Title 31. Rotational Motion Med 2 Answer.pdf ✅

1. (c) d dt  = 1.5 + 4t  = 9.5 rad/s 2. (d) f = i + t 60 = 0 + (5)  = 12  = 1 2 at2 = 1 2 × 12 × (5)2 = 150 rad 3. (b)Linear speed V = r V depends on radius 4. (b)  = t + 1 2 t 2 = 10 rad 5. (b) 0 = 3000 rad/min 0 = 3000 60 rad/sec = (50 rad/sec) t = 10 sec f = 0 f = 0 + t 0 = 50 –  (10)  = 5 rad/sec2 0 = o t + 1 2  + 2 0 = (50) (10) + 1 2 (–10) (10)2 500 – 250 = 250 rad 6. (c)V = R V = 10 × 0.2 = 2m /sec. 7. (b)Sphere is rotating about a diameter so, a = R but, R is zero for particles on the diameter. 8. (b)I = mR2 = 4 kgm2 9. (b)M.I of both spheres about common tangent  0 = 2 2 14 2 2 mR mR 5 5   + =     mR2  0 = 7  10. (b) x +  y =  z z axes is perpendicular to plane of body. 11. (d) 12. (d) 13. (a) 14. (b) 15. (d)  = 2 MR 2 + 3 2 2 MR 2       = 7 2 MR 2 16. (b) 17. (c) 18. (b)  = 2 2 dm R MR =  19. (b) =  CM + Md2  = 2 2 ML L M 12 2   +      = 2 ML 3 20. (a)  = 2 2 2 ML ML 2ML 3 3 3 + = 21. (c)  = m 2 3a 2          = 2 3ma 4 22. (b)  = 2 2 2 ML L ML M 12 6 9   + =     23. (b) 2 1 1 1 1 2 2 2 1 2 m R m 4 2 m R m 1 1  = =  =  1 2 m m = 1 2
24. (a)  = 4 2 2 ML L 4 2 M ML 12 2 3       + =         25. (a)  = 2 2 2 R 2 R 2 M M M(2R) 5 2 5 2       + +                = 21 5 MR2 26. (d)Moment of inertia of a disc about its diameter is  d = 1 2 MR 4 Now, according to perpendicular axis theorem moment of inertia of disc about a tangent passing through rim and in the plane of disc is  =  d + MR2 = 1 2 MR 4 + MR2 = 5 2 MR 4 27. (a) = 5 4 MR2 ’ = 3 2 MR2 = 6 5  28. (c)  = 7 5 MR2  MK2 = 7 5 MR2  K = 7 5 R 29. (a) = 4 × m 2 L 2       = 2 mL2 mK2 = 2 mL2 K = 2 L 30. (b) 2 ring 2 Disc MR MR 2  =  = 2 31. (a)  = 2 MR 2 K = R 2 = 3.54 cm 32. (b)Moment of inertia of solid sphere about an axis passing through its centre of gravity. r = 2 5 mr2 Where m = mass of sphere R = radius of sphere From theorem of parallel axis, moment of inertia about its tangential axis I = I’ + mR2 = 2 5 MR2 + MR2 = 7 5 MR2 33. (a)The moment of inertia in rotational motion is equivalent to mass as in linear motion. 34. (d) For disc, I = 1 2 ma2 For ring, I = ma2 For square of side 2a = M 12 [(2a)2+ (2a)2 ] = 2 3 Ma2 For square of rod of length 2a I = 4 ( ) 2 2 2a M Ma 12     +     = 16 3 Ma2 Hence, moment of inertia is maximum for square of four rods. 35. (b)The moment of inertial of the given system that contains 5 particles each of mass 2 kg on the rim of circular disc of radius 0.1 m and of negligible mass is given by = MI of disc + MI of particles Since, the mass of the disc is negligible therefore, MI of the system = MI of particles = 5 × 2 × (0.1)2 = 0.1 kg m2 36. (a)Moment of inertia of disc about a tangent and parallel to its plane,
 = 2 MR 2 MR 4 + ...(i) Moment of inertia of disc about a tangent and perpendicular to its plane, = 2 MR 3 2 2 MR MR 2 2 + = 2 perpendicular 2 5 MR 4 5 3 6 MR 2  = =    perpendicular = 6 5  37. (a)  = 2 dmR   = R2 dm  = R2 M = MR2 38. (c)IB = I A + Md2 Then I B > I A 39. (c)  0 =  1 +  2  0 = ( ) 2 m / 2 2 3       + ( ) 2 m / 2 2 3       = ( ) 2 m/ 12 40. (c)  x +  y =  z 2 x =  z   I = 2 × 200 = 400 gm cm2 . 41. (d)Perpendicular axis theorem  2 =  x +  y = 2 mr 2 from symmetry  x =  y   x = 2 mr 4 Perallel axis theorem  oo ́ =  x + mr2 = 2 mr 4 + mr2 = 5 4 mr2 42. (d)  =  1 +  2 +  3  1 =  2 = 3 2 mr2  3 = 2 mr 2   =  1 +  2 +  3 = 7 2 mr2 Moment of inertia = 3mk2 where k is radius of gyration. 3mk2 = 7 2 mr2  k = r 7 6 43. (c)Moment of Inertia about the shortest side BC is greater than the other two sides 44. (a)Moment of inertia depends on distribution of mass about axis of rotation. Density of iron is more than that of aluminium, therefore for moment of inertia to be maximum, the iron should be far away from the axis. Thus, aluminium should beat interior and iron surrounds it. 45. (a)By perpendicular axis theorem moment of inertia about any axis passing through centre and in the plane of plate will be  (by symmetry)
46. (a)We know that M.I. of a circular wire of mass M and radius R about its diameter is 2 MR 2 47. (c)If a body has mass M and radius of gyration is K, then  = MK2 Moment of inertia of a disc and circular ring about a tangential axis in their planes are respectively.  d = 2 d 5 M R 4   r = 2 r 3 MR 2 but  = MK2  K = M   d d r r r d K M K M  =   or 2 d d r 2 r d r (5 / 4)M R M 5 (3 / 2)M R M 6  =  =    d :  r = 5 : 6 48. (c)Moment of inertia of the system about AX is given by M = mA r 2 A + mB r 2 B + mC r 2 C M = m(0)2 + m() 2 + m( sin30°)2 = m 2 + 2 m 4 = 5 4 m 2 Alternative : Moment of inertia of a system about a line OC perpendicular to AB, in the plane of ABC is  CO = m× 0 + m × 2 2       + m × 2 2         CO = 2 2 2 m m m 4 4 2 + = According to parallel-axis theorem  AX =  CO + Mx2 where x = distance of AX from CO, M = total mass of system  AX = 2 2 m 3m 2 2   +      I AX = 2 2 m 3m 5 2 m 2 4 4 + = 49. (d)We should use parallel axis theorem Moment of inertia of disc passing through its centre of gravity and perpendicular to its plane is  AB = 1 2 MR 2 Using theorem of parallel axes, we have,  CD =  AB + MR2 = 1 2 2 MR MR 2 + = 3 2 MR 2 Note : The role of moment of inertia in the study of rotational motion is analogous to that of mass in study of linear motion. 50. (b) 51. (b) 52. (c)Let same mass and same outer radii of solid sphere and hollow sphere are M and R respectively. The moment the moment of inertia of hollow sphere (spherical shell) B about its diameter  A = 2 5 MR2 ...................(i) Similarly the moment of inertia of hollow sphere (spherical shell) B about its diameter  B = 2 5 MR2 ...................(ii) t is clear from eqs. (i) and (ii),  A <  B 53. (d)The mass of complete (circular) disc is