Physics-2.UNIT-6-Solution of Assignment

SOLUTION OF ASSIGNMENT UNIT - 06 1. (B) Angular momentum = (momentum) × (perpendicular distance of the line of action of momentum from the axis of rotation) Angular momentum about O … (i) Now, [ θ = 45°] ... (ii) From (i) and (ii) Also from (i) and (ii) 2. (A) Given that ⇒ vr = 4.4 × 1015 m2/s ⇒ ⇒ πr2 = 3. (A) Angular momentum. For an elliptical motion it can be shown that …….(i) Where = areal velocity. If an object moves under the action of a purely radial force then and are anti parallel and τ = 0. Since τ = . And τ = 0 ∴ ∴ L = consst. From eq. (1) 4. (D) . According to theorem of parallel axes. 5. (B) Taking moment of force about D ∴ 6. (C) Net torque about O = F × R FR = Iα ⇒ α = 7. (A) vel. of B along AB = vel. of A along AB. ⇒ v cos30 = ⇒ v = 8. (C) Consider F.B.D of Disc F = MA ⇒ , N = mg …..(2) F.B.D of disc from centre of disc. Net torque on disc about O. = FR. and FR = Iα ⇒ FR = ⇒ 9. (A) F – fr = MA .... (1) ⇒ fr R – rF = Iα .... (2) A = Rα .... (3) By soling given equation we will find A, fr and α. (Q.10-13) Fr = μmg μmg = mA ⇒ A = μg (10 : B) ....(1) FrR = μmg R = α = (11 : C) ....(2) Let at t = t, rolling starts. ⇒ V = V0 – A t = Vo – μ gt ....(3) ω = ω0 + αt = ....(4) At the time of rolling. V = Rω ⇒ Vo – μgt = 2μgt 2g t = V = Vo – μgt = Vo – (12 : D) ω = So at the time of rolling , (13 : A) 14. (C) If the system is remain in equilibrium then all torque on spool will balance and hence line of action of F must passes through c. Consider 15. (D) Friction at CN will act on backward direction ⇒ N + F sinθ – mg . . . (i) Fcosθ = fr . . . (ii) fr = μN . . . (iii) By solving (i), (ii) and (iii) we will get 16. (D) A = Rα . . . (i) Equation of torque about ⇒ f, R – Fr = I0α . . . (ii) Fcosθ – fr = MA . . . (iii) By solving (i), (ii) and (iii) we will get backward 17. (B) at θ = 180°, the figure will as follows Let rolling is taking place. So equation of torque about C. ⇒ F (R + r) = IC α …..(i) ….(ii) MA = net force in direction of F = So f r = in forward f r max = μN = f r required > f r max So f r acting = f r max = 18. (B) At θ = π / 2 N + Mg = Mg ⇒ N = 0 f r max = 0 So no horizontal force will act so horizontal acceleration = 0 f r = 0 and torque about O = Iα ⇒ Mg r = Iα ⇒ 19. (B) Acceleration = = 20. (A) Fr max = μN = and Fr sequence = Fr max Hence will be 21. (C) The applied force shifts the normal reaction to one corner as shown. At this situation, the cubical block starts topping about O. Taking torque about O F × L = mg × ⇒ 22. (A) Change in angular momentum of the system = Angular Impulse given to the system (Angular momentum)f – (Angular momentum)I = Let the system start rotating with the angular velocity ω. Moment of Inertia of the system about its axis of rotation [centre of mass of the system] From (i) ⇒ 23. (A) Given that ⇒ From cross–product rule, is always perpendicular to the plane containing and By the dot product definition Differentiating with respect to time ⇒ Since is perpendicular to ⇒ = 0 ⇒ L = constant 24. (B) Since v is changing (decreasing), L is not conserved in magnitude. Since it is given that a particle is confined to rotate in a circular path, it cannot have spiral path. Since the particle has two accelerations ac and at therefore the net acceleration is not towards the centre. The direction of remains same even when the speed decreases. 25. (C) Since the objects are placed gently, therefore no external torque is acting on the system. Therefore angular momentum is constant. i.e., I1ω1 = I2ω2 Mr2 × ω1 = (Mr2 + 2mr2)ω2 ∴ 26. (C) The moment of inertia of the system about axis of rotation O is I = I1 + I2 = 0.3 x2 + 0.7 (1.4 – x)2 = 0.3x2 + 0.7 (1.96 + x2 – 2.8π) = x2 + 1.372 – 1.96 x The work done is rotating the rod is converted into its Rotational Kinetic Energy. ∴ For work done to be minimum ⇒ 2x – 1.96 = 0 ⇒ 27. (B) Angular momentum of mass m moving with a constant velocity about origin is L = momentum × perpendicular distance of line of action of momentum from origin L = mv × y In the given condition mvy is a constant. Therefore angular momentum is constant. 28. (C) As the spheres are smooth, there will be no transfer of angular momentum. Thus A, after collision will remain with its initial angular momentum. 29. (C) The disc has two types of motion namely translational and rotational. Therefore there are two types of angular momentum and the total angular momentum is the sum of theses two L = LT + LR LT = angular momentum due to translational motion. LR = angular momentum due to rotational motion about C.M. L = MV × R + Icmω Icm = M.I. about centre of mass C. = M(Rω)R + (V = R ω in case of rolling motion and surface at rest) = 30. (A) or Net torque about O is zero. Therefore, angular momentum (L) about O will be conserved, or Li = Lf