alp solutions rbd physics eng

8/11/2019 ALP Solutions RBD Physics Eng

1/27

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

TARGET : JEE (IITs)


2/27


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.


3/27


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


6/27


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!


7/27


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).


8/27


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


9/27


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


10/27


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


11/27


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>!


12/27


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


13/27


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


14/27


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


15/27


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


16/27


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)


17/27


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)


18/27


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.


19/27


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.


20/27


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


21/27


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)


22/27


23/27


24/27


25/27


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


26/27


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!


27/27

=( )

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

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