
#$%
&+
R
a. ....(ii) (a = 8R)
For weight Amg T
1= ma
No slipping between pulleys and thread
a1= a + 8(2R) = (3a) ....(iii)
From equation (i), (ii) and (iii)
--.
--/
0
--1
--2
3
!"
#$%
& +44
49
2
1
Rm9M
g)m3M(3a
16. Velocity of end A at the moment it strikes ground = gh2
If velocity of COM of rod just after collision vPand angular velocityacquired by the rod is >clockwise as shown then using equationfor coefficient of restitutionvelocity of approach = velocity of sep. (applied at point A).
gh2 = vP+2
L>cos? .............(1)
Angular momentum can be conserved about A justbefore collision & after collision as only impulsiveforce will be acting at A only.
gh2 M2L cos?= +cm>MvP 2
L cos? .............(2)
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RIGID BODY DYNAMICS - 18
Putting value of >= ( gh2 vP)?cosL
2
from (1)
gh2 M.2
Lcos?=
12
ML2( gh2 vP)
?cosL2
MvP2
Lcos?
?cos6L
vP+ 2 vcosL P? = ?cos6 gh2L 2gh2 Lcos?
vPCCD
E
FFG
H
??4
cos6
cos31 2
=?
?5cos6
)cos31( 2
gh2
vP= !!"
#$$%
&
4?5
cos31
cos1 2
gh2
COM will of at maximum height when its velocity becomes zero during upward motion.O = vP22g H
H =g2
v
2
P =
2
2
2
cos31cos31 !!
"#$$
%&
?4?5 h.
[ Ans.: H = 1 3
1 3
2
2
2
5
4
&
%$
#
"!
cos
cos
?
?h; h =
49
144
7!]
17. NC+ N
B= 250
NBx = 250 3
NB=
x
750
f1=
x750 I
f2= IC
D
EFG
H4 25
x
750
workdone against friction
W = B Q4Q dx)( 21 = dx5.73.0x1500
5.4
3
B !"#
$%
&46 = 450 ! n
2
3+ 7.5 (4.5 3)
= 450 0.41 + 7.5 1.5
2
1mv2 = 400 1.5 195.75
v2= (600 195.75) 5.2
2
=161.7 2 = 323.4v = 18.52 m/sec.
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RIGID BODY DYNAMICS - 19
18. ? is very small? R 0
T
w/2
AN2
PN1
TIN1
A?IN2
b
a
Force balance in horizontal direction
N1= N
2
balancing torque about point P
For ? to be very small we can directly write
T.b + IN2b 2Wb
N2a = 0
Force in y direction if acceleration of windows is A
w IN1IN
2T =
g
wA... (ii)
For block
T 2
W=
g2
WA
T = !!"
#$$%
&4
g2
WA
2
W.... (iii)
Put equation (iii) in equation (i)
2
Wb +
g2
WAb + IN
1b = N
1a +
2
Wb
g2
WAb= N
1(aIb)
N1=
./0
123
I5 )ba(g2WAb
..... (iv)
Put N1
and T in equation (ii)
W 2I !!"
#$$%
&
I5 )ba(g2WAb
2
W g2
WA= g
WA
2
W )ba(g
WAb
I5I
= g2
WA3
1 )ba(g
Ab2
I5I
=g
A3
g (a Ib) = (2Ib + 3a 3Ib)A
A = )ba3(
g)ba(
I5
I5Ans.
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RIGID BODY DYNAMICS - 20
19. After collision, let COM move by velocity vP andsystem starts rotating by angular velocity >about COM.Using cons. of linear momentum
mv0= 3mvP * vP=
3
v0
conserving angular momentum about COM
mv0.
32
a= +> = !
!
"
#
$$
%
&6 3
3
ma2
.>
= ma2>
>=a32
v0
(a) Time to complete half revolut ion.
t =>7
=0v
a32 7
(b) Particle B completes half cycle during this duration. It s position const. COM in
shown.
Disp. of B in x-direction = Disp. due to linear motion of COM
+ Disp. due to Angular motion.
xB=
3
v0.t + MN
=3
v0.
0v
a32 7+
3
a2. cos30 =
3
2a7+ a
Disp. in Y-direction
YB=
3
a2cos60 =
3
a
Total displacement = 2B
2B yx 4
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RIGID BODY DYNAMICS - 21
20. >
f
A
f = Img
Torque about A
R(Img) = 8:2
mR2
89IR
g2
8966
R
1025.02
8= !"
#$%
&R
5
at constant angular speed
>= !"
#$%
&R
v
)n2(2R
v2
789!"
#$%
&
n =76
69!
!"
#$$%
&
87 .R.54
1818
R4
v2
2
n = !
"
#$
%
&
7
6
.R.20
1818=
7666
653
107520
1818
n =766
667520
101818 3= !
!"
#$$%
&
67666
2020
10186 3
n = !"
#$%
&67
66
2
4186
n =77636
=7
216
Number of revolutions executed by the disk before it comes at constant angular velocity n = !
"
#$
%
&
7
216.
21.a
2
f2
m
8
Friction on plate due to ground f1= 7.5 0.2 10 = 15
25 15 f2= 1.5 a
1
f2= 6a
2
10 = 1.5 a1+ 6a
2....(i)
f2. r = mr . 8* f2= ma
2...(ii)
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RIGID BODY DYNAMICS - 25
or MgR !"
#$%
&4
Mg !
"
#$%
&2
R=
8
Mv2+
2
1!!"
#$$%
&66
442
1 2RM
2
R/2
v!"
#$%
&
or8
7MgR =
16
3Mv2
= v = 3Rg14
26. When F is maximum equation. of rotational equilibrium.F.R. = (N
1+ N
2) R .............(1)
For equilibrium in horizontal direction
f1= N
2= N
1............(2)
In vertical directionF + N
1= mg
F = [(mg F) + (mg F)]
2
1 C
D
EFG
H545 )Fmg(
2
1)Fmg( C
D
EFG
H9
2
1putting C
D
EFG
H9 ijj[kus
2
1
F CD
EFG
H44
2
1
2
11 =
4
3mg
F =8
3mg =
8
3w
[ Ans.: 3w/8 ]
27. As fly moves to other end C.M must remains at same position so straw shifts left.
Torque about AB is balanced
2mg !"
#$%
&3
!= (m + m
A)g !
"
#$%
&6
!
4m = m + mA
mA= 3m
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RIGID BODY DYNAMICS - 26
28. (a) 2mg 2
Lmg
2
LT
4
L= 0
T = 2mg Ans.
(b) NP= 6mg Ans.
(c) 2mg 2
Lmg
2
L
= (2m 4
2!
+ m 4
2!
+12
m2)8x
= mg2
!=
x22
12
m
4
m38!
!"
#$$%
&4
!!
=2
mg!=
12
m10 2!8
8=!10
g6=
!5
g3Ans.
(d) 22
12
m10
2
1>6
!= 2mg
2
!mg
2
!
=2
2
12
m10
2
1>6
!=
2
mg!, >2=
!10
g12
!5
g6
>=!5
g6 , v =
2
!> =
!
!
5
g6
2Ans.
29. System is free to rotate but not free to translate. During collision, net torque of the system ( rod A + rod B +mass m ) about point P is zero.
Therefore, angular momentum of system before collision = Angular momentum of system just after collision.( About P ). Let >be the angular velocity of system just after collision, then
Li
= Lf
* mv (2l) = !>Here, ! = moment of inertia of system about P
= m (2!)2+ mA (!2 / 3 ) + m
B T
CCD
E
FFG
H4$
%
&4 2
2
)212
!!!
Given : != 0.6 m, m = 0.05 kg, mA = 0.01 kg and m
B= 0.02 kg
Substituting the values, we get
I = 0.09 kgm2
Therefore, from Eq. (1)
>=I
!2mv=
09.0
)6.0)()(05.0)(2( v
>= 0.67 v ........(2)Now after collision, mechanical energy will be conserved.
Therefore, decrease in rotational KE = increase in gravitational PE
or2
1I> U= mg (2!) + m
Ag !
"
#$%
&2
!+ m
Bg !
"
#$%
&4
2
!!
or >2 =Immg A )m34( B44!
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=( )
09.0
)02.0301.005.04(6.0)8.9( 6446
= 17.64 (rad /s)2
= >= 4.2 rad/s .........(3)Equating Eqs. (2) and (3) , we get
v = sm/
0.67
4.2
or v = 6.3 m/s
30. torque about Q point is balaned so
Pc = mg b
P = !"
#$%
&c
bmg
N = mg
f = P for no sliding
Img = PP
max= (Img)
Cmax
=P
bmg= mg
bmg
I = !!"
#$$%
&Ib
31. (i) In the limiting case normal reaction will pass through O. The cube will tip about O if torque of
Fabout O exceeds the torque of mg.
Hence !"
#$%
&V!
"
#$%
&2
amg
4
a3F
or mg3
2FV
therefore, minimum value of F is32 mg.
a/2
N
F
3a/4
O
mg
fr
(ii) In this case since it is not acting at COM, toppling can occur even after body started sliding
because of increasing the the torque of F about COM.hence Imin
= 0,
(iii) Now body is sliding before toppling, O is not I.A.R., torque equation can not be applied across
it. It can now be applied about COM.
F4
a= N
2
a................ (1)
N = mg .......................... (2)
from (1) and (2)
F = 2 mg
(iv) F >3
2mg ................... (1) (from sol. (i))
N = mg .......................(2)
F = sN =
smg ........... (3) from (1) and (2)
s=
3
2