Tiêu đề 06.ROTATIONAL MOTION.pdf ✅

Objective Physics Volume-I 843 YCT 06. Rotational Motion (a) Centre of Mass 1. Two objects P and Q initially at rest move towards each other under mutual force of attraction. At the instant when the velocity of P is v and that of Q is 2v, the velocity of centre of mass of the system is (a) v (b) 3v (c) 2v (d) 1.5v (e) zero Karnataka CET-2019 AIIMS-27.05.2018 (E) Kerala CEE - 2015 APEMCET (Medical) - 2010 AMU- 2009 EAMCET-2008 Ans. (e) : Force on particle P is given by, FP = mpap = m vP t ....(i) Similarly, FQ = mQaQ = m 2v Q t × ....(ii) Now, FP = FQ m vP m 2v Q t t × = mP = 2mQ Velocity of centre of mass of the system, P P Q Q P Q m v m v v m m − = + Q Q Q Q 2m v m 2v v 2m m − × = + v = 0 ∴ Velocity of centre of mass of the system is zero. 2. A circular disc of radius R is removed from one end of a bigger circular disc of radius 2R. The centre of mass of the new disc is at a distance αR from the centre of the bigger disc. The value of α is (a) 1 2 (b) 1 3 (c) 1 4 (d) 1 6 AP EAMCET (20.04.2019) Shift-1 JCECE-2008 UPSEE-2007 Ans. (b) : Let 'O' is the centre of circular disc of radius 2R and mass M, and C1 is the centre of removed disc. Let, σ = Surface mass density. m1 = σ A1 m1 = σ.4πR2 , (Where σ = Surface mass density) m2 = σA2 m2 = σπR2 ∴ 1 1 2 2 CM 1 2 m x m x x m m − = − 2 2 CM 2 2 4 R 0 R R x 4 R R σ⋅ π × − σπ ⋅ = σ π − σ⋅ π 3 2 R 3 R −σπ = σπ CM R x 3 − = Due to symmetry about x-axis yCM = 0 Then, CM CM CM r x i y j = + ɵ ɵ CM R r i 0 3 − = + ɵ CM R r R 3 = = α On comparing, 1 3 α = 3. Two bodies of mass 1 kg and 3 kg have position vectors ˆ ˆ ˆ ˆ ˆ ˆ i + 2j + k and - 3i - 2j + k respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector: (a) ˆ ˆ ˆ i 2j + k − (b) ˆ ˆ ˆ − − 3i 2j + k (c) ˆ ˆ − 2j + 2k (d) ˆ ˆ ˆ − − 2i j + 2k JEE Main-29.07.2022, Shift-I Ans. (a) : Given, m1 = 1kg, m2 = 3kg ( ) ( ) 1 2 ˆ ˆ ˆ ˆ ˆ ˆ r i 2j k , r 3i 2j k = + + = − − + Centre of mass 1 1 2 2 cm 1 2 m r m r r m m + = +
Objective Physics Volume-I 844 YCT ( ) ( ) cm ˆ ˆ ˆ ˆ ˆ ˆ 1 i 2j k 3 3i 2j k r 1 3 + + + − − + = + ˆ ˆ ˆ ˆ ˆ ˆ i 2 j k 9i 6j 3k 4 + + − − + = ˆ ˆ ˆ 8i 4 j 4k 4 − − + = cmr ˆ ˆ ˆ = − − + 2i j k cmr = ( ) ( ) ( ) 2 2 2 2 1 1 6 + + = (a) ˆ ˆ ˆ i 2j + k − 2 2 2 = + + = 1 2 1 6 (b) ˆ ˆ ˆ − − 3i 2j + k 2 2 2 = + + = 3 2 1 14 (c) ˆ ˆ − 2j + 2k 2 2 = + = 2 2 2 2 (d) ˆ ˆ ˆ − − 2i j + 2k 2 2 2 = + + = 2 1 2 3 The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector is ˆ ˆ ˆ i 2j + k − . 4. A circular portion of radius R2 has been removed from one edge of a circular disc of radius R1. The correct expression for the centre of mass for the remaining portion of the disc is– (a) 2 2 1 2 R R R − + (b) 2 2 1 2 R R R − + (c) 2 2 1 2 R R R+ (d) 2 1 1 2 R R R − + AP EAMCET-08.07.2022, Shift-II Ans. (a) : Let, M0 = mass of circular plate M0 = ρπr2 = ρπ(R1) 2 Mr = removed mass of circular plate. Mr = ρπr2 = ρπ(R1) 2 . Mass of remaining portion = M0 − Mr = ρπ(R1) 2 − ρπ(R2) 2 . ( ) ( ) ( ) ( ) ( ) 2 2 1 2 1 2 cm 2 2 1 2 R 0 R R R x R R     ρπ × − ρπ × −     = ρ −     ( ) ( ) ( )( ) 2 2 1 2 cm 1 2 1 2 R R R x R R R R − − = − + 2 2 cm 1 2 R x R R − = + 5. A solid circular disc of mass 100 kg rolls along a horizontal floor so that center of mass of the disc has a speed of 0.2 m s–1. The absolute value of work done on the disc to stop it is (a) 2 J (b) 3 J (c) 2.5 J (d) 4 J AP EAMCET-12.07.2022, Shift-I Ans. (b) : Given, mass of solid circular disc m = 100 kg v = 0.2 m/s Total energy of the disc = Translational kinetic energy + Rotational kinetic energy 1 1 2 2 mv I 2 2 = + ω Moment of inertia of solid circular disc- 1 2 I mr 2 = Total Energy 1 1 1 2 2 2 mv mr 2 2 2 = + × ω = 2 2 2 2 1 1 v mv mr 2 4 r + 2 2 2 v r   ω =     1 1 2 mv 1 2 2   = +     = 3 2 mv 4 = 3 100 0.2 0.2 4 × × × = 3 J Hence, work done on the disc to stop it is equal to total energy work done 3J = 6. The distance of centre of mass from end A of a one dimensional rod (AB) having mass density 2 0 2 x 1 L   ρ = ρ −     kg/m and length L (in meter) is 3L α m. The value of α is .......(where x is the distance form end A) JEE Main-28.07.2022, Shift-II Ans. (8) : Given, 2 0 2 x 1 L   ρ = ρ −     kg/m L 2 0 2 0 cm L 2 0 2 0 x A 1 xdx L x x A 1 dx L   ρ −     =   ρ −     ∫ ∫ L 3 2 0 cm L 2 2 0 x x dx L x x 1 dx L     −   =     −   ∫ ∫ L 2 4 2 0 cm L 3 2 0 x x 2 4L x x x 3L   −     =   −    
Objective Physics Volume-I 845 YCT 2 4 2 cm 3 2 L L 0 0 2 4L x L L 0 0 3L   − − −     =   − − −     2 2 2 cm L L L 2 4 4 3L x L 2L 8 L 3 3   −     = = =   −     According to question centre of mass is given 3L α on comparing with calculated value of centre of mass we get, α = 8 7. Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and (x, 0) cm respectively. The block of 10 kg is moved on the same line though a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is (a) 4 cm towards the 10 kg block (b) 2 cm away from the 10 kg block (c) 2 cm towards the 10 kg block (d) 4 cm away from the 10 kg block JEE Main-27.06.2022, Shift-I Ans. (c) : Two masses m1 and m2 are kept in a straight line m1 = 10 kg, m2 = 30 kg The fist block is moved on the same line a distance of 6 cm. 1 1 2 2 cm 1 2 m x m x x m m + = + 2 cm 10 6 30 x x 10 30 × + × = + According to question centre of mass of the system remain unchanged xcm = 0 2 cm 60 30x x 40 + = 30x2 = − 60 x2 = –2 cm So, the 30kg of block should be displaced 2cm to move towards 10 kg block. 8. A meter scale is balanced on a knife edge at its centre. When two coins, each of mass 10 g are put one on the top of the other at the 10.0 cm mark the scale is found to be balanced at 40.0 cm mark. The mass of the meter scale is found to be x×10-2kg. The value of x is ____. JEE Main-24.06.2022, Shift-I Ans. (6) Mass of coins added. m = 2 × 10 = 20 g d1 = 40 – 10 = 30 cm d2 = 50 – 40 = 10 cm Equilibrium- m × d1 = M × d2 20 × 30 = M × 10 M= 60 g According to the question, M = x ×10–2 kg 60g = x ×10–2 kg 60g = x ×10–2 × 1000 g x = 6 9. Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be xm. The value of x is JEE Main-25.07.2022, Shift-II Ans. (2) : According to the question, we can draw a figure which is given below: ( ) cm M (0i 0j) (3i 0j) 0i 3j ˆ ˆ ˆ ˆ ˆ ˆ r 3M   + + + + +   = cm ˆ ˆ 3 i j r 3   +   = cm ˆ ˆ r i j = + cm |r | 2 x = = x 2 = 10. Four masses are arranged along a circle of radius 1 m as shown in the figure. The center of mass of this system of masses is at (a) 1 1 i j 5 5 − −ɵ ɵ (b) 1 i j 5 ɵ ɵ+ (c) 1 i j 5 ɵ ɵ − (d) 1 1 i j 5 5 ɵ ɵ + TS EAMCET 18.07.2022, Shift-I