Niraj Sir. circular motion;; SOLUTIONS TO CONCEPTS CHAPTER 7

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Tabitha Holmes
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1 SOLUIONS O CONCEPS CHAPE 7 cicula otion;;. Distance between Eath & Moon k days (7.) sec sec v /sec v (05.4) a 0.007/sec.7 0 /sec Diaete of eath 800k adius 6400k V /sec V (46.585) a 0.08/sec V t, c a) adial acceleation at t sec. v a 4c/sec b) angential acceleation at t sec. dv d a (t) c/sec dt dt c) Magnitude of acceleation at t sec a 4 0 c/sec 4. Given that 50kg, v 6k/h 0/sec, 0 Hoizontal foce needed is 5. in the diaga cos..(i) Niaj Si v 50 (0) 0 v sin..(ii) Dividing equation (i) with equation (ii) v an v g v 6k/h 0/sec, v an g (/) tan (/) 6. adius of Pak 0 speed of vehicle 8k/h 5 /sec v Angle of banking tan g N 0 v / v 5 tan g tan tan (/4) 7.

2 Chapte 7 7. he oad is hoizontal (no banking) v N and N g v / So v v 5/sec, g Angle of banking 0 adius 50 v tan g v v tan 0 g v v g g /sec. 9. Electon evolves aound the poton in a cicle having poton at the cente. Centipetal foce is povided by coulob attaction. 5. t 0 ass of electon 9. 0 kg. chage of electon c. v q k v kq v v /sec 0. At the highest point of a vetical cicle v v g v g Niaj Si. A celling fan has a diaete 0c. adius 60c 0/6 Mass of paticle on the oute end of a blade is g. n 500 ev/in 5 ev/sec n Foce of the paticle on the blade M (0.00) 0.6 (57.4) 4.8N he fan uns at a full speed in cicula path. his exets the foce on the paticle (inetia). he paticle also exets a foce of 4.8N on the blade along its suface.. A osquito is sitting on an L.P. ecod disc & otating on a tun table at p. n p ps 60 0 n ad/sec c 0., g 0/sec g 09 7.

3 Chapte 7. A pendulu is suspended fo the ceiling of a ca taking a tun 0, v 6k/h 0 /sec, g 0/sec v Fo the figue sin cos sin cos 4. At the lowest pt...(i)..(ii) v v tan tan g v g tan 0 0 tan () 45 v + Hee g /0 kg,, v.4 /sec + v (. 4 ) N 5. Bob has a velocity.4/sec, when the sting akes an angle of 0. adian. g 0.kg,, v.4/sec. Fo the diaga, cos v 0. (. 4 ) v + cos ( 0. ) (.) ( cos fo sall ) Niaj Si (0.98) (0.98) N.6 N 6. At the extee position, velocity of the pendulu is zeo. So thee is no centifugal foce. So cos o 7. a) Net foce on the sping balance. So, faction less than the tue weight () is ( ) g b) When the balance eading is half the tue weight, ( ) / g/ Duation of the day is g 0 ad/sec sec sec h h v / v / sin cos sin cos / 7.

4 Chapte 7 8. Given, v 6k/h 0/s, 0, 0.4 he oad is banked with an angle, tan g tan 0 0 o tan 0.5 When the ca tavels at ax. speed so that it slips upwad, acts downwad as shown in Fig. v So, cos sin 0..(i) v And + sin cos 0..(ii) Solving the equation we get, V tan g tan /s 4.7 k/h. So, the possible speeds ae between 4.7 k/h and 54k/h. 9. adius of the bidge L total length of the ove bidge a) At the highest pt. v v g v g b) Given, v g v suppose it loses contact at B. So, at B, cos v g cos v g g cos g cos cos / 60 / Niaj Si l So, it will lose contact at distance fo highest point c) Let the unifo speed on the bidge be v. he chances of losing contact is axiu at the end of the bidge fo which So, v L cos v g cos L. 0. Since the otion is nonunifo, the acceleation has both adial & tangential coponent a v v / v / v / L/ v / L/ v / L/ dv a t a dt v / v / esultant agnitude v a dv/dt Now N v a v 4 ( g a ) v [( g a ) ] /4 v a g 4 v a N 7.4

5 Chapte 7. a) When the ule akes unifo cicula otion in the hoizontal plane, (fig a) L g L b) When the ule akes unifoly acceleated cicula otion,(fig b) L) (L ) ( 4 +. adius of the cuves Weight kg Velocity 8k/h 5/sec L g g L Niaj Si 7.5 / 4 (When viewed fo top) v 5 a) at B N N ( 0) N v At d, N N A b) At B & D the cycle has no tendency to slide. So at B & D, fictional foce is zeo. At C, sin F F 0 707N v v c) (i) Befoe C cos N N cos N v v (ii) N cos N + cos N d) o find out the iniu desied coeff. of fiction, we have to conside a point just befoe C. (whee N is iniu) Now, N sin So,.07. d.5 distance fo the cente to one of the kids N 0 ev pe in 0/60 / ev pe sec F / 5kg 5kg 5kg Fictional foce F () 5 (.5) 5 (0.5) Fictional foce on one of the kids is 0 4. If the bowl otates at axiu angula speed, the block tends to slip upwads. So, the fictional foce acts downwad. Hee, sin Fo FBD cos ( sin ) sin 0..(i) [because sin ] and sin ( sin ) cos 0..(ii) Substituting the value of fo Eq (i) in Eq(ii), it can be found out that / g(sin cos ) sin (cos sin ) Again, fo iniu speed, the fictional foce acts upwad. Fo FBD, it can be poved that, (FBD ) L B (FBD ) L C v / B N L (Fig a) L D (Fig b) E

6 Chapte 7 / g(sin cos ) sin (cos sin ) the ange of speed is between and 5. Paticle is pojected with speed u at an angle. At the highest pt. the vetical coponent of velocity is 0 So, at that point, velocity u cos v / centipetal foce u cos At highest pt. v u cos g 6. Let u the velocity at the pt whee it akes an angle / with hoizontal. he hoizontal coponent eains unchanged ucos So, v cos / cos v...(i) cos Fo figue v v cos (/) gcos / putting the value of v fo equn(i) u cos gcos ( / ) 7. A block of ass oves on a hoizontal cicle against the wall of a cylindical oo of adius Fiction coefficient between wall & the block is. a) Noal eaction by the wall on the block is b) Fictional foce by wall c) v dv d) Now, dt s v a a dv v v ds In V + c v v Niaj Si (Deceleation) ds dv v u sin cos v / / u cos cos/ v / At s 0, v v 0 heefoe, c In V0 v / so, s In v v 0 v v 0 e s/ Fo, one otation s, so v v 0 e 8. he cabin otates with angula velocity & adius he paticle expeiences a foce. he coponent of along the goove povides the equied foce to the paticle to ove along AB. cos a a cos B length of goove L L ut + ½ at L ½ cos t A t L L t cos cos 7.6 v /

7 9. v Velocity of ca 6k/h 0 /s adius of cicula path 50 ass of sall body g 0.kg. Fiction coefficient between plate & body 0.58 a) he noal contact foce exeted by the plate on the block Chapte 7 v 0. N 0.N 50 b) he plate is tuned so the angle between the noal to the plate & the adius of the oad slowly inceases N v cos..(i) v N sin..(ii) Putting value of N fo (i) v v cos sin tan tan tan (0.58) 0 0. Let the bigge ass acceleates towads ight with a. Fo the fee body diagas, a 0 (i) + a 0 (ii) Eq (i) Eq (ii) a a Substituting the value of a in Equation (i), we get 4/. * * * * a a Niaj Si a 7.7

CHAPTER 5: Circular Motion; Gravitation

CHAPTER 5: Circular Motion; Gravitation CHAPER 5: Cicula Motion; Gavitation Solution Guide to WebAssign Pobles 5.1 [1] (a) Find the centipetal acceleation fo Eq. 5-1.. a R v ( 1.5 s) 1.10 1.4 s (b) he net hoizontal foce is causing the centipetal