# Jee 2014 Booklet3 Hwt Solutions Rotation Motion

August 28, 2017 | Author: varunkohliin | Category: Rotation Around A Fixed Axis, Torque, Classical Mechanics, Mechanics, Physical Quantities

#### Short Description

Jee 2014 Booklet3 Hwt Solutions Rotation Motion...

#### Description

Vidyamandir Classes

Rotational Motion 1.(C)

HWT - 1

For the object to have only translational motion the force must be applied at the centre of mass. Let p = mass per unit length For Part A – B centre of mass = midpoint of A – B = Point D For the other part Centre of mass = mid point of CD =  distance from D Now centre of mass of the whole structure p  o  2 p    from point D p  2 p =

2  form point D 3

2 4   3 3 Let M be the mass of original disc mass of disc removed = M/4  mass remaining = 3M/4 From symmetry, the centre of mass of new part fromed is on line passing through C1C2 Let x be the distance of centre of mass of new part (towards left) from C1

 distance of p from C = 2 

2.(A)

0

M M  R 4 4 M R 1 x   3 3

x  3

3.(D)

w0  

w2  w02  2   w02  w02  2   72 4



 

3 2 1 w0  4 144

Again using equation w2f  wi2  2   0 

4.(D)

w02  2  4

  24

w02 144  4  8 3 w02

 = 12 rotations

M.I. of disc about tangential axis in the plane of disc

5 MR 2 5 R  MR 2  MR 2  rod of gyration = 2 4 4 M.I. of ring about tangential axis in the plane of the disc c

  ratio 

MR 2 3  MR 2  MR 2  rod of gyration = 2 2

3 R; 2

5 2 5   2 3 6

VMC/Rotational Motion

20

HWT-Solutions/Physics

Vidyamandir Classes 5.(B)

M.I. of a ball about (assuming solid sphere) its C0.M 2  MR 2 5 About centre of ring 2  MR 2  Mx 2 5 M.I. of all the balls about centre of ring  2  M i Ri2  M i xi2 5

2 2 R 5

 M   x  M 2

i

i

2  0.12  10  12  10 5

= 10.04 kg  m 2 6.(B)

Radius of sphere is increased M.I. will increase We don’t apply any external torque force angular velocity will decrease I1w1  I 2 w2

 I 2  I1  w2  w1

Angular momentum will remain constant L = Iw K.E. will decrease 7.(A)

F = 200 N

R = 25 cm   Torque = r  F  200  0.25  50 N  m

8.(A)

I = 5kg – m2 

  40 Nm

 I  

w  w0   t 24 = 0 + 8t

9.(D)

1 mV 2  mgh 2

10.(B)

  Ib

40  8rad / s 2 5



t = 3s

h

V2  5m 2g

MR 2 .b 2 2  Mg cos  b MR 2 2  g cos  b R N  Mg cos 

   NR   

Ma  Mg sin    Mg cos 

For rolling without sliding a  Rb Mg sin    Mg cos   2  Mg cos 

1 tan  3 2 a  g sin  3

 

VMC/Rotational Motion

21

HWT-Solutions/Physics

Vidyamandir Classes

Rotational Motion

1.(B)

m x m m y  m

i i

Co-ordinate xcm 

1

634 x 64

x  2

1

6344 64

y  2

i

i i

ycm

HWT - 2

i

Similarly z   2 2.(C)

V cm 

m V m i

i

 

6 si  2 j  10k  4 10i  2 j  5k

10

i

V cm  7i  2 j  8k 3.(D)

 

w2  w02  2  

 300  100 2 10

 40

4 2 .3002  4 2 .1002  2  40 

4.(A)

 

4 2  80 000 80 000 1 rad  4 2  . rotation 80 80 2 = 2000 rotation

M, I, R 2 I  MR 2 5

8 droplets are formed

 m = 1/8 M rad R1 = 1/2 R I new 

5.(C)

2 1 R2 1 2 I  . M   MR 2      5 8 4 32 5 32

M I about O on a axis perpendicular to plane of the square = 4m From symmetry

Using  ar axes theorem

MI about BD  2m2 

2  2 m 2 2

1  m 2 2

Using parallel axes theorem

2

   MI about P  m2  4m   3ml 2  2 

6.(D)

I 0  MR 2 I new  MR 2  4mR 2 

No momentum is transferred I 0 w0  I new ω new

ω new 

MR 2 ω MR  4mR 2

2

Mω M  4m

7.(C)

A couple produces zero net force and non zero torque which produces purely rotational motion.

8.(B)

M.R

M is kept const. R is increased force No external acts torque

VMC/Rotational Motion

;

angular momentum is constant

22

HWT-Solutions/Physics

Vidyamandir Classes 9.(C)

M, R, V, W V = W R

[rolling without slipping]

K.E. = 1/2 MV2

Translational

2 1 2 1 2 V 2 K.E. = 1/2 Iw2  2  5 MR R 2  5 MV

Rotational

1 MV 2 2 5   ratio = 1 MV 2 5 2 10.(A)

N  mg cos  mg sin    N  ma

a  g sin    g cos 

 NR  I 

R Mg cos  

1 MR 2 2

2  g cos  R For No slipping a  R 

 

tan   3

Rotational Motion 1.(B)

m2 g  T  m2 a

. . . . .(i)

T  m1 g  m1a

. . . . .(ii)

HWT - 3

(i) + (ii)

 m2  m1   m1  m2  

g a

2m1m2 g  m1  m2 

T 

Consider the blocks pulley and string as a system Net force downwards =  m1  m2  g  2T

acc  g  2 

2m1m2

 m1  m2 2

g

4  3 g  = g 1    16  4

2.(C)

m1d  m2 x

3.(B)

v  1 .5 m / s

ω

x

m1d m2

1.5  3rad / s 0.5

VMC/Rotational Motion

23

HWT-Solutions/Physics

Vidyamandir Classes 4.(A)

If mass hadn’t been cut mass of disc = M

2

 r2

MI of annular disc

1 M R2 1 Mr 2 1 2 . R  . r2  M R2  r2 2 2 2 2 2 2 R r 2 R r

5.(B)

R

R2

M.I. of rod A = 0 M.I. of rod B = ML2/12

6.(D)

Ii 

R1  R / 4

I i wi  I f w f

2 MR 2 5

If 

2 1 2  MR12  MR 2   5 16  5

w f  16 wi

24 t   1.5 h 16

K 

1 2 Iw 2

7.(B)

L  Iw

8.(B)

Horizontal vel = 4 cos  At every point torque acts downwards 2u sin  time  g

w

L I

K 

1 L2 I 2 I2

I 

L2 2K

At any time torque = xmg

Total torque = mg  1  10  1 2 Iw 2

H 2

102 1   50 N  m 10 2 2  360 I   1.8 kg m 2 20  20

9.(B)

K .E. 

10.(A)

Frictional force coverts translational energy to rotational energy.

Rotational Motion 1.(D)

2.(A)

xcm 

1  2 3 2 3

ycm 

1 2  3 2 3

HWT - 4

Let M be the mass of disc initially Let x be the shift towards left of the CM 8 1 M  x  M  3.2 9 9 x = 0.4 cm 

VMC/Rotational Motion

24

HWT-Solutions/Physics

Vidyamandir Classes 3.(A)

Earth – 1 rotation in 1 day Hours hand – 1 rotation in 12 hrs. Second hand – 1 rotation in 1 minute Fly wheel 1 rotation in 0.2 second

 increasing order of angular velocity 1, 2, 3, 4 4.(D)

M.I. about centre and  ar to plane  About AB 

5.(B)

ma 2 6

ma 2 ma 2 2   ma 2 6 a 3

1 2 r 2k mv . 2  2 v w K.E. is halved and angular velocity doubled L L  4

L  mvr 

 

6.(D)

L

mv

7.(A)

I 

1 MR 2 2

2

h

I 

1 MR 2  mr 2 2

0.5  2  0.22

w  I w / I  

0.5  2  0.22  0.25  0.2 2 = 24 rds /s

8.(A)

MI. about C.M. = ML2/2 Iw = J L/2 = MV L/2 

9.(D)

w.ML2/2 = MvL/2 

3v 2 1 1 v2  mv 2  mk 2 2 4g 2 2 R

1 K2  2 R2 10.(D)

v L

1 2 1 mv  Iw2 2 2

mgh 

mg 

w

KR/ 2

 disc

Using conservation of energy ` 1 1 mgh  mv 2  Iw2 2 2

mgh 

1 2 1 v2 mv  mR 2 . 2 2 2 R

VMC/Rotational Motion

v

gh

25

HWT-Solutions/Physics

Vidyamandir Classes

Rotational Motion 1.(A)

HWT - 5

No external force on the system momentum is conserved C.M. stays rest.

2.(A) 3.(C)

I decreases and no external torque acts on the system L = constant

 4.(B)

w increases till it reaches centre and then decreases to its initial value

MI about pt. A  2  ML2 / 96  2 

= 5.(C)

 

t  6.(B)

ML2 ML2 ML2   48 16 12

 10   4 rads 2 I 2.5

w f  wi 

20  5s 4

Duration of year will decrease

Zero

1 2 1 2 1 2 1 2 v 7 mv  Iw  mv   Mr 2 . 2  mv 2 2 2 2 2 5 10 r

K .E. 

9.(B)

K .Erot  40% K .Etrans

7.(D) 2

8.(A)

10.

ML2 2  16

1 1 Mk 2 w2  0.4  Mv 2 . 2 2 body is solid sphere

K 2  0.4 R 2

 K2 

2 2 R 5

Question cancelled.

Rotational Motion

HWT - 6

x

1.(A)

2.(B)

Centre of mass doesn’t shift

3.(A)

I  10kgm 2

  Torque    40t  10t  N  Angular acc.     4t  t  rad / s I

2.5  16  10cm 2.5  1.5

F  20t  st 2 N R  2m 2

2

2

 changes its direction at t = 4sec. dw  4t  t 2 dt

w

VMC/Rotational Motion

 0

t

dw 

  4t  t  dt  2t 2

2

 t3 / 3

0

26

HWT-Solutions/Physics

Vidyamandir Classes w changes sign at t = 6 sec. d  w  2t 2  t 3 / 3 dt  

2 3 t4 t  3 12

2 63  6  63   216  2 / 3  1 / 2   36 rad 3 12

No. of rotations = 4.(A)

 

2 r  L

36 18  2 

r 

L 2

ML2

I 

4 2

5.(A)

For no rotation M L 3M    x 4 2 4

 6.(D)

I AC 

8.(D)

 

mg  / 2 2

m / 3

L L L   from other end 2 6 3

3 g 2 

Using conservation of energy principle 1 2 Iw  mgh 2

m = 2kg

v  wr  F 

h

 2 w2 6g

r = 0.25 m

K .E rot  4 J

10.(D)

L 6

L  Iw  1  2  2  4 kg m 2 s 1

1  m 2  2  3  w  mgh 2 9.(B)

x

m 2 m 2 m 2   12 4 3

  mg  / 2  7.(B)

man is at

1 1  mr 2 w2  4 2 2

16  2 2 ms 1 2

mv 2 r

v1 R  1 v2 R2

F1 V12 R R   1 F2 R1 V22 R2

VMC/Rotational Motion

27

HWT-Solutions/Physics

Vidyamandir Classes

Rotational Motion 1.(B)

Centre of mass doesn’t change

2.(B)

Centre of mass of horizontal part is at 0 its area = 16m2

HWT - 7

Centre of mass of vertical part = 3 m down from P = 4 m down from O Its area = 12m2

C.M . from O 

16  12  4  1.7 m 16  12

ML2 ML2 2  M  L / 6  12 9

3.(B)

I 

4.(D)

Let M = mass of each rod 2 2  2  Mx 2  3    x    Mx I   M   2  M  x   2  12  2    12   

 Mx 2  Mx 2 Mx 2  3Mx 2  18Mx 2 =     2 = 12 4  4   12  12  x2  =  3M     2 5.(C)

I 

x 2

MR 2 2 mg  T  ma

. . . . (i)

Torque (  ) = TR I   TR MR 2 .   TR 2 

T 

MR 2

. . . . (ii)

Using (ii) in (i) 

mg 

MR  ma 2

6.(B)

 r   i  j  F   F k

7.(B)

m = 5 units

 

a R

mg 

   r  F   F j  i

speed = 3 2 m

MR 9  ma 2 R

a 

2m g 2m  M

y  x 4

Angular momentum = mv  r = 5  3 2  2 2 units = 60 units

8.(A)

360 

1 2 I  30 2

VMC/Rotational Motion

I = 0.8

28

HWT-Solutions/Physics

Vidyamandir Classes 9.(B)

mg  T  ma

. . . . . (i)

Tr 

mr 2  2

Tr 

mr 2 a . 2 r

T

ma 2

. . . . . (ii)

Using (ii) in (i) 2 a  g  3 10.(A)

(Consider solid cylinder) mg sin 30  f  ma f 

. . . . .(i)

ma 2

. . . . . (ii)

Using (ii) in (i) g a  3

Rotational Motion 1.(D)

HWT - 8

Statement 1 : False (VCM remain constant if Fnet = 0 Statement 2 : True

2.(A)

   5i  5 j  2   a cm    m / s V cm  2i  2 j m / s  2 

acm = constant 3.(A)

4.(BC)

straight line motion

1 mR 2 4 1 3 I   mR 2  mR 2  mR 2 2 2 I   6I I 

I EF 

m 2  I AC  I BD 12

I AB  I BC  I CD  I AD 

m 2 3

5.(D)

1 .6  1  F  0 .4

F = 4N

6.(C)

I1w   I1  I 2  w 

w 

7.(D)

I1

 I1  I 2 

w

L = mvr Frequency is doubled 1 2 r   1   2 r  1   mv 2  .   mv 2  .  . 2  v 2   v  

L  K .E 

1 1  freq. 

VMC/Rotational Motion

L 

L 4

29

HWT-Solutions/Physics

Vidyamandir Classes 8.(C)

9.(B)

10.(A)

I A  IB

LA  LB

I AWA  I BWB

1 1 1 1 I AWA2 .  I BWB2 . 2 WA 2 WB

I A  IB

WA  WB

KA W  A  KA WB

KA 1 KB

1 1 mV 2  Iw2  mgh 2 2 1 1 1 mV 2  . mR 2 . w2  mgh 2 2 5 1 2 1 2 V  V  gh  V  10m / s 2 5

Velocity of  = resultant of the two velocities Vcm and wr Similarly velocity of P is also resultant of wr and Vcm But at pt. P the angle between the two vectors is greater than that at  

V  VP

We can’t compare VC with VP V  VC 

VC  V  VP

Rotational Motion 1.(B)

F0 

Gm 2 4r

2

2.(A)

F 

3.(B)

v

mv 2 r Gm centre of mass is at rest 4R

1 R2 acc due to gravity =

For the planet g  

G  27 M 

 3R  2

4.(A)

g at equator will increase.

5.(C)

d  g  (at dept d) = g 1    R

GM R2

 g

 3g

 g  is equal at both points

2h   g  (at height h) = g 1   R 

6.(A) 8.(B)

HWT - 9

 d  2h

7.(C) K.Eremaining

2GM  R

Vc  2 gR 

2G 10 M  R / 10

2GM  10 = 110 km/s R

1 1  mV 2   mVe2 2 2

1 1 1 mVr2  mV p2  mVe2 2 2 2

9Vc2  V p2  Vc2 

VMC/Rotational Motion

V p  2 2VC  2.828 VC  31.7 km / s

30

HWT-Solutions/Physics

Vidyamandir Classes 9.(A)

10.(D)

T 

4 2 3 r GM

T 

4 2 3 r1 GM

K .E. 

1GMm 2 r

P.E. 

Tgeo  24hr. T3

T1  T2

3.53

T2 

2 2

24 2 2

h  6 2h

GMm r 1 GMm  E 2 r

Energy required =

Rotational Motion 1.(C)

2.(A)

3.(A)

F 

2Gm

F 

3

r2

5.(C)

3

cos 30 (towards centroid)

Gm 2 r2

mV 2 Gm 2   3 2 r r GM g g    2 3  n  1 R  n  12 g 

GM r

4.(C)

L

Distance of centroid from vertices = 2

2

GM

g 1  9 9

GM R2

g GM  2 r 2R2 Gm 4Gm  ; x2  r  x 2

g 

2

HWT - 10

Potential =

2Gm  r

V 

3GM L

g 2   n  1 g

 r  3R

r

2R

 r  x 2

 4 x2

height from earths surface = 2R

h

2 1 R

x  r/3

Gm G  4m  9Gm   r / 3  2 r / 3 r 2G  8M 

2GM  2 = 2  11  22km / s R

6.(C)

Ve 

7.(D)

E 

8.(B)

T 2  r3

9.(C)

The planet sweeps equal area in equal time speed will be maximum at A 

10.(A)

V 

2R

1GMm 1  GMm  GMm    2 3R 2  2R  12 R

1 r

V1  V2 

VMC/Rotational Motion

T1  T2

R 4R

velocity

r13 r23

 2 R 3  4 R 3

1 8

V1 1  V2 2 V 1  3V 2

31

3 V  V 2

HWT-Solutions/Physics