SOLUTIONS TO CONCEPTS CHPTE 0. 0 0 ; 00 rev/s ; ; 00 rd/s 0 t t (00 )/4 50 rd /s or 5 rev/s 0 t + / t 8 50 400 rd 50 rd/s or 5 rev/s s 400 rd.. 00 ; t 5 sec / t 00 / 5 8 5 40 rd/s 0 rev/s 8 rd/s 4 rev/s 40 rd/s 0 rev/s.. re under the curve will decide the totl ngle rotted xiu ngulr velocity 4 0 40 rd/s Therefore, re under the curve / 0 40 + 40 0 + / 40 0 800 rd Totl ngle rotted 800 rd. 4. rd/s, 0 5 rd/s; 5 rd/s w w 0 + t t ( 0 )/ (5 5)/ 0 sec lso, 0 t + / t 5 0 + / 00 00 rd. 5. 5 rev, rev/s, 0 0 ;? ( ) 5 5 rev/s. or 0 rd, 4 rd/s, 0 0,? 4 0 4 5 rd/s 5 rev/s. 6. disc of rdius 0 c 0. ngulr velocity 0 rd/s Liner velocity on the ri r 0 0. /s Liner velocity t the iddle of rdius r/ 0 (0.)/ /s. 7. t sec, r c 0.0 4 rd/s Therefore t 4 rd/s Therefore rdil ccelertion, n r 0.6 /s 6 c/s Therefore tngentil ccelertion, r r 0.04 /s 4 c/s. 8. The Block is oving the ri of the pulley The pulley is oving t 0 rd/s Therefore the rdius of the pulley 0 c Therefore liner velocity on the ri tngentil velocity r 0 0 00 c/s /s. 0 0 0 T 0.
9. Therefore, the distnce fro the xis (D) / 0 5 c. Therefore oent of inerti bout the xis BC will be I r 00 K ( 5 ) 00 5 5000 g c.5 0 kg. b) The xis of rottion let pss through nd to the plne of tringle I D Therefore the torque will be produced by ss B nd C B C Therefore net oent of inerti I r + r 00 0 40000 g-c 4 0 kg-. 0. Msses of g, g 00 g re kept t the rks c, c, 000 c on he x xis respectively. perpendiculr xis is pssed t the 50 th prticle. Therefore on the L.H.S. side of the xis there will be 49 prticles nd on the.h.s. side there re 50 prticles. Consider the two prticles t the position 49 c nd 5 c. Moent inertil due to these two prticle will be 49 + 5 + 00 g-c 0 0 0 40 50 60 70 80 90 00 Siilrly if we consider 48 th nd 5 nd ter we will get 00 g-c Therefore we will get 49 such set nd one lone prticle t 00 c. Therefore totl oent of inerti 00 { + + + + 49 } + 00(50). 00 (50 5 0)/6 49500 g-c 0.49 kg- 0.4 kg-. 48 49 5 5. The two bodies of ss nd rdius r re oving long the coon tngent. x Therefore oent of inerti of the first body bout XY tngent. r + /5 r Moent of inerti of the second body XY tngent r + /5 r 7/5 r r r Therefore, net oent of inerti 7/5 r + 7/5 r 4/5 r units. y. Length of the rod, ss of the rod 0.5 kg Let t distnce d fro the center the rod is oving pplying prllel xis theore : d The oent of inertil bout tht point (L / ) + d 0.0 (0.5 )/ + 0.5 d 0.0 d 0. 0.08 0.8 d 0.4 fro the centre.. Moent of inerti t the centre nd perpendiculr to the plne of the ring. l / (l /)+d So, bout point on the ri of the ring nd the xis to the plne of the ring, the oent of inerti + (prllel xis theore) K (K rdius of the gyrtion) K. 4. The oent of inerti bout the center nd to the plne of the disc of rdius r nd ss is r. ccording to the question the rdius of gyrtion of the disc bout point rdius of the disc. Therefore k ½ r + d (K rdius of gyrtion bout ccelertion point, d distnce of tht point fro the centre) K r / + d r r / + d ( K r) r / d d r /. k r d / r / r +d 0.
5. Let sll cross sectionl re is t distnce x fro xx xis. Therefore ss of tht sll section / x dx Therefore oent of inerti bout xx xis / I xx ( / ) (dx) x ( ( / )(x 0 / )] / Therefore I xx I xx + I yy * /) /6 Since the two digonls re to ech other Therefore I zz I x x + I y y /6 I x x ( becuse I x x I y y ) I x x / D x y 6. The surfce density of circulr disc of rdius depends upon the distnce fro the centre s P(r) + Br Therefore the ss of the ring of rdius r will be ( + Br) r dr r Therefore oent of inerti bout the centre will be 4 ( Br)r dr r dr Br dr 0 0 0 / 0 (r 4 /4) + B(r 5 /5) ] 0 4 [(/4) + (B/5)]. 7. t the highest point totl force cting on the prticle id its weight cting downwrd. nge of the prticle u sin / g Therefore force is t distnce, (totl rnge)/ (v sin )/g (Fro the initil point) Therefore F r ( ngle of projection) v sin /g (v initil velocity) (v sin) / v sin / v sin cos. 8. siple of pendulu of length l is suspended fro rigid support. bob of weight W is hnging on the other point. When the bob is t n ngle with the verticl, then totl torque cting on the point of I suspension i F r W r sin W l sin B t the lowest point of suspension the torque will be zero s the force cting on the body psses through the point of suspension. 9. force of 6 N cting t n ngle of 0 is just ble to loosen the wrench t distnce 8 c fro it. Therefore totl torque cting t bout the point 0 6 sin 0 (8/00) Therefore totl torque required t B bout the point 0 F 6/00 F 6/00 6 sin 0 8/00 F (8 ) / 6.5 N. 0. Torque bout point Totl force perpendiculr distnce fro the point to tht force. Let nticlockwise torque + ve nd clockwise cting torque ve Force cting t the point B is 5 N 0N E B Therefore torque t O due to this force 5 6 0 5N sin 7 8/5 5 6 0 6c /5 0.54 N- (nticlock wise) C 4c c Force cting t the point C is 0 N 4c 0 0 c Therefore, torque t O due to this force 0 4 0 0.4 N- (clockwise) 5N 0N Force cting t the point is 0 N Therefore, Torque t O due to this force 0 4 0 sin0 0 4 0 / 0.4 N- (nticlockwise) Therefore resultnt torque cting t O 0.54 0.4 + 0.4 0.54 N-. 0. y x y O r B x x Cy dx
. The force cting on the body hs two coponents sin nd cos nd the body will exert norl rection. Let Since nd cos pss through the centre of the cube, there will be no torque due to nd cos. The only torque will be produced by sin. i F r (r /) ( ges of the cube) i sin / sin cos / sin.. rod of ss nd length L, lying horizontlly, is free to rotte bout verticl xis pssing through its centre. force F is cting perpendiculr to the rod t distnce L/4 fro the centre. Therefore torque bout the centre due to this force i i F r FL/4. This torque will produce ngulr ccelertion. t sec Therefore c I c i c (L / ) (I c of rod L / ) B F i/4 (L / ) F/l F /4 Therefore / t (initilly t rest) B / (F / l)t (F/l)t.. squre plte of ss 0 g nd edge 5 c rottes bout one of the edge. Let tke sll re of the squre of width dx nd length which is t distnce x fro the xis of rottion. Therefore ss of tht sll re / dx ( ss of the squre ; side of the plte) I ( / ) x dx ( / )(x 0 / )] 0 / Therefore torque produced I ( /) {(0 0 5 0 4 )/} 0. 0. 0 4 0 5 N-. 4. Moent of inertil of squre plte bout its digonl is / ( ss of the squre plte) edges of the squre Therefore torque produced ( /) x x {(0 0 5 0 4 )/ 0. 0.5 0 5 N-. 5. flywheel of oent of inerti 5 kg is rotted t speed of 60 rd/s. The flywheel coes to rest due to the friction t the xle fter 5 inutes. Therefore, the ngulr decelertion produced due to frictionl force 0 + t 0 t ( 0+ (60/5 60) /5 rd/s. ) Therefore totl workdone in stopping the wheel by frictionl force /iw W / i / 5 (60 60) 9000 Joule 9 KJ. b) Therefore torque produced by the frictionl force () is I I 5 ( /5) IN opposite to the rottion of wheel. c) ngulr velocity fter 4 inutes 0 + t 60 40/5 rd/s Therefore ngulr oentu bout the centre 5 60 kg- /s. sin x cos dx 0.4
6. The erth s ngulr speed decreses by 0.006 rd/dy in 00 yers. Therefore the torque produced by the ocen wter in decresing erth s ngulr velocity I /5 r ( 0 )/t /6 6 0 4 64 0 0 [0.006 /(6400 00 65)] ( yer 65 dys 65 56400 sec) 5.678 0 0 N-. 7. wheel rotting t speed of 600 rp. 0 600 rp 0 revolutions per second. T 0 sec. (In 0 sec. it coes to rest) 0 Therefore 0 t 0/0 rev/s 0 + t 0 5 5 rev/s. Therefore ngulr deccelertion rev/s nd ngulr velocity of fter 5 sec is 5 rev/s. 8. 00 rev/in 5/8 rev/s 0/ rd/s 0 rev 0 rd, r 0. fter 0 revolutions the wheel will coe to rest by tngentil force Therefore the ngulr deccelertion produced by the force / Therefore the torque by which the wheel will coe to n rest I c F r I c F 0. / r [(0/) / ( 0)] F / 0 0. 00 / (9 0) 5 / 8 5.7/8 0.87 N. 9. cylinder is oving with n ngulr velocity 50 rev/s brought in contct with nother identicl cylinder in rest. The first nd second cylinder hs coon ccelertion nd deccelertion s rd/s respectively. Let fter t sec their ngulr velocity will be se. For the first cylinder 50 t t ( 50)/ 50 rev/s nd for the nd cylinder t t /I So, ( 50)/ 50 5 rev/s. t 5/ sec 5 sec. 0. Initil ngulr velocity 0 rd/s Therefore rd/s t / 0/ 0 sec Therefore 0 sec it will coe to rest. Since the se torque is continues to ct on the body it will produce se ngulr ccelertion nd since the initil kinetic energy the kinetic energy t instnt. So initil ngulr velocity ngulr velocity t tht instnt Therefore tie require to coe to tht ngulr velocity, t / 0/ 0 sec therefore tie required t + t 0 sec.. I net I net F r F r (r r ) 0 0.5 5 0 0.5 (5 (/) + (/) ) 5 7/4 60/7 8.57 rd/s.. In this proble the rod hs ss kg ) net I net 5 0 0.5 0 0.5 (5 (/) + (/) + /) kg 5 kg kg 5 kg 0.5
5 (.75 + 0.084) 500/(75 + 8.4) 500/8.4 8. rd/s (g 0) 8.0 rd/s (if g 9.8) b) T g T + g ( + g) (r + g) (8 0.5 + 9.8) 7.6 N on the first body. In the second body g T T g T 5(g ) 5(9.8 8 0.5) 9 N.. ccording to the question Mg T M () T () (T T ) /r () [becuse r] [T.r I(/r)] If we dd the eqution nd we will get Mg + (T T ) M + (4) Mg I/r M + (M + + I/r ) Mg Mg/(M + + I/r ) 4. I 0.0 kg- (Bigger pulley) r 0 c 0., sller pulley is light ss of the block, kg therefore T () T I/r () ( + I/r ) >( 9.8) / [ + (0./0.0)] 9.6 / 0.89 /s Therefore, ccelertion of the block 0.89 /s. 5. kg, i 0.0 kg-, r 5 c 0.05 i 0.0 kg-, r 0 c 0. Therefore T () (T T )r I () T r I () Substituting the vlue of T in the eqution (), we get (t I /r )r I (T I /r ) I /r T [(I /r ) + I /r )] Substituting the vlue of T in the eqution (), we get [(I /r ) + I /r )] [(I / r ) (I / r )] 9.8 0.6 /s (0./ 0.005) (0. / 0.0) T I /r 0.0 0.6 6. N. 0.0 6. ccording to the question Mg T M () (T T ) I/ (T T ) I/ () (T T ) I/ () T (4) By dding eqution () nd () we will get, (T T ) I/ (5) By dding eqution () nd (4) we will get 0.6 T T T M T T T kg 5 kg g g T T T T kg kg T T T T T T T Mg 0c M I r T T
+ Mg + (T T ) M + (6) Substituting the vlue for T T we will get Mg M + + I/ (M )G (M I/ ) 7. is light pulley nd B is the descending pulley hving I 0.0 kg nd r 0. Mss of the block kg ccording to the eqution T () (T T )r I () T g / T + T () T T T I/ 5/ nd T (becuse /) / B T 7/ g / + 7/ + I / r g I/r / + 9/ (/ r I) 98 5 + 4.5 98/9.5 0. s 8. g sin T () (T T ) I/r () T g sin () T T dding the equtions () nd () we will get g sin + (T T ) g sin ( + ) kg 4 kg ( )g sin ( + + /r ) ( )gsin 0.48 0.5 s 45 45. ( / r ) 9. 4 kg, kg Frictionl co-efficient between kg block nd surfce 0.5 T T 0 c 0. T T I 0.5 kg 4 kg g sin T () T ( g sin + g cos ) () cos (T T ) I/r dding eqution () nd () we will get g sin ( g sin + g cos ) + (T T ) + 45 45 4 9.8 ( / ) {( 9.8 ( / ) + 0.5 9.8 ( / ) } (4 + + 0.5/0.0) 7.80 (.90 + 6.95) 65 0.5 s. 40. ccording to the question 00 g, I, 0 g Therefore, (T r ) (T r ) ( f r g) 0 T 0.7 T 0. 0. g 0 7T T.9 () T + T 0. 9.8 + 0.0 9.8.56 () Fro the eqution () nd () we will get 0 T 0. T.08 N.04 N Therefore T.56.08.8. N. 4., 6g + 60 g 745 N 0 cos 7 6g 5 sin 7 + 60 g 8 sin 7 8 48g + 88 g 6g/8 4 N f Therefore / 4/745 0.55. T T 70c 00g 0g f 00kg 7 60g 6g 0.7
4. 0.54, 6g + ; 0 cos 7 6g 5 sin 7 + 8 sin 7 8 48g + 4/5 48g 4 / 5 8 0.54 4.0g 4 40 4 6g + 6 5 8 0.54 40 0.54 44 kg. 4. 60 kg, ldder length 6.5, height of the wll 6 Therefore torque due to the weight of the body ) 600 6.5 / sin i 600 6.5 / [ (6 / 6.5) ] 75 N-. b) 60 9.8 6.5 cos 60g sin 6.5/ 60 g tn 60 g (.5/)[becuse tn.5/6] (5/) g.5 N. 44. ccording to the question 8g F + F ; N N Since, Therefore F F F 8 g F 40 Let us tke torque bout the point B, we will get N 4 8 g 0.75. N (80 ) / (4 4) 5 N Therefore N 40 5 ( F 4.7 4 N. 45. od hs length L It kes n ngle with the floor The verticl wll hs height h cos () sin () cos (h/tn ) + sin h / cos (cos / sin )h + sin h / cos L / cos {(cos / sin )h sin h} cos sin / L / cos sin {(cos / sin )h sin h} L / L / cos.sin sin (cos h sin h) L cos sin cos.sin {(cos / sin)h sin h)} / cos L cos sin h L cos sin 46. unifor rod of ss 00 grs nd length 50 c rottes with n unifor ngulr velocity rd/s bout n xis perpendiculr to the rod through n end. ) L I I t the end L / (0. 0.5 }/ 0.05 kg- 0.05 0.05 kg /s b) Speed of the centre of the rod V r w (50/) 50 c/s 0.5 /s. c) Its kinetic energy / I (/) 0.05 0.05 Joule. f 7 60g 6g 6.5 F N B sin N 600 8g cos h 0.8
47. I 0.0 N-; 0 c 0. ; kg Therefore ( /) 0.0 N- 60 rd/s Therefore 0 + t 60 5 00 rd/s Therefore ngulr oentu I (0.0 / 60) 00 0.50 kg- /s nd 0 kinetic energy / I / (0.0 / 60) 00 75 Joules. 48. ngulr oentu of the erth bout its xis is /5 r ( / 85400) (becuse, I /5 r ) ngulr oentu of the erth bout sun s xis ( / 86400 65) (becuse, I ) / 5r ( / 86400) Therefore, rtio of the ngulr oentu /(86400 65) (r 65) / 5 (.990 0 0 ) / (.5 0 7 ).65 0 7. 49. ngulr oentu due to the ss t the centre of syste is r. r () ( ) Siilrly the ngulr oentu due to the ss t the centre of syste is r r ( ) Therefore net ngulr oentu ) () r ( ) ( )r r r ( ( ) 50. I F r (r + r ) 5 0.5 r.5 0 0.5 0.05 0.5 0 0 rd/s, t 0.0 sec, 0 + t 0 + 00 0 0 + rd/s. 5. wheel hs I 0.500 Kg-, r 0., 0 rd/s Sttionry prticle 0. kg Therefore I I (since externl torque 0) 0.5 0 (0.5 + 0. 0. ) 0/0.508 9.69 9.7 rd/s 5. I 6 kg-, rd/s, I 5 kg- Since externl torque 0 Therefore I I (6 ) / 5.4 rd/s 5. 0 rp 0 ( / 60) 4 rd /s. I 6 kg, I kg Since two blls re inside the syste Therefore, totl externl torque 0 Therefore, I I 6 4 rd/s 6 rev/s 60 rev/inute. 0.9 r ( ) (proved) r + l0.0n- r + 0.5kg 0.5kg r
54. I 0 kg- ; I 0 kg- ; rd/s Fro the erth reference the ubrell hs ngulr velocity ( ) nd the ngulr velocity of the n will be Therefore I ( ) I 0 ( ) 0 5 4 0.8 rd/s. 55. Wheel () hs I 0.0 kg-, 60 rev/in Wheel () hs I? ; 00 rev/in Given tht fter they re coupled, 00 rev/in Therefore if we tke the two wheels to ben isolted syste Totl externl torque 0 Therefore, I + I (I + I ) 0.0 60 + I 00 (0.0 + I ) 00 5I 0.8 I 0.04 kg-. 56. kid of ss M stnds t the edge of pltfor of rdius which hs oent of inerti I. bll of thrown to hi nd horizontl velocity of the bll v when he ctches it. Therefore if we tke the totl bodies s syste Therefore v {I + (M + ) } (The oent of inerti of the kid nd bll bout the xis (M + ) ) v. (M ) 57. Initil ngulr oentu Finl ngulr oentu (the totl externl torque 0) Initil ngulr oentu v ( ss of the bll, v velocity of the bll, rdius of pltfor) Therefore ngulr oentu I + M Therefore V I + M V. ( M ) 58. Fro inertil fre of reference when we see the (n wheel) syste, we cn find tht the wheel oving t speed of nd the n with ( + V/) fter the n hs strted wlking. ( ngulr velocity fter wlking, ngulr velocity of the wheel before wlking. Since I 0 Extended torque 0 Therefore ( + M ) I + ( + V/) (I + ) + I + + V V. ( ) 59. unifor rod of ss length l is struck t n end by force F. to the rod for short tie t ) Speed of the centre of ss v Ft v Ft Erth reference w fro erth V/ of n w.r.t. the pltfor b) The ngulr speed of the rod bout the centre of ss l r p (l / ) (/) v l / (/) l 6Ft / l c) K.E. (/) v + (/) l (/) (Ft / ) (/) l (/) ( F t / ) + (/) (l /) (6 ( F t / l )) 0.0
F t / + / (F t ) / F t / d) ngulr oentu bout the centre of ss :- L vr Ft / (/) F l t / 60. Let the ss of the prticle & the ss of the rod M Let the prticle strikes the rod with velocity V. If we tke the two body to be syste, Therefore the net externl torque & net externl force 0 Therefore pplying lws of conservtion of liner oentu MV V (V velocity of the rod fter striking) V / V / M gin pplying lws of conservtion of ngulr oentu M V l V M M t t V Therefore distnce trvelled :- M M V t V M 6. ) If we tke the two bodies s syste therefore totl externl force 0 pplying L.C.L.M :- V (M + ) v v v M b) Let the velocity of the prticle w.r.t. the centre of ss V 0 Mv Mv v v M M c) If the body oves towrds the rod with velocity of v, i.e. the rod is oving with velocity v towrds the prticle. Therefore the velocity of the rod w.r.t. the centre of ss V V M O v v M M d) The distnce of the centre of ss fro the prticle M l / O M l / (M ) (M ) Therefore ngulr oentu of the prticle before the collision l Mr c { l/) / (M + )} V/ (l/) (M vl) / (M + ) Distnce of the centre of ss fro the centre of ss of the rod M 0 (l / ) (l / ) c (M ) (M ) Therefore ngulr oentu of the rod bout the centre of ss MV c c M {( v) / (M + )} {(l/) / (M + )} M lv (M ) M lv (M ) (If we consider the nitude only) e) Moent of inerti of the syste M.I. due to rod + M.I. due to prticle v M, v 0.
Ml M(l / ) (M ) Ml (M 4). (M ) (Ml / s) (M ) f) Velocity of the centre of ss V M 0 V V (M ) (M ) (Velocity of centre of ss of the syste before the collision Velocity of centre of ss of the syste fter the collision) (Becuse Externl force 0) ngulr velocity of the syste bout the centre of ss, P c I c MV r v r I M c v l Mv Ml M (M ) (M ) (M ) (M ) Ml (M 4) (M ) M vl M (M ) vl Ml (M 4) (M ) M /(M ) Ml (M ) (M ) (M ) 6v (M 4) l 6. Since externl torque 0 Therefore I I l 4 l 4 l 4 l 4 l l 4 l I Therefore I l 4 6. Two blls & B, ech of ss re joined rigidly to the ends of light of rod of length L. The syste oves in velocity v 0 in direction to the rod. prticle P of ss kept t rest on the surfce sticks to the bll s the bll collides with it. ) The light rod will exert force on the bll B only long its length. So collision will not ffect its velocity. B hs velocity v 0 If we consider the three bodies to be syste v o pplying L.C.L.M. v L Therefore v 0 v v 0 v o L v Therefore hs velocity 0 b) if we consider the three bodies to be syste Therefore, net externl force 0 Therefore V c v v0 0 v 0 v 0 v 0 (long the initil velocity s before collision) 0.
c) The velocity of ( + P) w.r.t. the centre of ss 0 v 0 v v0 v The velocity of B w.r.t. the centre of ss v0 0 [Only nitude hs been tken] Distnce of the ( + P) fro centre of ss l/ & for B it is l/. Therefore P c l c v 0 6 v0 l 6v 0l 6l 8 9 v 0 l l 64. The syste is kept rest in the horizontl position nd prticle P flls fro height h nd collides with B nd sticks to it. Therefore, the velocity of the prticle before collision gh If we consider the two bodies P nd B to be syste. Net externl torque nd force 0 Therefore, v (gh) / gh v Therefore ngulr oentu of the rod just fter the collision (v r) ( g h ) / l / l (gh) / L l gh l ( l / 4 l gh / 4 ) l 8gh l b) When the ss will t the top ost position nd the ss t the lowest point, they will utoticlly rotte. In this position the totl gin in potentil energy (l/) (l/) (l/) Therefore l/ l/ l l/ (/ l ) / 4 (8gh / 9gl ) h l/. 65. ccording to the question 0.4g T 0.4 () T 0.g 0. () (T T )r I/r () Fro eqution, nd ( 0. 4 0. ) g g / 5 ( 0. 4 0.. 6 / 0. 4 ) v 0 6 & y x p,o v B 5 Therefore (b) V h ( gl 0.5) ( g / 5 ) ( 9. 8 / 5 ).4 /s. ) Totl kinetic energy of the syste / V + / V + / 8 (/ 0.4.4 ) + (/ 0..4 ) + (/ (.6/4).4 ) 0.98 Joule. 66. l 0. kg-, r 0., K 50 N/, kg, g 0 s, h 0. Therefore pplying lws of conservtion of energy h / v + / kx / V + / 0. V /0.04 + (/) 50 0.0 (x h) 0.5 v +.5 v + /4 v /4 v / 0.5 /s 0. T T 00g T T 400g
67. Let the ss of the rod Therefore pplying lws of conservtion of energy / l l/ / M l / / g / l g / l 5.4 rd/s. 68. / I 0 0. 0 0 For collision 0. 0 + 0 [(0.4/) + (0.) ] 0 /[0.(0.8)] 0 / g l ( cos ) g l/ ( cos ) 0. 0 ( cos ) 0.4 0 0.5 ( cos ) / 0.8 (0/.4).( cos ) ( cos ) /(..8) cos 0.5 cos 0.5 0.748 cos (0.748) 4. 69. Let l length of the rod, nd ss of the rod. pplying energy principle (/) l O (/) (cos 7 cos 60 ) l 4 t 5 9g g 0.9 0 l l l gin sin 7 l 5 g 0.9 ngulr ccelertion. l So, to find out the force on the prticle t the tip of the rod F i centrifugl force (d) l 0.9 (d) g F t tngentil force (d) l 0.9 (d) g So, totl force F i Ft F 0.9 (d) g 70. cylinder rolls in horizontl plne hving centre velocity 5 /s. t its ge the velocity is due to its rottion s well s due to its lenir otion & this two velocities re se nd cts in the se direction (v r ) Therefore Net velocity t 5 /s + 5 /s 50 /s 7. sphere hving ss rolls on plne surfce. Let its rdius. Its centre oves with velocity v Therefore Kinetic energy (/) l + (/) v v v v v 5 0 7. Let the rdius of the disc Therefore ccording to the question & figure Mg T () & the torque bout the centre T I T (/) / 5 v 7 v 0 0 0.kg 7 60 O 5 /s 0.4
T (/) Putting this vlue in the eqution () we get (/) / g/ 7. sll sphericl bll is relesed fro point t height on rough trck & the sphere does not slip. Therefore potentil energy it hs gined w.r.t the surfce will be converted to ngulr kinetic energy bout the centre & liner kinetic energy. Therefore h (/) l + (/) v h h + v 5 gh 5 v + v v 0 0 gh v gh 7 7 74. disc is set rolling with velocity V fro right to left. Let it hs ttined height h. Therefore (/) V + (/) l h (/) V + (/) (/) h (/) V + /4 V gh (/4) V gh h V 4 g 75. sphere is rolling in inclined plne with inclintion Therefore ccording to the principle Mgl sin (/) l + (/) v l sin /5 v + (/) v Gl sin 7/0 0 v gl sin 7 76. hollow sphere is relesed fro top of n inclined plne of inclintion. To prevent sliding, the body will ke only perfect rolling. In this condition, sin f () & torque bout the centre f f () Putting this vlue in eqution () we get h L sin L cos sin sin 5 g sin sin f 5 sin f 5 sin cos 5 sin 5 tn b) 5 tn ( cos ) gsin 0 c g sin 5 g sin 5 4 sin 0.5
t s c l 4gsin 5 5l gsin gin, t K.E. (/) v + (/) l (/) (s) +(/) l ( t ) 4gsin l + 5 4l sin l sin 7 l sin 5 40 8 v 77. Totl norl force + r ( r) (/) l + (/) v ( r) v 5 v 9 00 7 v ( r) v 0 g( r) 0 7 g sin 5l g sin 0 g( r) 7 Therefore totl norl force + 0 7 + r 7 7 78. t the top ost point v r v g( r) Let the sphere is thrown with velocity v Therefore pplying lws of conservtion of energy (/) v + (/) l ( r) + (/) v + (/) l 0 7 v g ( r) + 0 7 v v 0 g ( r) + g ( r) 7 7 v g( r) 7 79. ) Totl kinetic energy y (/) v + (/) l Therefore ccording to the question H (/) v + (/) l + ( + cos ) H ( + cos ) (/) v + (/) l (/) v + (/) l (H sin ) b) to find the ccelertion coponents (/) v + (/) l (H sin ) 7/0 v (H sin ) v 0 H g sin rdicl ccelertion 7 H +v /( r) sin v 0 g (H ) sin 7 dv 0 d v g cos dt 7 dt dv 5 d g cos dt 7 dt dv 5 g cos tngentil ccelertion dt 7 0.6
c) Norl force t 0 v 70 0 0. 6 0. 0 5N 000 7 0. Frictionl force :- 5 70 50 f - (g ) (0 0) 0.07 7 7 0 0.N 00 80. Let the cue strikes t height h bove the centre, for pure rolling, V c pplying lw of conservtion of ngulr oentu t point, v c h l 0 v c v c h h 8. unifor wheel of rdius is set into rottion bout its xis (cse-i) t n ngulr speed This rotting wheel is now plced on rough horizontl. Becuse of its friction t contct, the wheel ccelertes forwrd nd its rottion decelertes. s the rottion decelertes the frictionl force will ct bckwrd. If we consider the net oent t then it is zero. Therefore the net ngulr oentu before pure rolling & fter pure rolling reins constnt Before rolling the wheel ws only rotting round its xis. Therefore ngulr oentu l (/) M () ( st cse) fter pure rolling the velocity of the wheel let v Therefore ngulr oentu l c + (V ) (/) (V/) + V / V () v Becuse, Eq() nd () re equl Therefore, / V ½ ( nd cse) V / 8. The shell will ove with velocity nerly equl to v due to this otion frictionl force well ct in the bckground direction, for which fter soe tie the shell ttins pure rolling. If we consider oent bout, then it will be zero. Therefore, Net ngulr oentu v v bout before pure rolling net ngulr oentu fter pure rolling. Now, ngulr oentu before pure rolling bout M (V ) nd ngulr oentu fter pure rolling :- ( st cse) (/) M (V 0 / ) + M V 0 (V 0 velocity fter pure rolling) v o MV / MV 0 + MV 0 (5/) V 0 V V 0 V/ 5 ( nd cse) 8. Tking oent bout the centre of hollow sphere we will get x h v c F M F M gin, (/) t (Fro 0 t + (/) t ) t 8M F F c F I X (/) c t (/) 4 0.7
84. If we tke oent bout the centre, then F l f F /5 + () gin, F c () F c Putting the vlue c in eq() we get F 5 /5 (F + ) + F F 0.5 0 0.5 0 5 5 7 F 4 0 5 7 7 F 5 0. N 85. ) if we tke oent t then externl torque will be zero Therefore, the initil ngulr oentu the ngulr oentu fter rottion stops (i.e. only lenir velocity exits) MV l MV O MV /5 M V / MV O V O V/5 wv/ v b) gin, fter soe tie pure rolling strts therefore M v o (/5) M (V/) + MV (V/5) (/5) MV + MV V V/7 86. When the solid sphere collides with the wll, it rebounds with velocity v towrds left but it continues to rotte in the clockwise direction. So, the ngulr oentu v (/5) v/ fter rebounding, when pure rolling strts let the velocity be v v nd the corresponding ngulr velocity is v / Therefore ngulr oentu v + (/5) (v/) So, v (/5), v/ v + (/5) (v/) v (/5) v (7/5) v v/7 So, the sphere will ove with velocity v/7. v V/ F I * * * * 0.8