Tiêu đề 7.SYSTEM OF PARTICLES AND ROTATIONAL MOTION - Explanations.pdf ✅

2 (d) Moment of inertia of system about YY′ I = I1 + I2 + I3 = 1 2 MR 2 + 3 2 MR 2 + 3 2 MR 2 = 7 2 MR 2 3 (a) v = √ 2gh 1 + K2 R2 = √ 2gh 1 + 2 5 = √ 10 7 gh 4 (d) According to conservation of momentum 5 × 10 = (955 + 5)v v = 50 100 = 1 20 ∴ % KE lost = K1−K2 K1 × 100 = 95.5% 5 (a) M.I. of system about the axis which passing through m1 Isystem = m1 (0) 2 + m2 ( a 2 ) 2 + m3 ( a 2 ) 2 Isystem = (m2 + m3 ) a 2 4 6 (c) Frequency of wheel, v = 300 60 = 5 rps. Angle described by wheel in one rotation=2π rad Therefore, angle described by wheel in 1s = 2π × 5 rad = 10π rad. 8 (b) Moment of inertia of triangle sheet ABC about AC = 1 2 moment of inertia of square ABCD about ABC 1 2 (2M) l 2 12 = Ml 2 12 9 (b) Initially rod stand vertically its potential energy = mg l 2 When it strikes the floor, its potential energy will convert into rotational kinetic energy mg ( l 2 ) = 1 2 Iω 2 [Where, I = ml 2 3 = M.I. of rod about point A] ∴ mg ( l 2 ) = 1 2 ( ml 2 3 ) ( vB l ) 2 ⇒ vB = √3gl 10 (c) Moment of inertia of whole disc about an axis through centre of disc and perpendicular to its plane is I = 1 2 mr 2 As one quarter of disc is removed, new mass, m′ = 3 4 m ∴ I ′ = 1 2 ( 3 4 m) r 2 = 3 8 mr 2 11 (b) m1v = (m1 + m2 )v/3 3m1v = m1v + m2v 3m1v − m1v = m2v 2m1v = m2v ∴ m2 m1 = 2 12 (b) In the absence external force, position of centre of mass remain same therefore they will meet at their centre of mass 13 (b) From the theorem of parallel axis, the moment of inertia I is equal to I = ICM + Ma 2 where ICM moment of inertia is about centre of mass and a the distance of axis from centre. ∴ I = MK 2 + M × (6) 2 MK1 2 = MK 2 + 36M ⇒ K1 2 = K 2 + 36 ⇒ (10) 2 = K 2 + 36 ⇒ K 2 = 100 − 36 = 64 ⇒ K = 8 cm 14 (b) τ = dL dt = m(u 2 cos2 θ) = (1)(10) 2 cos2 45° = 50 Nm 15 (a) Y Y 1 2 3 m1 a/2 a/2 m2 m3 a a C.G. B vB l/2 A
In the absence of external torque angular momentum remains constant 16 (d) xcm = ∫ x dm ∫ dm , xcm = ∫ x k ( x L ) n dx L 0 ∫ k ( x L ) n dx L 0 = ( n + 1 n + 2 ) L 17 (c) XCM = m1x1+m2x2 m1+m2 ∴ XCM = (12×0)+(16×1.13) 12+16 = 0.6457Å 18 (c) Kinetic energy K = J 2 2I where J is angular momentum and I the moment of inertia. ∴ K1 = J 2 2I , K2 = (J+ 10 100 J) 2 2I ∴ K1 K2 = (100) 2 (110)2 = 100 121 % change = K2−K1 K1 = K2 K1 − 1 = 121 100 − 1 = 21% 19 (b) Initial position of centre of mass rcm = m1x1+m2x2 m1+m2 ...(i) If the particles of mass m1 is pushed towards the centre of mass of the system through distance d and to keep the centre of mass at the original position let second particle be displaced through distance d′ away from the centre of mass Now rcm = m1 (x1+d)+m2(x2+d ′ ) m1+m2 ...(ii) Equating (i) and (ii) m1x1 + m2x2 m1 + m2 = m1 (x1 + d) + m2 (x2 + d ′) m1 + m2 By solving d ′ = − m1 m2 d Negative sign shows that particle m2 should be displaced towards the centre of mass of the system 20 (b) Let m1 = m, m2 = 2m, m3 = 3m, m4 = 4m r⃗1 = 0î+ 0ĵ r⃗2 = a cos 60î+ a sin 60ĵ= a 2 i + a√3 2 ĵ r⃗3 = (a + a cos 60)î+ a sin60ĵ= 3 2 aî+ a√3 2 ĵ r⃗4 = aî+ 0ĵ By substituting above value in the following formula r⃗ = m1r⃗1 + m2r⃗2 + m3r⃗3 + m4r⃗4 m1 + m2 + m3 + m4 = 0.95aî+ √3 4 aĵ So the location of centre of mass [0.95a, √3 4 a] 21 (d) When a particle is projected with a speed v at 45° with the horizontal then velocity of the projectile at maximum height. v ′ = v cos 45° = v √2 Angular momentum of the projectile about the point of projection = mv′h = m v √2 h = mvh √2 22 (a) As two solid spheres are equal in masses, so mA = mB ⇒ 4 3 πRA 3ρA = 4 3 πRB 3 ρB ⇒ RA RB = ( ρB ρA ) 1⁄3 The moment of inertia of sphere about diameter I = 2 5 mR 2 ⇒ IA IB = ( RA RB ) 2 (as mA = mB) ⇒ IA IB = ( ρB ρA ) 2⁄3 23 (b) Since there is no external torque, angular momentum will remain conserved. The moment of inertia will first decrease till the tortoise moves from A to C and then increase as it moves from C to D. Therefore ω will initially increase and then decrease m1 m2 x1 x2 d y a sin 60 a cos 60o 2m 3m 4m m x 60o a
Let R be the radius of platform m the mass of tortoise and M is the mass of platform Moment of inertia when the tortoise is at A I1 = mR 2 + MR 2 2 and moment of inertia when the tortoise is at B I2 = mr 2 + MR 2 2 Here r 2 = a 2 + [√R2 − a 2 − vt] 2 From conservation of angular momentum ω0I1 = ω(t) I2 Substituting the values we can found that the variation of ω(t) is non linear 24 (d) Centre of mass C1 of the point masses placed at corners P and S from point P = 1×0+2×a 1+2 = 2a 3 Therefore, distance of centre of mass C1 from point S = a − 2a 3 = a 3 Similarly, centre of mass C2 of point masses placed at corners Q and R, from point Q = 2a 3 Distance of centre of mass C2 from point R = a − 2a 3 = a 3 Centre of mass of all four point masses is at the mid point of the line joining C1 and C2, which is farthest from points P and Q. 25 (a) 15m + 10m = mv1 + mv2 25 = v1 + v2 And v2−v1 u1−u2 = 1 ⇒ v2 − v1 15 − 10 = 1 ⇒ v2 − v1 = 5 Solving Eqs. (i) and (ii), we have v2 = 15ms −1 , v1 = 10ms −1 27 (d) In the absence of external torque for a body revolving about any axis, the angular momentum remains constant. This is known as law of conservation of angular momentum, τ⃗ = dL⃗⃗ dt As τ⃗ = 0 ∴ dL⃗⃗ dt = 0 or L⃗⃗ = constant 28 (a) Due to centrifugal force 29 (b) From conservation of angular momentum (Iω= constant), angular velocity will remain half. As, K = 1 2 Iω 2 The rotational kinetic energy will become half. 30 (b) 1 2 MR 2 = MK 2 ⇒ K = R √2 = 2.5 √2 = 1.76cm 31 (b) Given, I = 1 kg − m2 , n = 2 rps ω = 2πn = 2π × 2 = 4π rads −1 L = Iω = 1(4π) = 12.57 kg − m2 s −1 32 (c) ω 2 = ω0 2 − 2αθ ⇒ 0 = 4π 2n 2 − 2αθ θ = 4π 2 ( 1200 60 ) 2 2 × 4 = 200π 2 rad ∴ 2πn = 200π 2 ⇒ n = 100π = 314 revolution 33 (c) Time taken in reaching bottom of incline is t = √ 2l(1 + K2/R2) g sin θ For solid cylinder (SC), K 2 = R 2 /2 For hollow cylinder (HC), K 2 = R 2 For solid sphere (S), K 2 = 2 5 R 2 34 (b) θ = ω0t + 1 2 αt 2 ⇒ θ = 100rad ∴ Number of revolution = 100 2π = 16 (approx.) 35 (c) The rolling sphere has rotational as well as translational kinetic energy ∴ Kinetic energy = 1 2 mu 2 + 1 2 Iω 2 = 1 2 mu 2 + 1 2 ( 2 5 mr 2 ) ω 2 = 1 2 mu 2 + mu 2 5 = 7 10 mu 2 Potential energy = kinetic energy ∴ mgh = 7 10 mu 2 ⇒ h = 7u 2 10g 36 (d) Angular momentum L is given as L = r⃗ × p⃗ = rp sin θ r⃗ = position vector of the particle w. r.t. origin,
p⃗ = its linear momentum r⃗ × p⃗ is maximum when p is perpendicular to r i. e. θ = 90° 37 (a) Since disc is rolling (without slipping) about point O Hence OQ > OC > OP ∴ v = rω ∴ vQ > vC > vP 39 (b) As E = 1 2 Iω2 ω = √ 2E I = √ 2 × 600 3.0 = 20 ω = 2π T = 20, T = 2π 20 = π 10 = 0.314 s 40 (c) Given, I = 2 5 MR 2 Using the theorem of parallel axes, moment of inertia of the sphere about a parallel axis tangential to the sphere is I ′ = I + MR 2 = 2 5 MR 2 + MR 2 = 7 5 MR 2 ∴ I ′ = MK 2 = 7 5 MR 2 , K = (√ 7 5 )R 41 (b) Form the triangle OAC d = OC sin45° = 4 × 1 √2 = 2√2 Angular momentum = Linear momentum x Perpendicular distance of line of action of linear momentum from the point of rotation L = p × d = mvd = 5 × 3√2 × 2√2 = 60 units 43 (a) L = Iω ∴ L ∝ ω (If I = constant) So graph between L and ω will be straight line with constant slope 44 (d) v = √ 2gh 1 + I mr 2 = √ 2 × 10 × 3 1 + mr 2 2 × mr 2 = √ 2 × 10 × 3 3 2 = √40 ⇒ v = rω ⇒ r = v ω = √40 2√2 = √ 40 8 = √5 m 45 (b) ICM = ML 2 12 (about middle point) ∴ I = ICM + Mx 2 = ML 2 12 + M ( L 6 ) 2 I = ML 2 9 46 (d) Melting of ice produces water which will spread over larger distance away from the axis of rotation. This increases the moment of inertia so angular velocity decreases 47 (b) Angular momentum = linear momentum × Perpendicular distance of line of action of linear momentum from the axis of rotation = mv × l 48 (d) Time taken by first block to reach second block = L v Since collision is 100% elastic, now first block comes to rest and 2nd block starts moving towards the 3rd block with a velocity v and takes time = L v to reach 3rd block and so on ∴ Total time = t + t+. . . (n − 1) time = (n − 1) L v Finally only the last nth block is in motion velocity v, hence final velocity of centre of mass vCM = mv nm = v n 49 (a) Here, m1 = 1 kg, v⃗⃗1 = 2î m2 = 2 kg, v⃗⃗2 = 2 cos 30 î− 2 sin30ĵ v⃗⃗CM = m1v⃗⃗1 + m2v⃗⃗2 m1 + m2 = 1 × 2î+ 2(2 cos 30°î− 2 sin30°ĵ) 1 + 2 = 2î+ 2√3î− 2ĵ 3 = ( 2 + 2√3 3 ) î− 2 3 ĵ 50 (d) O  P C Q d A B 4 O 4 y=x+4 p =mv y x C 45°