Title 6-rotational-motion-.pdf ✅

Rotational Motion 1. (AD) L ̅ = m(r⃗ × v⃗) L ̅ tot = L ̅ 1 + L ̅ 2 = mv(d2 − d1 ) ⊗ 2. (D) IB = 1 2 mr 2 and IA about the given axis by parallel axis theorem = 1 2 mr 2 + m(2r) 2 = 9 2 mr 2 ⇒ I = 1 2 mr 2 + 9 2 mr 2 = 5mr 2 3. (B) Angular momentum conservation about combined COM mv R 2 = (m ( R 2 ) 2 + mR 2 + m ( R 2 ) 2 ) w w = V 3R 4. (B) I1 about axis = I2 about axis I of ring about axis = I1 + I2 = 2I1 = 2I2 ⇒ I = I = 1 2 × I ring about axis (Diameter) = 1 2 × ( 1 2 × (2M)R 2) = 1 2 MR 2 ⇒ independent of θ. 5. (ABCD) In all the cases net torque about axis is zero. 6. (A) m′ = 9m R2 × ( R 3 ) 2 = m I0 = 9m R 2 2 − ( m 2 ( R 3 ) 2 + m ( 2R 3 ) 2 ) = 4mR 2 7. (BC) τ⃗ = (r⃗ × F⃗) Greater the distance from axis more is torque 8. (abc) Ix = 2 {m ( l 2 ) 2 + ml 2 12 } = Iy Iz = Ix + Iy Iside = Ix + 4m ( l 2 ) 2 = 5 3 ml 2 9. (BC) Let F is the force applied to the rope and assuming the cylinder rolls without siding. ⇒ F − f = ma fR2 − FR1 = mk 2α a = R2α N = mf | ⇒ a = F(1 − R1/R2 ) m(1 + K2/R2 2 ) And to avoid sliding f ≤ Nμ ⇒ F[K 2+R1R2] K2+R2 2 ≤ mgμS
F ≤ mgμS [K 2 + R2 2 ] K2 + R1R2 ⇒ F ≤ 50 × 10 × 0.1[0.3 2 + 0.6 2 ] 0.3 2 + (0.3)(0.6) ⇒ F ≤ 250 3 N But applied force is 200N ⇒ rolling with sliding and f = fK = NμK = mgμK = 50 × 10 × 0.08 = 40N [ α = FR1 − fR2 mk 2 clockwise = 200(0.3) − 40(0.6) 50(0.3) 2 = 8rad/s 2 ] a = F − f m towards left 200 − 40 50 = 3.2 m/s 2 10. (BD) By parallel axis theorem Iz ′ = Iz + ma 2 2 Perpendicular axis theorem can't be applied for 3d objects Ix = Iy by symmetry. Ix = Iy by symmetry. 11. (C) L = muxH = m u √2 u 2 /2 2g = mu 3 4√2g 12. (A) ⇒ Nμ = ma ⇒ NμR = 2 5 mR 2α ⇒ a = gμ α = 5gμ 2R ⇒ ω = 5gμ 2R t and v + Rω = v0 ⇒ gμt + 5 2 gμt = v0 ⇒ t = 2v0 7gμ s = 1 2 at 2 = 1 2 gμ 4v0 2 49g2μ2 = 2v0 2 49gμ ⇒ s = 2v0 2 49g(2/7) = v0 2 7g 13. (D) Tx = m l (l − x)w 2 (x + l − x 2 ) = mw 2 2l (l 2 − x 2 ) 14. (BCD) Net force on the particle in uniform circular motion passes through the centre of the circular motion. Angular momentum about O is perpendicular to the plane in which string and velocity vector of the particle lie i.e. its makes angle θ with horizontal and horizontal component remains constant in magnitude but vertical component remains constant.
15. (A) I = mv(3R)(−kˆ ) + 1 2 mR 2ωkˆ and V = Rω = − 5 2 mR 2ωkˆ 16. (D) τOf centripetal force is zero only about center of circle 17. (AD) vres = √v 2 + u 2 + 2u 2cos θ = 2ucos θ 2 18. (B) Angular momentum of the system is conserved. Ltor + Lplate = constant Moment of inertia of tortoise first increases and then decreases 19. (C) Moment of inertia, l = 1 2 MR 2 + mx 2 where m = mass of insect, and x = distance of insect from centre. Clearly, as the insect moves along the diameter of the disc. Moment of inertia first decreases momentum and then decreases, angular speed first increases, then decreases. 20. (A) Linear velocity of P is v + ⇒ linear acceleration rω = d dt (v + rω) = a + rα Angular velocity = ω ⇒ radial acceleration = ω 2 r 21. (AB) ω = V+3V 2r = 2V r also ω = V + V0 r ⇒ V + V0 r = 2V r ⇒ V0 = V 22. (D) Let J : impulse generated in the string ⇒ J = 1 2 mr 2ω ... for pulley For block: − J = mv2 − mv ... and v2 = rω Where v2 is velocity of the block just after the string becomes taut. ⇒ 0 = mv2 2 + mv2 − mv ⇒ v2 = 2v 3 = 10/3 m/s 23. (A) mg − N = mV 2 R and N = 0 ⇒ V = √Rg 24. (C) Form the principle of conservation of angular momentum, we have Σ/ω = constant where, l is moment law of angular momentum, By conservation law of angular momentum, about ground mr 2ω0 = mvr + mr 2 × v r ⇒ v = ω0r 2
25. (A) Since F will produces as anticlockwise torque. So it will cause slipping and hence friction will act towards left. And so will accelerate in left direction. 26. (d) F − f = ma fR = ICα ⋯ . (i) a = αR ⋯ (ii) ⇒ F = ma + ICα R ⇒ a = 5F 7M ⇒ f = 2 7 F ⋯ (iii) f ≤ fl ⇒ 2 7 F ≤ μmg F ≤ 7 2 μmg Fmax = 7 2 μmg 27. (C) Net angular momentum will be as shown in figure whose value will be constant (|L| = mvl). But its direction will change as shown in the figure. 28. (BD) K1 = 1 2 ( 1 2 mr 2 ) ω1 2 = 1 4 mr 2ω1 2 L = m(r × v) and 1 2 mr 2ω1 = ( 1 2 mr 2 + mr 2 ) ω2 ... . (ΔL = 0) ⇒ ω2 = ω1/3 ⇒ K2 = 1 2 { 1 2 mr 2 + mr 2 } ω2 2 = 1 2 mr 2ω1 2 = K1/3 ⇒ Energy lost = K1 − K2 = 2K1/3 29. (A) Given, force F = 20t − 5t 2 α = FR I = (20t − 5t 2 )2 10 = 4t − t 2 ⇒ dω dt = 4t − t 2 ⇒ ∫0 ω dω = ∫0 t (4t − t 2 )dt ⇒ ω = 2t 2 − t 3 3 When direction is reversed, ω = 0, i.e., t = 0,6 s Now, dθ = ωdt ⇒ ∫0 6 dθ = ∫0 6 (2t 2 − t 3 3 ) dt ⇒ θ = [ 2t 3 3 − t 4 12] 0 6 ⇒ θ = 144 − 108 = 36rad ∴ Number of rotations, n = θ 2π = 36 2π < 6 30. (A) For the mass m, mg − T = ma As we know, a = Rα So, mg − T = mRα As torque = T × R = mR 2 2 α From Equations (i) and (ii), we get : Hence, the acceleration body is 2g 3 31. (B) Moment of inertia of the system about the axis of rotation (through point P ) is I = m1x 2 + m2(L − x) 2 . By work energy theorem : work done to set the rod rotating with angular velocity ω0 = increase in rotational kinetic energy W = 1 2 Iω0 2 = 1 2 [m1x 2 + m2(L − x) 2 ]ω0 2 . For W to be minimum, dW dx = 0 i.e. 1 2 [2m1x + 2m2(L − x)(−1)]ω0 2 = 0 or m1x − m2(L − x) = 0 or (m1 + m2 )x = m1L