CIRCULAR MOTION EXERCISE 1 1. d = rate of change of angle

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1 CICULA MOTION EXECISE. d = rate of change of angle as they both complete angle in same time.. c m mg N r m N mg r Since r A r B N A N B. a Force is always perpendicular to displacement work done = 0 4. b 5. a f = mg a a a resultan t c tan g m / s N mg mr mg g r 6. c If the coin just slips at a distance of 4r from centre mg = m4r () If angular elocity is doubled mg = m () () From () and () = r 7. d T = mr 0 () T = mr () 0 5 rpm 8. c Since force is always perpendicular to elocity particle moes in circle. It s speed is constant and elocity ariable. 9. c

2 m max mg r max gr m / s 08 km / hr 0. c m Fnet ma rad. r. b T = mr = =.6N. c Centripetal force is proided by friction. a m NA mg r m m NA mg, NB mg r r A m NC mg rc 4. a eal forces are mg and T only 5. a Bead starts slipping, when N = ml ml = ml = (0 + t) t EXECISE. (a, c) h 5 Time to fall = s g 0 Distance coered in y direction = t = = m Since there is no elocity along x-direction x is always m.. (a, b) For no wear and tear friction is zero o tan5 g o g tan5 8. m / s B

3 max g( tan ) tan 8.m / s. (b, c) dr ; dr dt aag 0 t (i.e. along the tangent) 4. (a, b) ds K S dt S S Kt S ds S 0 0 K t 4 ds K t dt t Kdt 5. (a, b, c) T m,, F ert = 0 6. (a, b, d) constant, = t F y = F sin = F sin t m sin t V r = sin = sin t x-coordinate = cos t 7. (b, d) If = 0., f max = = 0.5N eq. centripetal force = mr = =.5N f N, Tension zero 8 If, fmax N 0 0 f N, Tension zero 8 If, fmax N 40 40

4 f N, Tension 0 8 EXECISE. (a) K rt r (gien) (a) Centripetal force = mk rt d (b) Tangential force = m mkr dt (c) Power of centripetal force F centripetal. 0 (d) Power of tangential force F t. Ft m.k r t. (b) Assuming no friction between m and m a = T - m a = T - m a a Friction on upper block acts towards left and on lower block towards right.. (a) Let the required angular elocity be Then T + m g = m () T = m g + m () m g = (m m ) 4 m g (m m ) 6. rad / s 4. (b) T = m g + m = = 0N 5. (b) design tan ; g 6. (a) f m gsin cos 00 5 m / s

5 7. (a) f m cos g sin m / s tan 8. (c) mg sin = m () L m( gl) m mgl( sin ) () sin ; Also gl 9. (c) 0 cos hmax L( sin ) g 40L 7 0. (b) m( gl) mg h max h max L EXECISE 4. Here frictional force will proide the required tangential and centripetal force for the circular motion of car. Force of friction will act along the direction of net acceleration. a net a f m a Car will skid when m a a m a a t 4 4 ( g a ) t 4 a /4 5

6 Till this time distance traeled by car is D at. Putting alue of t we get. The insect will slide when mg sin becomes equal to fmax N limiting friction. At eery instant the insect is in equilibrium. So, N = mg cos N = mg sin mgcos mg cos = mg sin = tan = tan = tan (/) mg g a D. a N mg sin. Applying Newton s law towards the centre of circle we get N = m Let be the minimum angular speed for which man is not falling. At this instant its weight will be balanced by limiting friction acting upwards. i.e, N = mg = 70 0 = 4.7 rad/sec. B A / 6 T T P Mg 4. Applying Newton s law along ertical we get cos cos T T mg () 6 mg T Applying Newton s law along horizontal. We get T cos T cos m m 6 g P Q O 5. Particle P and Q will be at same angular position wheneer 5t = t + n. n t (n = integer) () 6 Similarly, particle P and will be at same angular position. Wheneer 5t = t = + m (m = integer) t m () 6

7 All three particles will be at same angular position when () = () n m 6 n = + m Smallest integral alue of m & n satisfying the aboe equation is m =, n =. Putting these alues in () & () we get t =.5 sec. So, they will meet for the first time at t =.5 sec. 6. According to question, ac a t t d d dt dt t 0 V(t) t T dt 0 T dx 0 ( n( t 0) 0 t T ( e ) 0 7. Lets assume that the particles meet at time t. Distance traeled by A = distance traeled + by B 7 t t t 5 5 t 6 dis tan ce traelled Angle traced by A 6 at 7 Angular elocity 5 0 B a t A EXECISE 5. (c, d) m F P = 0 kinetic energy = constant; F is constant (gien) = constant.. (a) adius of curature in (a) is minimum. (a) mg sin = mg cos cot = constant 4. (c) a a a and a tangential normal tangential is downward. 7

8 mg 5 5. Kx cos 0 + mg cos 0 = ma t. As x and K, a t g ; 4 8 mg N + Kx cos 60 = mg cos 60 N 8 8

9 . D WOK POWE & ENEGY EXECISE xx x5 W Fdx (7 x x )dx (7x x x ) = 5 J xx x0 x5 y x0. A This is the statement of Work Kinetic Energy Theorem. B Since the force acting on the particle is perpendicular to the displacement eerywhere, the work done iz zero 4. B Instantaneous power, P = F. (0i ˆ 0j ˆ 0k) ˆ. (5iˆ j ˆ 6k) ˆ = 40 W 5. D Mass of the hanging part = M ; when the hanging part of the chain is paralled on the table, its centre of mass is raised by L 6. The work done = rise in potential energy = M L MgL g B m mg g 7. B K longer l longer = K original l original k Klonger k ( / ) 8. C W F = increase in potential energy = mgl ( cos ) 9. D 0. B Mean power of graity = work done by graity 0 time elapsed 9

10 Acceleration, a = kx F = Kx loss of KE : gain of potential energy x.. C t x t Now, W = KE = m( f ) = 6J. C KE max = Maximum loss of KE = Mgl ( cos ). A Centre of mass of the rope is lifted by h and the back by h. Therefore, 0 W = Mgh + mg h M m gh 4. D d x mg = m E x 5. D mg W d 4 6. C B F W F. dr A x xg dx dx x0 0 m d 7. C dw t P W 4J m / s dt 8. C W : W : W = atio of corresponding displacements = : ( ) : ( ) = : : 5 9. C /6 d a 6h 0 a x 7 U a b b min dx x x b (a / b) (a / b) 4a b Minimum energy required = 4a

11 0. D mg max = P max =. C mg W 6. J. C The KE intercepted P mg. D T mg = m 5g T mg 6mg 4. A mg(h + x) = kx 980x 9.8 (0.4 + x) = 0 50x x 0.4 = 0 (0x ) (5x + 0.4) = 0 x = 0. m = 0 cm 5. D. c, d ya W (kx ˆj). (ai) ˆ k(yi ˆ aj) ˆ. dy ˆj ka y0 EXECISE Since no work is done by the force speed is constant not elocity. circle.. b, c W.d by all forces = K.E. W g + W N = K.E f K.E i mgh + 0 = m 0 gh. b, c P Q where h is the initial height of both blocks from ground. 4. b, c w.d = F. d ( ˆi j ˆ k) ˆ. j ˆ = 6 J a r along the centre of

12 w.d. = F. d ( ˆi j ˆ k) ˆ. (j ˆ 4k) ˆ 8J 5. a, b, c W.d = Area enclosed by the F-x graph 6. a, d Fnet a m / s m (t = 4s) = = m/s s (t = 4s) = 4 4 m w.d by net force = K.E m = 44 J w.d. by applied force = 0 4 = 40 J w.d by friction = F. d = 4 4 = 96 J 7. b, c 8. b, c, d 9. a, b, d w.d by all forces = K.E. Fx mgx = 40 J 0x = 40 J x = m w.d. graity = mgx = 0 = 40 J w.d. tension = Fx = 40 = 80 J 0. a, d Power = F. F Fat or F ax Since a and F are constants Power aries linearly with time and parabolically with displacement. a, c Hint: Direction of spring force and displacement are same in (a) & (c). b, c Pmg F. mg( ˆj).[u cos ˆi (u sin gt) ˆj] = mg (u sin gt) u sin P 0 for t and P 0 for g u sin u sin t g g

13 . a, c, d F U ˆ U a i ˆj iˆ 4j ˆ m mx my (at x 0) u as = 0 m/s x x 0 ut at ˆ ˆ 6i 4j 0 ( i ˆ 4j) ˆ 4.5i ˆ j ˆ 4. c, d w.d by all force = increase in spring energy Fx 0 + mgx 0 = x0 F k mg mg k(x 0 ) k k k 5. b, c, d w.d. by F J w.d. by F J w.d. by F F cos90 rd F is conseratie in nature as it is always directed towards P. 6. b, d m At highest point Fnet m mg mg g Consering energy elocity at lowest point 7g 7. b, d w.d. = F. d (6i ˆ 6j) ˆ. ( i ˆ 4j) ˆ = 8 4 = 4 J Had there be no initial elocity particle must hae moed along straight line making an angle of 45 with x-axis.

14 8. a, c, d P F. ie angle is acute = ie angle is obtuse Area under graph = K.E. = 0J 9. (a, c) m = M, where and are speeds of mass m and M, as seen from ground. The elocity of m relatie to M is = ( ). Hence, or t V V V or + = l/t. m l M 0. (a, b) du A B A F at equilibrium F = or, r dr r r B A B B At infinity U = 0 r, U U. B 4A 4A. (a, c) From conseration of linear momentum ( + ) = (6 x ) + ( ) = 4m/s (of both the blocks) From work energy therom i.e., W total = KE on kg block, Wf (4 6 ) 0J on kg block Wf (4 ) 7J. Net work done by friction is J.. (b, d) In region OA particle is acccelerated, in region AB particle has uniform elocity while in region BD particle is deceleration., Therefore, work done is positie in region OA, zero in region AB and negatie in region BC.. (a, c) g at B acceleration of block g 4. (a, d) EXECISE III. (A q) Work energy theorem w.d. by all forces is equal to change in K.e. (B s) Negatie of work done by conseratie force is equal to change in potential energy (C r) W.d ext. + W non cons. = K.E. W cons. = K.E. + U = T.M.E. (A r) 4

15 mu mg mb B 7g (B - q) mu mg m C 5g (C p) mb TB 7 mg (D t) mc TC mg m5g TC mg 4mg. (A q) (B p) (C r) C 4 4 kx w.d kx dx k[4 ] ie kx w.d kx dx k[ 4 ] ie 4 4 kx w.d kx dx 0 4. (A t), (B p), (C s), (D q) S (4) 6m w.d graity = mg 6 = 0 6 = 60 J w.d normal reaction = N cos S = m(g+a) cos S = 44J w.d friction = f S sin = m(g+a) sin S = 48 J w.d. forces = K.E. m(at) ( 4) J 5. c work done by both against graity = mgh 6. b Aerage Power = mgh W t 0 7. b 5

16 Chemical energy expended by the physicist ends up increasing the potential energy. 8. b As the physicist falls, graitational potential gets conerted into kinetic energy, increasing his speed. After he hits the cushion, this kinetic energy gets conerted into heat. 9. d K.E. = K.E f K.E i = K.E f 0 = w.d. by mg = mgh 5 = J 0. a. c. a From conseration of energy at A and B, we hae At B the string becomes slack. Therefore m mgsin B () After passing through B, the ball goes in a projectile sin t cos () B and B cos t gt sin (4) On soling,, & 4 = 0 7g g and B. c ( - x) + hg dt dt = ( - (x + dx)] ( + d), where is mass per unit length hg dt = ( - x) d hg x dx x x0 0 hg ln x gh ln h at B d 4. a KE x ( x) gh ln x ma mb mg( sin ) () 6

17 It is maximum, when e x mgh KEmax hg e e 5. b Heat generated = ( - h) gh. h. gh ln h mgh h h ln h 6. b 7. a 8. b 9. c U(x) = 0 + (x ) At x = 0, T.M.E = U + K.E = = 49 J At extreme positions, K.E = 0 U = 49J 0 + (x ) = 49 x = 9 x 9 i.e.,.4 and 7.4 m K.E. max = T.M.E Min. potential energy = 49 0 (at x = ) = 9 J Body is in equilibrium at min. potential energy i.e., at x = 0. W mg is path and rate independent. Total energy is consered when there is no external and no internal non conseratie force.. Work done for conseratie forces are path independent. W mg = mgh 4. F. r F. (r r ) 5. sin t is time of flight and ertical displacement is zero. g 7

18 EXECISE 4. W F = Fh = 80J; W weight = (mg)h = 40 J. mm m m. Tension, T g; acceleration a = g m m m m mm (m m ) W = T. at g t (m m ) 00 J x W ( x) dx.5j x F F N mg; N F F mg F (a) F WF S 7.J (b) W friction = W F = 7.J (c) W graity = 0 mg.6 5. W = Area under the cure = = 0J 6. W F = increase in potential energy = mg ( cos ) 7. dw = mg( cos + sin ) ds = mg ( d dh) W = mg( h) 8. W = KE = () 0 400J 9. W KE m 0. a) (m)g x m = k x 4mg m xm k b) xm xm (m) K (m) g g m k c) x m g mg k (m) a a 4. mag m K mg ma K. Let x be the extension of the spring and the angle that the spring makes with the ertical at break off.

19 K x cos = mg x. 0.4 x x = 0.m; The of B = the slide of (m) Kx mg h =.54 m/s m d d m dt 7. (i) t0 0 t dt A (0.5) (0.4) 0. metres = h (say) t 0 0 ; S dt ln t t t Fb( sin ) m sin Fb( ) m, Fmax mg Consering mechanical energy: 0 = ( 5 ) =.9 m/s (4,6) W F. ds (x ˆi yj) ˆ. (dxiˆ dy ˆj dzk) ˆ (4,6) (,) (,) (x dx y dy) x y = 8J (ii) (iii) c W (xy ˆi xy ˆj). (dx ˆi dy ˆj) 0 (alongoc) 4 ˆ ˆ ˆ x (i j). dx i x dx J 0 0 A 4 W xy (i ˆ ˆj). dx ˆi xy (i ˆ ˆj). dy ˆj 0 (along OA) A (along AC) y 0 y dy J y0 k W 0 x dx J k0 8. (a) dmg h mgh (b) x x 0 m dx gx mg 6 C 9

20 / (c) m mg d g ( cos ) sin 0 9. (a) u cos = 0. m m (b) m m / s 4 D mg d k m 4 4 d g k d 6m. mg h min mg r m ( gr) 5r h min ; mg(5r) mg r = m m Now, Fresultan t 6mg r. m mg ( cos ) = () mg mg ( cos ) m () From () and () g ; cos m mg ( cos 60 o ) 0 g m / s 0 m( 5g ) mg( cos ) m () sin sin Also, t flight () cos g From () and () = 0 or = 60 m mg. m, where is elocity at the highest point. u 4g Now, t flight = 4 u 4g g 0

21 u g ; 4 x min min tflight gh g x x0 x ( dx) g r sin x ( r ) r gr mg cos m 4 () mg cos = m () cos 6 n 5 n n mg Mg n 0 m n M n (F mg) m F g m xl 0 x0 m (Gx)(mg) dx k L k 4ag m. Stretch x = 0.4 (sec 0 - ) = 0.4 ; kx sin 0 = (mg kx cos 0 ) mg kx o o sin 0 cos0 Now, W = U = kx 0.09J. a) mg ( cos ) = m m F + mg cos = From () and () F = mg ( cos ) () ()

22 N = Mg F cos, = Mg mg ( cos ) cos which is minimum when = cos b) N = 0 m M EXECISE - V. (b) The centripetal acceleration a c k rt or k rt r = krt d So, tangential acceleration, a t kr dt Work is done by tangential force. Power = F t.. cos 0º = (m a t ) (krt) = (mkr) (krt) = mk r t. (b) The force constant of a spring is inersely proportional to the length of the spring. Let the original length of spring be L and spring constant is K (gien) Therefore, L K L K ' K ' K. (d) du (x) = Fdx x U x Fdx 0 4 kx ax 4 k U = 0 at x = 0 and at x ; we hae potential energy zero twice (out of which one is at a origin). Also, when we put x = 0 in the function, du We get F= 0. But F dx du At x 0; 0 i.e. the slope of the graph should be zero. These characteristics are dx represents by (d).

23 4. (b) Let the maximum extension of the spring be x as shown in the figure. Work is done by the graitational and the spring force. There is no change in the kinetic energy between the initial and final position of the mass. From Work-energy theorem; W g + W s = 0 Where W g = work done by graity And W = work done by spring g s Mgx kx 0 Mg x k 5. (b) In a conseratie field work done does not depend on the path. The graitational field is a conseratie field. W = W = W 6. (b) We know that x x U Fdx or U k xdx 0 0 kx U x U 0 Gien U (0) = 0 kx U x 7. (d) 5gL () gh () h = L( cos ) () Soling Eqs. (), () and (), we get 7 7 o cos or cos (c) When the block B is displaced towards wall, only spring S is compressed and S is in its natural state. This happens because the other end of S is not attached to the wall but is free. Therefore the energy stored in the system k x. When the block is released, it will come back to the equilibrium position, gain momentum, oershoot to equilibrium position and moe towards wall. As this happens, the spring S comes to its natural length and S gets compressed. As there are no frictional forces inoled, the P.E. stored in the spring S gets stored as the P.E. of spring S when the block B reaches its extreme position after compressing S by y.

24 kx k y kx 4ky x = 4y y x 9. (b) The forces acting on the bead as seen by the obserer in the accelerated frame are: (a) N; (b) mg; (c) ma (Pseudo force). Let is the angle which the tangent at P makes with the X- axis. As the bead is in equilibrium with respect to the wire, therefore N sin = ma and N cos = mg a tan... i g But y = kx. Therefore, dy kx tan.. ii dx From (i) & (ii) a a kx x g kg 0. Let M strikes with speed. Then, elocity of m at this instant will be cos or 5. Further M will fall a distance of m while m will rise up by ( 5 )m. From energy conseration: decrease in potential energy of M = increase in potential energy of m + increase in kinetic energy of both the blocks. or () (9.8) () = (0.5) (9.8) ( 5 ) Soling this equation, we get =.9 m/s

25 . Let the string slacks at point Q as shown in figure. From P to Q path is circular and beyond Q path is parabolic. At point C, elocity of particle becomes horizontal, therefore, QD = half the range of the projectile. Now, we hae following equations () m T Q = 0. Therefore, mg sin = L (i) () = u gh = u gl ( + sin ) (ii) () QD = (ange) o L sin (90 ) sin L cos 8 g g Eq. (iii) canbe written as cos sin cos 8 gl Substituting alue of sin from eq. (i), we get gl cos sin ( cos ) cos 8 or cos cos cos 8 o cos or cos or 60 8 From Eq. (i) = gl sin = gl sin 60 or gl Substituting this alue of in Eq. (ii) u = + gl ( + sin ) gl gl (iii) gl gl gl u gl. The ball is moing in a circular motion. The necessary centripetal force is proided by (mg cos N). Therefore, 5

26 m mg sin N A... i d According to energy conseration d m mg cos... ii From (i) and (ii) N A = mg ( cos ) (iii) The aboe equation shows that as increases N A decreases. At a particular alue of, N A will become zero and the ball will lose contact with sphere A. This condition can be found by putting N A = 0 in eq. (iii) 0 = mg(cos ) cos The graph between N A and cos From equation (iii) when = 0, N A = mg. When cos The graph is a straight line as shown when cos 6

27 N B m mg cos d m B d Using energy conseration d d m mg cos m mg cos... d From (i) and (), we get N B + mg cos = mg mg cos N B = mg( cos) N mg cos... i When cos, NB 0 When cos =, N B = 5 mg. Gien m = 0.6 kg, M = 0.7 kg. The figure shows the forces on m and M. When the system is released, let the acceleration be a. Then T mg = ma Mg T = Ma M mg a g / M m and T = 4mg/ For block m: u = 0, a = g/, t =, s=? g s ut at 0 g / 6 Work done by the string on m is mg T. s Ts 4 8J 6 6 a M T Mg M T Mg a 7