Title 31.ROtational motion hard Ans.pdf ✅

1.(c) Let carbon atom is at the origin and the oxygen atom is placed at x-axis m1 = 12 , m2 = 16 , r i j r i j ˆ 0 ˆ and 1.1 ˆ 0 ˆ 1 = 0 + 2 = + → → 1 2 1 1 2 2 m m m r m r r + + = → → → i ˆ 28 16  1.1 = r i ˆ = 0.63 → i.e. 0.63 Å from carbon atom. 2.(a) Velocity of centre of mass 1 2 3 1 1 2 2 3 3 m m m m v m v m v vcm + + + + = 100 ˆ 50 10 ˆ 30 10 ˆ 20  10i +  j +  k = i j k ˆ 5 ˆ 3 ˆ = 2 + + . 3.(b) Let corner A of square ABCD is at the origin and the mass 8 kg is placed at this corner (given in problem) Diagonal of square d = a 2 = 80 cm  a = 40 2cm 8 , 1 m = kg 2 , 2 m = kg 4 , 3 m = kg m4 = 2kg Let 1 2 3 4 r , r , r , r are the position vectors of respective masses r i j ˆ 0 ˆ 0 1 = + , r ai j ˆ 0 ˆ 2 = + , r ai aj ˆ ˆ 3 = + , r i aj ˆ ˆ 4 = 0 + From the formula of centre of mass 1 2 3 4 1 1 2 2 3 3 4 4 m m m m m r m r m r m r r + + + + + + = i j ˆ = 15 2 + 15 2  co-ordinates of centre of mass = (15 2,15 2) and co- ordination of the corner = (0, 0) From the formula of distance between two points ( , ) 1 1 x y and ( , ) 2 2 x y distance 2 2 1 2 2 1 = (x − x ) +(y − y ) = 2 2 (15 2 − 0) + (15 2 − 0) = 900 = 30cm 4.(c) m1 = 7gm , m 2 = 4 gm , m3 = 10 gm and ), (2 5 7 ), ˆ 3 ˆ 5 ˆ ( 1 2 r = i + j − k r = i + j + k ) ˆ ˆ 3 ˆ r3 = (3i + j −k Position vector of center mass 7 4 10 ) ˆ ˆ 3 ˆ ) 10(3 ˆ 7 ˆ 5 ˆ ) 4(2 ˆ 3 ˆ 5 ˆ 7( + + + − + + + + + − = i j k i j k i j k r 21 ) ˆ 3 ˆ 85 ˆ (45i + j − k =  r i j k ˆ 7 1 ˆ 21 85 ˆ 7 15 = + − . So coordinates of centre of mass       − 7 1 , 21 85 , 7 15 . 5.(b) We know that second's hand completes its revolution (2) in 60 sec  rad sec t / 60 30  2   = = = 6.(d) Angular acceleration () = rate of change of angular speed t 2 (n n ) 2 − 1 =  10 60 4500 1200 2       − =  2 2 360 10 60 3300 2 sec degree   =  2 = 1980 degree / sec . 7.(d) Angular acceleration (at bt ct ) dt d dt d 2 3 2 2 2 2 = + +   = = 2b + 6ct 8.(a) Distance covered by wheel in 1 rotation = 2r = D (Where D= 2r = diameter of wheel)  Distance covered in 2000 rotation = 2000 D = m 3 9.5 10 (given)  D = 1.5 meter 9.(d) Angular acceleration 2 1 2 12 / 5 60 0 rad sec t = − = − =    Now from 2 1 2 1  =  t +  t = (12)(5) 150 . 2 1 0 2 + = rad 10. (c) Angular displacement in first one second 2 (1) 2 1 2 1   =  = ......(i) [From 2 1 2 1  =  t +  t ] y B 8kg x 2kg 2kg 4kg D C A (0, 0) (a, 0) (0, a) (a, a) 402 C.M. y m1 m2 C O x
Now again we will consider motion from the rest and angular displacement in total two seconds    (2) 2 2 1 2 1 + 2 = = ......(ii) Solving (i) and (ii) we get 2 1   = and 2 3 2   =  3 1 2 =   . 11. (d) Number of revolution = Area between the graph and time axis = Area of trapezium = (2.5 5) 3000 2 1  +  = 11250 revolution. 12. (c) Linear displacement (S) = Radius (r) × Angular displacement () S  r (if  = constant) = Distance travelled by mass ( ) Distance travelled by mass ( ) B y A x Radius of pulley concerned with mass (2 ) Radius of pulley concerned with mass ( ) B r A r 2 1 =  y = 2x . 13. (a) v = r = ) ˆ 2 ˆ ˆ ) (0 ˆ 0 ˆ 4 ˆ (3i + j + k  i + j + k i j k i j k ˆ 3 ˆ 6 ˆ 8 0 1 2 3 4 0 ˆ ˆ ˆ = = − + 14. (b) We will not consider the moment of inertia of disc because it doesn't have any mass so moment of inertia of five particle system 2 2 I = 5 mr = 5  2 (0.1) 2 = 0.1 kg-m . 15. (a) Moment of Inertia of disc I = 2 2 1 MR 2 2 ( ) 2 1 = R t R 4 2 1 = t R [As M = V   = R t 2 where t = thickness,  = density]  4         = x y x y x y R R t t I I [If  = constant]  (4) 64 4 1 4 = = x y I I [Given Ry = 4Rx , 4 x y t t = ]  y x I = 64 I 16. (b) Moment of inertia of disc about a diameter = MR = I 2 4 1 (given)  MR 4I 2 = Now moment of inertia of disc about an axis perpendicular to its plane and passing through a point on its rim = MR (4 I) 6I 2 3 2 3 2 = = . 17. (a) Moment of inertia of rod AB about point 2 12 1 P = Ml M.I. of rod AB about point O 2 2 12 2       = + l M Ml 2 3 1 = Ml [by the theorem of parallel axis] and the system consists of 4 rods of similar type so by the symmetry 2 3 4 I Ml System = . 18. (b) M.I of system about YY ' 1 2 3 I = I + I + I where I1 = moment of inertia of ring about diameter, I2 = I3 = M.I. of inertia of ring about a tangent in a plane  2 2 2 2 3 2 3 2 1 I = mR + mR + mR 2 2 7 = mR 19. (a) Let Z I is the moment of inertia of square plate about the axis which is passing through the centre and perpendicular to the plane. Z AB A'B' CD C'D' I = I + I = I + I [By the theorem of perpendicular axis] Z 2 AB 2 A'B' 2 CD 2 C'D' I = I = I = I = I [As AB, A' B' and CD, C' D' are symmetric axis] Hence I I l CD = AB = 20. (a) Moment of inertia of the system about z-axis can be find out by calculating the moment of inertia of individual rod about z-axis  A B A B C D C D
3 2 1 2 ML I = I = because z-axis is the edge of rod 1 and 2 and I3 = 0 because rod in lying on z-axis  system 1 2 3 I = I + I + I 0 3 3 2 2 = + + ML ML 3 2 2 ML = . 21. (c) The moment of inertia of system about AB side of triangle A B C I = I + I + I 2 = 0 + 0 + mx 2 2 3         = a m 2 4 3 = ma 22. (c) Moment of inertia of rod AB about its centre and perpendicular to the length = 12 2 ml = I  ml 12I 2 = Now moment of inertia of the rod about the axis which is passing through O and perpendicular to the plane of hexagon Irod= 2 2 12 mx ml + [From the theorem of parallel axes] 6 5 2 3 12 2 2 2 ml m l ml =         = + Now the moment of inertia of system Isystem = 6 5 6 6 2 rod ml  I =  2 = 5ml Isystem = 5 (12 I) = 60 I [As ml 12I 2 = ] 23. (b) If 1 r and 2 r are the respective distances of particles m1 and m2 from the centre of mass then 1 1 2 2 m r = m r  1  x = 35.5 (L − x)  x = 35.5 (1 − x)  x = 0.973 Å and L − x = 0.027 Å Moment of inertia of the system about centre of mass 2 2 2 1 I = m x + m (L − x) 2 2 I = 1amu (0.973 Å) + 35.5 amu (0.027 Å) Substituting 1 a.m.u. = 1.67  10–27 kg and 1 Å = 10–10 mI 47 2 1.62 10 kgm − =  24. (c) Since the circular frame is massless so we will consider moment of inertia of four masses only. 2 2 2 2 2 I = ma + 2ma + 3ma + 2ma = 8ma .....(i) Now from the definition of radius of gyration 2 I = 8mk ....(ii) comparing (i) and (ii) radius of gyration k = a . 25. (b) M. I. of sphere A about its diameter 2 ' 5 2 IO = Mr Now M.I. of sphere A about an axis perpendicular to the plane of square and passing through its centre will be 2 ' 2         = + R IO IO M 5 2 2 2 2 MR = Mr + [by the theorem of parallel axis] Moment of inertia of system (i.e. four sphere)= O 4I         = + 5 2 2 4 2 2 MR Mr   2 2 4 5 5 2 = M r + R 26. (c) Moment of inertia of sphere about it diameter 2 5 2 I = MR 3 2 3 4 5 2 R R      =   [As M = V =   3 3 4 R ] O O R / 2 A B D C H m1 Cl m2 C.M. x L – x A B O x l l l B C m A a a x a m m x z y 3 2 1
I =   5 15 8 R   5 5 105 176 15 7 8 22 R = R   = 27. (c) Moment of inertia of circular disc about an axis passing through centre and normal to the circular face 2 2 1 I = MR         =  t M M 2 1 [As M = V  R t 2 =   t  M R = 2 ]   t M I 2 2 = or  1 I  If mass and thickness are constant. So, in the problem A B B A d d I I = A B I  I [As d A  d B ] 28. (b) F i j k)N ˆ 2 ˆ 4 ˆ = (2 − + and ) ˆ r = (3i + 2 − 4k meter Torque  = r  F 2 4 2 3 2 4 ˆ ˆ ˆ − = − i j k  i j k ˆ 16 ˆ 14 ˆ  = −12 − − and 2 2 2 |  | = (−12) + (−14) + (−16) = 24.4 N-m 29. (c) By taking moment of forces about point R, 5  PR − 3  RQ = 0  PR RQ 5 3 = . 30. (d) Let the mass of the rod is M  Weight (W) = Mg Initially for the equilibrium F + F = Mg  F = Mg / 2 When one man withdraws, the torque on the rod 2 l  = I = Mg  3 2 2 l Mg Ml  = [As I = Ml 2 / 3]  Angular acceleration l g 2 3  = and linear acceleration 4 3 2 l g a =  = Now if the new normal force at A is F' then Mg − F' = Ma  4 3 ' Mg F = Mg − Ma = Mg − 4 Mg = 4 W = . 31. (a) Initial angular momentum of the system about point O = Linear momentum × Perpendicular distance of linear momentum from the axis of rotation =       2 L Mv ....(i) Final angular momentum of the system about point O = I1 + I2 = (I1 + I2 )                 +      = 2 2 2 2 L M L M ....(ii) Applying the law of conservation of angular momentum   2 2 2 2        =      L M L Mv  L v  = 32. (a) Initial angular momentum of ring   2 = I = MR If four object each of mass m, and kept gently to the opposite ends of two perpendicular diameters of the ring then final angular momentum = ( 4 ) ' 2 2 MR + mR  By the conservation of angular momentum Initial angular momentum = Final angular momentum ( 4 ) ' 2 2 2 MR  = MR + mR          + = M m M 4 ' . 33. (b) The angular momentum (L) of the system is conserved i.e. L = I = constant When the tortoise walks along a chord, it first moves closer to the centre and then away from the centre. Hence, M.I. first decreases and then increases. As a result,  will first increase and then decrease. Also the change in  will be non-linear function of time. 34. (a) =  = → → → L r p 3 4 2 1 2 1 ˆ ˆ ˆ − − i j k i j k j k ˆ 2 ˆ ˆ 2 ˆ ˆ = 0 − − = − − and the X- axis is given by i j k ˆ 0 ˆ + 0 + Dot product of these two vectors is zero i.e. angular momentum is perpendicular to X-axis. 35. (c) Mg A F  B B  Mg A F F B