Title Notes_Exercise_07.Circular Motion.pdf ✅

7.1 SOLUTIONS TO CONCEPTS circular motion;; CHAPTER 7 1. Distance between Earth & Moon r = 3.85 × 105 km = 3.85 × 108 m T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 106 sec v = T 2r = 6 8 .2 36 10 2 .3 14 .3 85 10     = 1025.42m/sec a = r v2 = 8 2 .3 85 10 (1025.42)  = 0.00273m/sec2 = 2.73 × 10–3m/sec2 2. Diameter of earth = 12800km Radius R = 6400km = 64 × 105 m V = T 2R = 24 3600 2 .3 14 64 105     m/sec = 465.185 a = R V2 = 5 2 64 10 (46.5185)  = 0.0338m/sec2 3. V = 2t, r = 1cm a) Radial acceleration at t = 1 sec. a = r v2 = 1 22 = 4cm/sec2 b) Tangential acceleration at t = 1sec. a = dt dv = )t2( dt d = 2cm/sec2 c) Magnitude of acceleration at t = 1sec a = 2 2 4  2 = 20 cm/sec2 4. Given that m = 150kg, v= 36km/hr = 10m/sec, r = 30m Horizontal force needed is r mv2 = 30 150 (10) 2  = 30 150 100 = 500N 5. in the diagram R cos  = mg ..(i) R sin  = r mv2 ..(ii) Dividing equation (i) with equation (ii) Tan  = rmg mv2 = rg v2 v = 36km/hr = 10m/sec, r = 30m Tan  = rg v2 = 30 10 100  = (1/3)   = tan–1(1/3) 6. Radius of Park = r = 10m speed of vehicle = 18km/hr = 5 m/sec Angle of banking tan = rg v2   = tan –1 rg v2 = tan–1 100 25 = tan –1(1/4) R mv2 /R mg Page 1 CIRCULAR MOTION
Chapter 7 7.2 7. The road is horizontal (no banking) R mv2 = N and N = mg So R mv 2 =  mg v = 5m/sec, R = 10m  10 25 = g   = 100 25 = 0.25 8. Angle of banking =  = 30° Radius = r = 50m tan  = rg v2  tan 30° = rg v2  3 1 = rg v2  v2 = 3 rg = 3 50 10  v = 3 500 = 17m/sec. 9. Electron revolves around the proton in a circle having proton at the centre. Centripetal force is provided by coulomb attraction. r = 5.3 t 10–11m m = mass of electron = 9.1 × 10–3kg. charge of electron = 1.6 × 10–19c. r mv2 = 2 2 r q k  v 2 = rm kq2 = 11 31 9 38 3.5 10 1.9 10 9 10 6.1 6.1 10           = 13 10 48.23 23.04   v 2 = 0.477 × 1013 = 4.7 × 1012  v = 12 7.4 10 = 2.2 × 106 m/sec 10. At the highest point of a vertical circle R mv 2 = mg  v 2 = Rg  v = Rg 11. A celling fan has a diameter = 120cm. Radius = r = 60cm = 0/6m Mass of particle on the outer end of a blade is 1g. n = 1500 rev/min = 25 rev/sec  = 2 n = 2  ×25 = 157.14 Force of the particle on the blade = Mr2 = (0.001) × 0.6 × (157.14) = 14.8N The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle also exerts a force of 14.8N on the blade along its surface. 12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at 3 1 33 rpm. n = 3 1 33 rpm = 3 60 100  rps  = 2  n = 2  × 180 100 = 9 10 rad/sec r = 10cm =0.1m, g = 10m/sec2 mg  mr2   = g r 2   10 9 10 1.0 2            81 2  R mg g mv2 /R Page 2 CIRCULAR MOTION
Chapter 7 7.3 13. A pendulum is suspended from the ceiling of a car taking a turn r = 10m, v = 36km/hr = 10 m/sec, g = 10m/sec2 From the figure T sin  = r mv2 ..(i) T cos  = mg ..(ii)    cos sin = rmg mv2  tan  = rg v2   = tan –1         rg v2 = tan–1 10 10 100  = tan–1(1)   = 45° 14. At the lowest pt. T = mg + r mv 2 Here m = 100g = 1/10 kg, r = 1m, v = 1.4 m/sec T = mg + r mv2 = 10 )4.1( 8.9 10 1 2   = 0.98 + 0.196 = 1.176 = 1.2 N 15. Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian. m = 100g = 0.1kg, r = 1m, v = 1.4m/sec. From the diagram, T – mg cos  = R mv 2  T = R mv2 + mg cos   T =               2 )1.0( 8.9 1 1 1.0 )4.1( 2 2  T = 0.196 + 9.8 ×          2 (. )2 1 2 ( cos  = 2 1 2   for small )  T = 0.196 + (0.98) × (0.98) = 0.196 + 0.964 = 1.156N  1.16 N 16. At the extreme position, velocity of the pendulum is zero. So there is no centrifugal force. So T = mg cos o 17. a) Net force on the spring balance. R = mg – m2 r So, fraction less than the true weight (3mg) is = mg mg (mg m )r2    = g 2  = 10 6400 10 24 3600 2 2 3           = 3.5 × 10–3 b) When the balance reading is half the true weight, mg mg (mg m )r2    = 1/2 2 r = g/2   r2 g  3 2 6400 10 10   rad/sec  Duration of the day is T =  2 = 8.9 2 6400 10 2 3    sec = 49 64 10 2 6   sec = 7 3600 2 8000   hr = 2hr mv2 /R mg  mg mv2 /r T mg sin  mg cos  T mg sin  mg cos  T m2 /R mg R Page 3 CIRCULAR MOTION
Chapter 7 7.4 18. Given, v = 36km/hr = 10m/s, r = 20m,  = 0.4 The road is banked with an angle,  = tan –1         rg v2 = tan –1       20 10 100 = tan–1       2 1 or tan  = 0.5 When the car travels at max. speed so that it slips upward, R1 acts downward as shown in Fig.1 So, R1 – mg cos  – r mv 2 1 sin  = 0 ..(i) And R1 + mg sin  – r mv 2 1 cos  = 0 ..(ii) Solving the equation we get, V1 =       1 tan tan rg = 2.1 1.0 20 10  = 4.082 m/s = 14.7 km/hr So, the possible speeds are between 14.7 km/hr and 54km/hr. 19. R = radius of the bridge L = total length of the over bridge a) At the highest pt. mg = R mv 2  v 2 = Rg  v = Rg b) Given, v = Rg 2 1 suppose it loses contact at B. So, at B, mg cos  = R mv 2  v 2 = Rg cos   2 2 Rv         = Rg cos   2 Rg = Rg cos   cos  = 1/2   = 60° = /3  = r   l = r = 3 R So, it will lose contact at distance 3 R from highest point c) Let the uniform speed on the bridge be v. The chances of losing contact is maximum at the end of the bridge for which  = 2R L . So, R mv2 = mg cos   v =       2R L gRcos 20. Since the motion is nonuniform, the acceleration has both radial & tangential component ar = r v 2 at = dt dv =a Resultant magnitude = 2 2 2 a r v          Now N = m 2 2 2 a r v            mg = m 2 2 2 a r v           2 g2 = 2 4 a 2r v          v 4 = (2 g2 – a2 ) r2  v = [(2 g2 – a2 ) r2 ] 1/4  R1  mv1 2 /r mg R1  mv2 2 /r R2  mg R2  2 mv2 /R mg 2= L/R  mv2 /R mg 2 2= L/R  mv2 /R 2 2= L/R mv2  mg /R m dv/dt m N mv2 /R m Page 4 CIRCULAR MOTION