Tiêu đề 31. Rotational Motion Med 1 Ans.pdf ✅

1. (c) The axis of a spinning top moves around the vertical through its point of contact with the ground sweeping out a cone as shown in the figure. This movement of the axis of the top around the vertical is known as precession. 2. (d) According to the definition of centre of mass, we can imagine one particle of mass (1+2+3) kg at (1,2,3) ; another particle of mass (2 + 3 ) kg at (-1, 3, -2). Let the third particle of mass 5 kg put at ( 3 3 3 x , y ,z ) i.e., m 6kg,(x , y , z ) (1,2,3) 1 = 1 1 1 = m 5kg,(x , y , z ) ( 1,3, 2) 2 = 2 2 2 = − − m 5kg,(x , y ,z ) 3 = 3 3 3 =? Given, (X ,Y ,Z ) (1,2,3) CM CM CM = Using 1 2 3 1 1 2 2 3 3 CM m m m m x m x m x X + + + + = 6 5 5 6 1 5 ( 1) 5x 1 3 + +  +  − + = 5x3 =16−1=15or x3 = 3 Similarly, y3 =1 and z3 = 3 3. (c) As the mass density of the cube is uniform therefore its centre of mass lies at its geometric centre. 4. (c) Reduced mass 3 2m m 2m m 2m = +  = 5. (a) Choosing the x and y axes as shown in the figure. The coordinates of the vertices of the L – shaped lamina is as shown in the figure. Divide the L-shaped lamina into three squares each of side 1m and mass 1 kg (  the lamina is uniform) by symmetry, the centres of mass 1 C2 C , and C3 of the squares are their geometric centres and have coordinates                   2 3 , 2 1 and C 2 1 , 2 3 ,C 2 1 , 2 1 C1 2 3 respectively. The coordinates of the centre of mass of the L-shaped lamina is 1 2 3 1 1 2 2 3 3 CM m m m m x m x m x X + + + + = m 6 5 1 1 1 2 1 1 2 3 1 2 1 1 = + +  +  +  = 1 2 3 1 1 2 2 3 3 CM m m m m y m y m y Y + + + + = m 6 5 1 1 1 2 3 1 2 1 1 2 1 1 = + +  +  +  = 6. (a) Refer figure. The distances of centre of mass CM form masses m1 and m2 are 1 2 2 1 1 2 1 2 1 2 2 1 m m d d m m m d and d m m m d d  = + = + = 7. (d) May lie within or outside the body. 8. (b) The forces leading to the explosion are internal force. They contribute noting to the motions of the centre of mass. The total external force, namely the force of gravity acting on the shell is the same before and after the explosion. Therefore, the centre of mass of the fragments of the shell continues along the same of the fragments of the shell continues along the same parabolic as it would have followed. If there were no explosion as shown in figures 9. (d) The speed of the centre of mass of the system remains unchanged (equal to v) because no external forces acts on the system. The force involved in running on the trolley are internal to this system. 10. (b) The relation between linear velocity v and angular velocity  is v = r 11. (d) The direction of the angular velocity is along the axis of rotations. 12. (a) Here, k ˆ j z ˆ i y ˆ r = x + + P K ˆ j ˆ i P ˆ P = Px + y + z Let L K ˆ j L ˆ i L L ˆ = x + y + z ...... (i) As L r p  = 
k(xp yp ) ˆ j(zp xp ) ˆ i(yp zp ) ˆ = Z − y + x − Z + y − x .(ii) Comparing the coefficients of j ˆ i, ˆ and k ˆ in Eqs. (i) and (ii), we get x Z Y L = yp − zp y x xpz L = zp − Lz = xpy − ypx 13. (b) Remains constant 14. (d) As L = r  p Differentiate both sides with respect to time, we get (r p) dt d dt dL =  dt dp p r dt dr =  +             =    p = 0 dt dr dt dp r  0 dt dp r dt dL −  = 15. (a) Here, k ˆ j ˆ i ˆ r = − + k ˆ j 5 ˆ i 3 ˆ F = 7 + − Torque,  = r F k(3 ( 7)) ˆ j(7 ( 5)) ˆ i(5 3) ˆ 7 3 5 1 1 1 k ˆ j ˆ i ˆ = − + − − + − − −  = − or k ˆ j 10 ˆ i 12 ˆ  = 2 + + 16. (b) A couple produces rotation without translation. 17. (d) When the total external torque acting on the system is zero, then the total angular momentum of the system is conserved. 18. (d) Moment of a couple depends on the point about which we take the moment. 19. (c) Moment of couple is called torque. 20. (a) Torque,  = rF  Power associated with the torque is P = . = (r  F.) 21. (a) Here, k ˆ j 2 ˆ i 4 ˆ k,p 3 ˆ j ˆ i 2 ˆ r = + − = + −  3 4 2 1 2 1 k ˆ j ˆ i ˆ L r P − =  = − k ˆ j 2 ˆ i 1 ˆ k(4 6) 0 ˆ j( 3 2) ˆ i( 4 4) ˆ = − + + − + + − = − − L. has components along – y axis and –z axis but it has no component along in the x-axis. The angular momentum L. is in yz-plane. i.e., perpendicular to x – axis. 22. (d) When the total external force acting on the system is zero, the velocity of centre of mass remains constant. 23. (b) Torque is always perpendicular to F as well as r. r. = 0 as well as F. = 0 24. (c) AB is the rod, K1 and K2 are the two knife edges. Since the rod is uniform, therefore its weight acts at its centre of gravity G. Let R1 and R2 be reactions at the knife edges. For the translational equilibrium of the rod. R1 + R2 − 60N − 40N = 0 R1 + R2 = 60N + 40N =100N .....(i) For the rotational equilibrium, takings moments about G, we get −R1 (40) + 60(20) + R2 (40) = 0 30N 40 1200 R1 − R2 = = ..... (ii) Adding (i) and (ii), we get 2R1 =130N or R1 = 65N Substituting this value in Eq. (i), We get R2 = 35N 25. (b) Let T1 and T2 be the tensions in two strings as shown in the figures. For translations equilibrium along the horizontal directions, we get
1 1 2 2 T sin = T sin ..... (i) For rotational equilibrium about G − T1 cos1d + T2 cos2 (L − d) = 0 T cos d T cos (L d) 1 1 = 2 2 − Dividing equations (i) by (ii), we get 1 1 2 2 tan tan d L d or (L d) tan d tan   = − −  =          +    =    =      − 1 tan tan d L tan tan 1 d L 1 2 1 2 1 2 1 tan tan tan d L   +  =          +   = 1 2 1 tan tan tan d L 26. (a) Statements (1) and (4) are true. 27. (c) A rigid body is said to be in partial equilibrium when it is in translational equilibrium but not in rotational equilibrium or when it is in rotational equilibrium but not in translational equilibrium. 28. (c) From figure, moment of force F about A, F a, 1 =  anticlockwise. Moment of weight Mg of cube about A. , 2 a Mg 2 =  Clockwise. The cube will not exhibit any motions, If 1 2  =  Or 2 Mg orF 2 a F a = Mg   The cube will rotate only, when 1 2    2 Mg orF 2 a F a  Mg  If we assume that normal reactions is effective at a/3 from A, then block would turn if . 3 Mg F a orf 3 a Mg  =  = When , 3 Mg 4 Mg F =  there will be no motion. Hence, we conclude A − q;B− r;C− p;D −s. 29. (c) Mechanical advantage Effort Load = Load arm × Load = Effort arm × Effort 30. (b) Let m be the mas of the metre stick concentrated at C, the 50 cm mark as shown in the figure. In equilibrium, taking moments of forces about C’ we get 10g (45-12) = mg (50-45) 10 g × 33 = mg × 5 66g 5 10 33 m =  = 31. (a) Let AB be ladder  AB = 3m Its foot A is a distance 1m from the wall.  AC = 1m And BC (AB) (AC) (3) (1) 2 2m 2 2 2 2 = − = − = The various force acting on the ladder are (i) Weight W acting at its center of gravity G. (ii) Reactions force R1 of the wall acting Perpendicular to the wall (  the wall is frictionless). (iii) Reactions force R2 of the floor. This force can be resolved into two components, the normal reactions N and the force of frictions f. For translator equilibrium in the horizontal directions, 1 R1 f − R = 0 orf = .....(i) For translator equilibrium in the vertical directions, N – W = 20g = 20 × 10 = 200 N ..... (ii) For rotational equilibrium, taking moment of the forces about A, we get ( ) 0 2 1 R1 2 2 W  =      − 25 2N 4 2 200 4 2 W R1 = = = ..... (iii) From (ii), f = R1 = 25 2N R N f (200N) (25 2N) 203N 2 2 2 2 2 = + = + = 32. (a) By sign convention anticlockwise moment (or torque) is taken as positive while clockwise moment (or torque) is taken as negative. In case I, net torque about its centre is  = FL + FL = 2FL
In case II, net torque about its centre is FL 2 5 F L 2 1  = FL + F +  = In case III, net torque about its centre is F L 0 2 L F 2 L  = −FL + F −  +  = 33. (a) Analogue of mass in rotational motions is moment of inertia. 34. (a) When the person withdraws his hands to his chest. His moment of inertia decreases. No external torque is acting on the system. So to conserve angular momentum, the angular velocity increases. 35. (d) Moment of inertia of a body depends on the mass of the body, its shape and size, distribution of mass about the axis of rotation, and the position and orientations of the axis of rotation. 36. (d) Moment of inertia of the solid sphere of mass M and radius R about any diameter is 2 diamete MR 5 2 I = According to theorem of parallel axes 2 Itan gent = Idiameter + MR , MR2 ILi' kZT; k = IO;kl + 2 2 2 MR 5 7 MR MR 5 2 = + = 37. (c) Moment of inertia of a the solid cylinder about its axis is 2 MR I 2 =  Rotational kinetic energy of the cylinder is 2 2 2 2 2 MR 4 1 2 MR 2 1 I 2 1 K =  =  =  38. (d) For translations equilibrium, F = 0A – R For rotational equilibrium,  = 0 B – S Moment of inertial of a body = 2 MK C – P Torque is required to produce angular accelerations. D – q 39. (b) The moment of inertia of the system about the given axis is. 2 ML 4 ML 4 ML 2 L M 2 L I M 2 2 2 2 2  = + =       +      = 40. (c): Moment of inertia ofrodof mass Mand length l about its axis passing through one of its ends andperpendicular to it is. 2 Ml 3 1 I = As: Mk2 I = where k is the radius of the gyration 2 2 Ml 3 1 Mk = or 3 1 k = 41. (b): As z axis is parallel to z’ axisand distance between them 2 a 2 a 2 = = By the theorem of parallel axes, 2 ma I 2 a I' I m 2 z 2 z z  = +      = + Hence, option (a) is correct. z and z” are notj parallel axes. Hence, option (b) is incorrect. By symmetry, x y' I = I hence, option (c) is correct 42. (c): Let M and R be the mass and radius respectively. Moment ofinertia of a ring about any of its diameter is. 2 2 1 Iring = MR Moment of inertia of a disc about any of its diameter is 2 4 1 Idisc = MR Moment of inertia of a hollow sphere about any of its diameter is. 2 3 2 Ihollowsphere = MR Moment of inertia of a solid sphere about any of its diameter is. 2 5 2 Isolidsphere = MR Thus, hollow sphere I is largest.