PHYSICS SOLUTIONS ADVANCE ADV ANCE LEVEL PROBLEMS PR OBLEMS TARGET : JEE (IITs)
TOPIC : RIGID BODY DYNAMICS PART - I 1.
Let m 1 = mass of the square plate of side 'a' and m2 = mass of the square of side 'a/2'
& a # Then m 1 = ' $ ! % 2 "
2
; m 2 = ' (a )2
; ( ' being the areal density)
and m2 – m – m 1 = M.
*
=
+
m 2a 2 3 - m1 (a / 2)2 m & a # 52 4 1$ ! = 6 6 % 4 " -
1
1 1 0 31 ' a4 2 5 5 / 1 6 16 6 6 4 6 16 .
2
0/= -.
2 2 3- ' (a / 2) 4 & a # & a # 052 4 ' $ ! .$ ! / 6 % 2 " % 4 " -. -1 6
' a4
3 ( 2 6 16 ) 5 2 5 3 0 / 1 16 6 12 .
4 = 'a 2
0 4 3 27 + = ' a 212 6 16 / 1 . 4 M & 1 # 2 & 4 M # 4 $ ! .a * + Also ; M = ' $1 5 ! a * ' = = 3 a2 % 4 " % 3 a 2 " 2.
M.I. about ‘O’ is
3 M a2 * + = 16
MR 2 2
By parallel-axis theorem
& 4R # MR 2 . 2! = +cm + M $ % 3 7 " 2
MR 2 & 4R # 5M $ 2. ! * +cm = 2 37 " % 3.
3 27 0 2 / 112 6 16 .
2
2
From FBD Equation in horizontal direction T = Nx For Rotational equation about P T. 2 = 1.5 × 300 T = 225 N
. . .. .. . . . . . . .. .( 1 )
N x = 225 N N y = 300 N And Ng = mg = 300 N RIGID BODY DYNAMICS - 1
4.
There is no slipping between pulley and thread. So, (a = 8r) . . . .. . . . .. ( i ) For point mass : mg – T – T = ma ...........(ii) Equation of torque for disc Tr = +.8
mr 2 .8 Tr = 2 T=
& mg # mr8 ! = $ % 2 " 2
mg –
mg = ma 2
mg =
3mg 2
a=
...........(iii)
2g . 3
2V
2V
R
5.
2R
V R
x 2R : 2 4R 9 2v x = 2v = g g
16Rv 2 g
F sin 8
6.
F r
)
R
8
Fcos 8
f mg
(F sin 8 + N = mg) F cos 8 increased linear speed for pure rolling friction force acting leftward direction and thread winds. 7.
As ;< = 0, angular momentum remains conseved :
= *
& 300R 2 # $ ! > = L = $0 4 ! 2 % " 0 150 >0 = 180 >
& 300R 2 # $ 30R 2 ! .> 4 $ ! % 2 " * > = 5/6 >0
Ans.
RIGID BODY DYNAMICS - 2
8.
Impulse = change in momentum
=
m! 2 P. = .> 2 12
*
> =
For
? =
!
(about centre of AB)
6P m!
7 2
7
;
2
= >t
*
t=
7m! 7 = 2 6 6p 2>
*
t=
7m! 12p
G 9. ?
g sin? a Circular pipe is rest so g sin ? = a 10.
Cons. of ang. momentum about P gives L ( 2M) (2 L )2 MV = w 2 12 w=
11.
3V 4L
Ans. (C)
Rotation energy = linear energy =
2L w V = 3 2
*
1 2 +> 2
+ = mK 2
1 mv 2 2
K = gyration radius
1 2 1 +> + mv 2 v = >R 2 2 Frictional of its total energy associated with rotation. Total energy =
1 2 +> mK 2 >2 2 = = 2 2 2 2 = 1 1 2 2 mK m R > 4 > mv 4 mv 2 2
12.*
& K 2 # $ ! $ R2 4 K 2 ! % "
@ @ @ < = A ×L
(A)
@ i.e
@ @ dL = A ×L dt
@ This relation implies that
@ @ dL is perpendicular to both A and L . Therefore option (A) is correct. dt
@ @
(C) Here L . L = L2 Differentiating with repeat to time, we get
RIGID BODY DYNAMICS - 3
@ @ @ dL dL @ dL + . L = 2L L . dt dt dt @
@
*
2L .
But
since
@ L
dL dL = 2L dt dt
.....(1)
@
A
dL dt
@ @ dL = 0 L
=
dt
Therefore, from equation (1)
dL = 0 dt
@
or magnitude of L i.e. L does not change with time. (B) So far we are confirm about two points : @ <
@ <
@
@ @ dL or and L A < dt
(1)
@ <
@
(2)
@ <
@
| L | or L not changing with time, therefore it is a case when direction of L is changing
@
@
but its magnitude is constant and < is perpendicular to L at all points. This can be written as :
@
If
L = ( a cos
@ ? ) i@ + ( a sin ? ) j Here a = positive constant
@ ˆ < = (a sin ? ) ˆi – ( a cos ? ) j @ @ @ @ so that L . < = 0 and L A < @ @ @ Now A is a constant vector and it is always perpendicular to < . Thus A can be written as @ = A kˆ then
A
@ @
@
@ A A also. @ @ @ @ Thus we can say that compound of L along A is zero or component of L along A is always We can see that L . A = 0.
i.e. L
constant.
@ @
@
Finally we conclude < , A and L are always mutually perpendicular..
13.
A
S
B >
constant Angular acceleration will disappear after point B because no any friction force is present so torque is zero. But linear acceleration is increased continuously. RIGID BODY DYNAMICS - 4
S=
1 2 aT 2
From A to B
1 * aT2 2 >ƒ2 = >in2 + 28? 27R =
>ƒ =
28? =
2.
27 =
2S T 2R
1 (8 T2) 2
.27 =
8S 7 RT 2
After point B
? = >ƒT = T
8S 7 RT 2
=
So disc will make one roation in
8S 7 = R
8.27R .K = 47 R
T time and in T time it will coveral distance greater than S. 2
14.
Angular Momentum about point O 1 22mR2>2 2 (>1 = 11>2) .......(i) Relative angular velocity about point O of man >r = >1 + >2 = (12 >2) Man covers 27 angle relative to disc mR2>1 =
27 = t
>r = 12 >2
(27 = 12 >2t)
& 7 # >2t = $ 6 ! % " Some time t angle taken by disc Ans. 60º East of South 60º East of South
RIGID BODY DYNAMICS - 5
PART - II > CM
1.
m !
+ CM
x=
x=
x
m!2 9 12
+ = +CM + mx 2
+ 5 + CM m
m! 2 +5 12 m
=
7 9 0.34 m 60
A dx 2.
(i)
x B
+AB =
B dmx
2
!
+AB =
B ax dx 3
& a! 4 # $ ! $ 4 !. % "
=
0
dm (ii)
dx
x !
B ax dx 2
xcm =
0 Q
& 2 # !! % 3 "
= $
B ax dx 0
A +AB
B
2!/3
+cm
dx
& 2 # +AB = +cm + m $ ! ! % 3 "
2
4m ! 2 +cm = +AB – 9 RIGID BODY DYNAMICS - 6
!
a! 2 ax dx m= = 2 0
B
a! 4 2a! 4 5 +cm = 4 9
H a! 4 E F+ cm 9 C 36 CD GF
3.
Ans.
(a)
For 2kg mass, T1 – 2g sin 45º = 2a For 4kg mass 4g sin45º – T2 = 4a For pulley, r(T2 – T1) = +8 = +(a/r)
......(i) ............(ii)
............(iii)
& mr 2 # $ ! (+ = $ 2 ! ) % "
From eq. (i),(ii) and (iii) a=
a=
( 4 5 2)g sin ?
+ # & $4 4 2 4 2 ! r " % ( 4 5 2) 6 10 6 1 / 52 . & 4 2 0.5 # $ 4 4 ! 0.01 " %
a = 0.248 = (0.25 m/s2). (b)
m1 = 4kg m2 = 2kg I = 0.2 (between inclined plane and 2kg block) + = 0.5 kg-m2 r = 0.1 m m1gsin? – T2 = m1a .........(i) T1 – (m2gsin? + Im2gcos?) = m2a .........(ii)
& a # r(T1 – T2) = +.8 = $ + ! % r "
.........(iii)
From eq. (i),(ii) and (iii) m1g sin? – (m2g sin? + Im2gsin?) +
+a r2
= m1a + m2a
Put values : 4g sin45º – (2g sin45º + 0.2 2g sin45º) +
*
27.80 – (13.69 + 6.95) = 56a
=
a=
0.5 a = 6a 0.01
7 = (0.125 m/s2). 56
RIGID BODY DYNAMICS - 7
4.
N2 sin 8 = f ——(i) N1 + N2 cos 8 = mg ——(ii) Torque about point A
& b # ! + N2 sin 8 b = mg a cos 8 ) 2 % tan 8 "
( N2 cos 8) $ N2 =
(mga cos 8 sin 8 ) 2b
N2 cos 8 =
(mga cos
2
8 sin 8 )
2b
From equn. ....(ii) mga cos 2 8 sin 8 N1 = mg – N2 cos 8 = mg – 2b ( 2b – a cos 2 8 sin 8 ) N1 = mg 2b N2 sin 8 = IN1
I =
N2 sin 8 N1
mg a cos 8 sin 2 8 2b I = mg(2b – a cos 2 8 sin 8 ) 2b
5.
Nx
o
Ny b
H a cos 8 · sin2 8 E * I = F 2b – a cos2 8 sin 8 C GF DC
C b
mg mb 2 mg b / 2 = + 8 , + = + m 6 !
& b # $$ !! % 2 "
2
1 # mb 2 mb 2 mb 2 & $1 4 ! + = I = 6 2 2 % 3 "
2mb 2 + = 3
RIGID BODY DYNAMICS - 8
2mb 2 = 2 3
mgb Hence
8 ! 8 =
3g 2 2b
3g 2 Acceleration of O in horizontal direction is zero So Nx = 0 Accn of corner C =
mg – Ny = m
= m
=
6.
Ny =
b
b2 4 b2
8 =
8
2
& 3g # $ ! = 3 mg $ 2 % 2 2 b "! 4
b
mg 4
(a)
1
2
kxsin30º
kxsin30º 30º
A
30º
B
L mg
(i)
Before cutting 2k xsin30º = mg kx = mg (T = kx = mg) After cutting (ii) Torque about COM (Tsin30º) x
!
2
= +8
mg! m! 2 .8 = 4 12
& 3g # 8 = $ ! ! (clockwise) % " (b) acceleration of point A max = T cos30º
mg 3 3g = = aAC @ 2m 2 mg – T sin30º = ma y ax =
mg –
mg = may 2 RIGID BODY DYNAMICS - 9
& g # 8! g 3g ! + = 5 + = (g) (J) % 2 " 2 2 2
aAy = $ 5
& 3 ˆ ˆ # $ ! aA = $ 2 i 4 j ! g % " (c)
aBx =
3 g@ 2
& g # 8! aBy = $ ! 4 = 2g ( K ) % 2 " 2
& 3 ˆ ˆ # $ ! aB = $ 2 i 5 2 j ! g % "
(ii)
kxsin30º
1
kxsin30º 2
30º
30º L/3
L/3
L/3
mg Before cutting mg = 2kx sin30º = kx = T T = mg. After cutting
(a)
Torque about COM
& ! #
(T sin30º) $ ! = +.8 % 6 "
& 1 # & ! # m! 2 .8 (mg) $ ! $ ! = % 2 " % 6 " 12 & g # 8 = $ ! ! (cw). % " (b)
(T cos30º) = max mg
3 = max 2
RIGID BODY DYNAMICS - 10
& 3g # $ ! $ 2 ! % "
ax =
aAx =
3g ˆ (5 i ) 2
mg –
ma = may 2
ay =
g ( 5 jˆ ) 2
aAy = (ay – 8 "
& g # g & ! # ) = $ ! – $ ! = 0 % 2 " ! % 2 " 2 !
& 3 ˆ # ! $ 2 gi ! % "
aA 9 $ 5
(c)
"
a cx 9 5
3 ˆ gi 2
! # & g 4 8 ! jˆ = – g jˆ 2 " % 2
"
a cy 9 5$
ac = acx ˆi + acy jˆ
=
& 3 ˆ ˆ # g i 4 g j ! . ! 2 % "
– $$
7.
Angular momentum about point A Li = m1vs! (us : Final velocity of ball after collision) Lƒ =
m2 ! 2 > + m 1us! 3
Li = Lƒ m 2 ! 2 .> (m1vs! = + m1us!) 3 2×5=
10 =
8 6 1. 2 6 > b + (2 × us) 3
32 > + 2us 10
............ (i)
Coefficient of restitution e=
>! 5 u s vs RIGID BODY DYNAMICS - 11
>! 5 u s
0.8 =
vs
> (1.2) 5 u s 4 = 5 5 4=
6> – us 5
& 6> – 20 # ! % 5 "
us = $
............ (ii)
Put equation (ii) in equation (i) 10 =
& 6> – 20 # 32> ! + 2 $ 10 % 5 "
10 =
32> 12> – 40 + 10 5
100 = 32> + 24> – 80
> =
45 14
Put > in equation (ii)
& 6> 5 20 # ! % 5 "
us = $
& 45 # 5 20 ! % 14 "
6$ us =
us =
5
& 1 # 270 – 280 10 = – = $ 5 ! 14 6 5 14 6 5 % 7 "
& 1 # So direction is (L) us $ ! % 7 "
8.
(a)
Let coordinates of instantaneous axis of rotation be P(x,y). then velocity of P w.r.t. C is zero. "
*
> 6 CP 4 v ˆi 9 0
* *
b t ( kˆ ) × [( x 5 v t ) ˆi 4 y jˆ] 4 v ˆi 9 0
x = vt M yt = V and from these eliminating t RIGID BODY DYNAMICS - 12
v2 My x . 91 or xy = v N M * locus of P is a Hyperbola.
(b)
& 1 2 # Here coordinate at point C = $ Nt , 0 ! % 2 "
=
> 6 CP 4 v ˆi 9 0
*
"
1 H E >kˆ 6 F( x 5 w t 2 )ˆi 4 y jˆC 4 w t ˆi 9 0 2 G D >y = w t
*
x=
1 2 wt 2
from these eliminating t, 2
*
1 & > # 2 x = w$ ! y 2 % w "
*
x=
>2 2w
y2
Eqn. of parabola. 9.
a = 8 R mg sin 30 0 – T = ma or
.........(1)
mg – T = ma 2
.........(2)
TR
8 =
!
I
= 1 MR 2 2
2T MR Solving Equations (1), (2) and (3) for T, we get
8=
T=
.........(3)
1 M mg 2 M 4 2m
Substituting the value, we get
& 1 # 3 (2)(0.5)(9 .8) 0 / = 1.63 N T= $ ! 2 % 2 " 1 2 4 (0.5)(2) . T = 1.63 N (ii) From Eq. (3) , angular retardation of drum (2)(1.63) 2T = = 8.15 rad/s2 (2)(0.2) MR or linear retardation of block a = 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 be v 0 = > 0 R = (10) (0.2) = 2 m/s Now, the distance (s) travelled by the block until it com es to rest will be given by
8 =
v 20 s= 2a (2) 2 = m 2(1.63)
[ Using v 2 = v02 – 2as with v = 0 ]
or
s = 1.22 m
RIGID BODY DYNAMICS - 13
10.
Angular momentum conservation about A
O 3R 10
x
& %
3 R # ! = 2mR2 > 10 "
mvR + mv $ R 5
v'=
1.7v 2
3 & 1.7 v #0 !/ 2>' 9 $ 2 R % " 1 .
*
By Energy conservation 1 1 1 1 mv 2 + +>2 = mg × 0.3 R + mv'2 + +>'2 2 2 2 2 2 2 mv = mg × 0.3 R + mv' 2
& 1.7 # 2 ! v v = g × 0.3 R + $ % 2 " 2
vmin = 11.
2 0.3gR 1.7
Minimum velocity required by block ‘m’ to complete the m otion in
5gR
conserving mech. energy 1 2 R MgR +> = Mg . * > = 2 2 + Cons. angular momentum wrt P before & after collision.
+> = m.R +
MgR
5gR
= mR
5gR
+ MgR + = m2R2 5gR ML2 putting + = 3 M = m Ans. :
15
M 9 15 m RIGID BODY DYNAMICS - 14
12.
(i)
L = Natural length of spring
(ii)
(a)
By energy conservation from (i) to (ii)
H1 2 1 2 1 2 & ! #E F 2 I> 4 2 kx 9 2 I>1 4 mg $ 2 5 x !C % "D G & ! # I = Icm + , $ 5 x ! % 2 "
............. (i)
2
& ! # m! 2 I= + m $ 5 x ! % 2 " 12
2
............. (ii)
2 2 & $ x 9 x 4 y 5 L #! % "
............. (iii)
Put equation (ii) and (iii) in equation (i) 2 1 & ! # # $ m! 4 m & $ 5 x ! ! >2 + 1 K & x2 $ ! 2 $ 12 2 % " " 2 % % 2
1 / 2
4 y 5 L #! " 2
2 2 1 & ! # #! $ m ! 4 m & $ 5 x! = 2 $ 12 % 2 " !
%
& ! # >12 + mg $ 8 5 x ! % " "
x = 150 mm, y = 20 mm, ! = 450 mm, K = 300 N/m m = 3 kg, > = 4 rad/sec
RIGID BODY DYNAMICS - 15
Put all the data
>1 = (b)
2 86 rad/sec 3
rotating to 180° condition is
This is like a initial condition so >2 = > >2 = 4 rad / sec 13.
Force moment relative to point O
"
dM " " N = dt = 2b t
"
"
Let the angle between M and N 8 = 45 º at t = t0 "
" "
Then
M .N = = M N 2
(a 4 bt 02 )
1
=
Solving, t0 =
2b 2 t 30 a 2 4 b 2 t 4 . 2bt 0
bt 20
=
a 2 4 b 2 t 04
a (as t0 cannot be nagative) b "
Therefore
a 2 4 b 2 t 04 2bt 0
"
"
N 9 2bt 0 9 2b
a b
8 = angular acceleration
14.
8 = angular acceleration For the plank F – ƒ = m1>1
....... (i) RIGID BODY DYNAMICS - 16
For sphere torque about point C fr = Ic8 =
2 m r28 5 2
....... (ii)
assuming >2 is the acceleration of COM of sphere at point A (>1 = >2 + 8r) From equation (i), (ii) and (iii)
>1 =
....... (iii)
F & m 2 m # $ 14 2! 7 " %
and
& 2 # >2 = $ 7 >1 ! % "
15.
For cylinder Mg + T1 – 2T = Ma Torque about axis of rotation
......(i)
& a # ! % R "
2TR + T1(2R) = +O8 = $ + .
For weight A mg – T1 = ma No slipping between pulleys and thread a1 = a + 8(2R) = (3a) From equation (i), (ii) and (iii)
3 -2a1 9 -1 16.
....(ii)
(a = 8R)
....(iii)
0 3(M 4 3m) g / & M 4 9m 4 + # $ ! R 2 " % .
Velocity of end A at the moment it strikes ground = 2gh If velocity of COM of rod just after collision vP and angular velocity acquired by the rod is > clockwise as shown then using equation for coefficient of restitution velocity of approach = velocity of sep. (applied at point A). L .............(1) > cos ? 2 Angular momentum can be conserved about A just before collision & after collision as only impulsive force will be a cting at A only. 2gh = v P +
2gh M
L L cos ? = +cm > – Mv P cos ? 2 2
.............(2)
RIGID BODY DYNAMICS - 17
Putting value of > = ( 2gh – vP )
2 L cos ?
from (1) 2gh M.
2 L L ML2 cos ? = ( 2gh – vP ) – MvP cos ? L cos ? 2 2 12
L L 2gh L cos ?vP vP + = 6 cos ? 2 6 cos ?
2gh 2
Lcos?
H1 4 3 cos2 ? E (1 5 3 cos 2 ?) vP F 6 cos ? C = 2gh FG CD 6 cos ? & 1 5 cos2 ? # $ ! vP = $ 1 4 3 cos ! % "
2gh
COM will of at maximum height when its velocity becomes zero during upward motion. O = vP 2 – 2g H 2
& 1 5 3 cos 2 ? # vP 2 ! H= = $$ 2 ! h. 2g 4 ? 1 3 cos % "
& 1 5 3 cos ? # [ Ans.: H = $ ! % 1 4 3 cos ? " 2 2
17.
2
49
h; h =
7 !
144
]
NC + N B = 250 NB – x = 250 × 3 NB = f 1 =
750 x
750 I x
H 750 E 4 25 C I f 2 = F G x D workdone against friction 4 .5
W=
B
( Q 1 4 Q 2 ) dx =
& 1500
B $% 3
# 6 0.3 4 7.5 ! dx = 450 x "
!
n
3 + 7.5 (4.5 – 3) 2
= 450 × 0.41 + 7.5 × 1.5 1 mv2 = 400 × 1.5 – 195.75 2 v2 = (600 – 195.75) ×
2 2.5
=161.7 × 2 = 323.4 v = 18.52 m/sec.
RIGID BODY DYNAMICS - 18
18.
? is very small ? R 0º T
IN1
b N1
T IN2
P
a
A
?
A
N2 w/2
Force balance in horizontal direction N1 = N2 balancing torque about point P ? to be very small we can directly write For Wb – N2a = 0 2 Force in y direction if acceleration of windows is A T.b + IN2b –
w – IN1 – IN2 – T =
wA g
... (ii)
For block T –
WA W = 2g 2
& W WA # !! 4 2 2 g % "
T = $$
.... (iii)
Put equation (iii) in equation (i) W W Wb b+ Ab + IN1b = N1a + 2g 2 2 WAb 2g = N1 (a – Ib)
3 WAb 0 / 1 2g (a 5 Ib) .
N1 = 2
..... (iv)
Put N1 and T in equation (ii)
& WAb # W WA WA !! – – = 2g g 2 % 2g (a 5 Ib) "
W – 2I $$
I WAb 3 WA W – = g (a 5 Ib) 2g 2 1 –
2I Ab 3A = g (a 5 Ib) g
g (a – Ib) = (2Ib + 3a – 3Ib)A A=
(a 5 Ib) g (3a 5 Ib)
Ans.
RIGID BODY DYNAMICS - 19
19.
After collision, let COM m ove by velocity vP and system starts rotating by angular velocity > about COM. Using cons. of linear momentum mv 0 = 3mvP
*
vP =
v0 3
conserving angular momentum about COM
mv 0.
a 2 3
& ma 2 # $ ! = $ 3 6 3 ! .> % "
= +> = m a2> v0
> = 2 3 a (a)
Time to complete half revolution. t=
(b)
2 3 a7 7 = v0 >
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 =
v0 .t + MN 3
v 0 2 3 a7 2a . + . cos30° v0 3 3 Disp. in Y-direction
=
YB =
2a 3
cos60°
Total displacement =
=
=
2 3
a7 + a
a 3
x B2 4 y B2
RIGID BODY DYNAMICS - 20
>
20.
f = I mg
A f Torque about A R(I mg) =
mR 2 :8 2
2Ig 98 R 2 6 0.25 6 10 98 R
& 5 # 8 = $ ! % R " at constant angular speed
& v # > = $ ! % R " 2
& v # $ ! 9 28 (27n) % R " n=
& v 2 # 18 6 18 $ ! $ 487R2 ! 9 4 6 5.R.7 % " 18 6 18 & 18 6 18 # ! = % 20.R.7 " 20 6 75 6 10 53 6 7
n= $
18 6 18 6 10 3 n= = 20 6 75 6 7
& 6 6 18 6 103 # $ ! $ 20 6 7 6 20 ! % "
& 6 6 18 6 4 # ! % 7 6 2 "
n= $ n=
36 76
7
=
216
7 & 216 # !. % 7 "
Number of revolutions executed by the disk before it comes at constant angular velocity n = $
21.
8
a2
m f2
Friction on plate due to ground f1 = 7.5 × 0.2 × 10 = 15 25 – 15 – f 2 = 1.5 a1 f 2 = 6a2 10 = 1.5 a 1 + 6a 2 ....(i) f 2 . r = mr . 8 * f 2 = ma2 ...(ii)
RIGID BODY DYNAMICS - 21
f 2 = ma 1 – ma 2 a2 + r 8 = a1
* *
*
a1 – a2 = a2
a2 = a1 – a2 a1 = 2a 2
10 – 3a 1 = 1.5 a1
*
100 20 = 45 9
a2 =
20 a1 = 18 2
v1 = a 1t =
20 5 3 × = (Plate) 9 3 4
v2 = a 2t =
20 5 3 × = (pipe). 18 6 4
>2 =
22.
a1 =
v 2 5 1000 9 x 9 10 .42 rad / s (pipe) r 6 160 2
4.8 Ma2 Assuming the lamina to be in xy plane. Then from the perpendicular axis theorem Ix + Iy = Iz but Ix = Iy ( by symmetry ) 2 and Iz = 1.6 Ma (Given ) Iz = 0.8 Ma2 2 Now from the Parrallel axis theorem IAB = Ix + M(2a)2 = 0.8 Ma2 + 4Ma2 = 4.8 Ma2
=
Ix =
y A
B
x
23.
Under the given conditions only possibility is that friction is upwards and it accelerates downwards as shown below :
RIGID BODY DYNAMICS - 22
The equations of motion are : a=
mg sin ? – f mg sin 30 º – f g f = = – m m 2 m
<
fR
= 8 = + = +
2f mR
......(1) ......(2)
For rolling (no slipping) a = R8 or g/2 – f/m = 2f/m
=
3f = g/2 or f = mg/6 m
(1)
f S fmax
S I mg cos 30º S
3 2
I mg
(2) Other possibilities which are not feasible are as follows : (a) Friction is downwards. In this case a and 8 will be as shown and rolling is not possible.
(b) Friction is upwards and the disc has linear acceleration in upward direction.
In this case also rolling is not possible. 24.
Between the time t = 0 to t = t0. There is forward sliding, so friction, f is leftwards and maximum i.e., I mg. For time t > t0, friction f will become zero, because now pure rolling has started i.e., there is no sliding (no relative motion) between the points of contact.
So, for time t < t0
Linear retardation, 8 =
f = Ig (f = I mg) m
fR 2Ig and angular acceleration, 8 = = I mR 2 = I R 2
<
Now let V be the linear velocity and >, the angular velocity of the disc at time t = t0 then V = V0 – at0 = V0 – Igt0 ......(1) RIGID BODY DYNAMICS - 23
and
> = 8 t 0 =
2Igt 0 R
......(2)
For pure rolling to take place V = R> i.e., V0 –2Ito = 2Ito V0
*
t0 = 3 I g
Substituting in Eq. (1), we have
& V #
0 V = V0 – Ig $$ 3 I g !! % "
2 V 0 3
V=
Work done by friction For t S t0, linear velocity of disc at any time t is V = V0 – Igt and angular velocity is > = 8t =
2Igt . From R
work-energy theorem, work done by friction upto time t = Kinetic energy of the disc at time t – Kinetic energy of the disc at time t = 0
=
W=
1 1 1 mV2 + I >2 – mV02 2 2 2
# 2Igt # 1 1 & 1 $ mR 2 ! & $ ! = m [V0 – Igt]2 + 2 2 % 2 " % 2 " =
2
–
1 mV02 2
1 [mV02 + mI2g2t2 – 2mV0Igt + 2mI2g2t2 – mV02] 2
or
W=
m Igt [3Igt – 2V0] 2
For t > to, friction force is zero i.e., work done by friction is zero. Hence, the energy will be conserved. Therefore, total work done by friction over a time t much longer then t0 is total work done upto time t0 (because beyond the work done by friction is zero) which is equal to W=
mIgt 0 [3Igt0 – 2Vo] 2
Substituting t0 = V0 /3Ig, we get W=
mV0 [V0 – 2V0] 6
W = – 25.
mV02 6
Let M be the mass of unwound carpet. Then , 2
& M # & R # ! 7 $ ! = M M = $ 2 % " R " % 2 " 4 ‘
M
M’ v
R
R/2
From conservation of mechanical energy : MgR – M ‘ g
R
2
=
1 & M # 2 $ ! v + 1 I > 2 2 % 4 " 2
RIGID BODY DYNAMICS - 24
& M # & R # Mv 2 1 or MgR – $ ! g $ ! = + % 4 " % 2 " 8 2 or
= 26.
& 1 M R 2 # & v # 2 $ 6 6 ! $ 2 4 4 ! $% R/2 "! % "
7 3Mv 2 MgR = 8 16
v=
14 Rg 3
When F is maxim um equation. of rotational equilibrium. F.R. = µ (N 1 + N 2) R .............(1) For equilibrium in horizontal direction f 1 = N2 = µN 1 ............(2) In vertical direction F + N1 = mg
F = µ [(mg – F) + µ (mg – F)]
1 H(mg 5 F) 4 1 (mg 5 F )E F C 2 D 2 G
H G
F F1 4
H Fputting µ 9 G
1E H 1 E µ 9 j[kus ij C C F 2D G 2 D
1 1E 4 C = 3 mg 2 2D 4
3 3 mg = w 8 8 [ Ans.: 3w/8 ]
F=
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 + mA)g $ 6 ! % " % " 4m = m + mA mA = 3m
RIGID BODY DYNAMICS - 25
28.
L L L – mg × – T × = 0 2 2 4 T = 2mg Ans. 2mg ×
(a)
NP = 6mg
(b)
2mg ×
(c)
Ans.
L L – mg × 2 2
= (2m ×
!
2
4
+m ×
!
2
4
m2 ) 8 x 12
+
& 3m! 2 m! 2 # x $ ! = mg = $ 4 4 12 ! 8 2 % " !
mg! 10m! 2 8 = = 2 12
8 = (d)
Ans.
! ! 1 10m! 2 2 6 > = 2mg × – mg × 2 2 2 12
=
12g 6g mg! 1 10m! 2 2 6 > = , >2 = – 10! 5! 2 2 12
> = 29.
6g 3g = 10 ! 5!
6g , 5!
v= >
!
2
=
!
2
6g 5!
Ans.
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
* Here,
!
L i = L f mv (2l) = !> = moment of inertia of system about P
= m (2!) 2 + m A (! 2 / 3 ) + m B
H ! 2 & ! 2E F 4 $ 4 !)T C GF12 % 2 DC
Given : ! = 0.6 m, m = 0.05 kg, mA = 0.01 kg and mB = 0.02 kg Substituting the values, we get I = 0.09 kg –m 2 Therefore, from Eq. (1) (2)(0.05 )(v )(0.6) 2mv ! = 0.09 I > = 0.67 v Now after collision, mechanical energy will be conserved.
> =
........(2)
Therefore, decrease in rotational KE = increase in gravitational PE or
or
& ! # & ! # 1 U I> = mg (2!) + mA g $ ! + m B g $ ! 4 ! % 2 " % 2 " 2
> 2 =
g !( 4m 4 m A
4 3 mB )
I
RIGID BODY DYNAMICS - 26
(9.8 )(0.6 )( 4 6 0.05 4 0.01 4 3 6 0.02) 0.09 = 17.64 (rad /s) 2 = > = 4.2 rad/s Equating Eqs. (2) and (3) , we get
=
.........(3)
4.2 m / s 0.67 v = 6.3 m/s
v= or 30.
torque about Q point is balaned so Pc = mg b
& mg b #
P= $ c ! % " N = mg f = P for no sliding I mg = P Pmax = (I mg) Cmax = 31.
(i)
mg b mg b = I mg = P
& b # $$ !! % 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.
& 3a # & a # ! V mg $ ! % 4 " % 2 "
Hence
F$
or
2 F V mg 3
2 therefore, minimum value of F is mg. 3
N a/2
fr
F 3a/4
O mg
(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. F×
a a = N × ................ (1) 4 2
N = mg .......................... (2) from (1) and (2) F = 2 mg
(iv)
F>
2 mg ................... (1) 3
(from sol. (i))
N = mg .......................(2) F = µ sN = µsmg ........... (3) from (1) and (2) µs =
2 3
RIGID BODY DYNAMICS - 27
