Page 1 7.1 SOLUTIONS TO CONCEPTS circular motion;; CHAPTER 7 1. Distance between Earth & Moon r = 3.85 × 10 5 km = 3.85 × 10 8 m T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10 6 sec v = T r 2 ? = 6 8 10 36 . 2 10 85 . 3 14 . 3 2 ? ? ? ? = 1025.42m/sec a = r v 2 = 8 2 10 85 . 3 ) 42 . 1025 ( ? = 0.00273m/sec 2 = 2.73 × 10 –3 m/sec 2 2. Diameter of earth = 12800km Radius R = 6400km = 64 × 10 5 m V = T R 2 ? = 3600 24 10 64 14 . 3 2 5 ? ? ? ? m/sec = 465.185 a = R V 2 = 5 2 10 64 ) 5185 . 46 ( ? = 0.0338m/sec 2 3. V = 2t, r = 1cm a) Radial acceleration at t = 1 sec. a = r v 2 = 1 2 2 = 4cm/sec 2 b) Tangential acceleration at t = 1sec. a = dt dv = ) t 2 ( dt d = 2cm/sec 2 c) Magnitude of acceleration at t = 1sec a = 2 2 2 4 ? = 20 cm/sec 2 4. Given that m = 150kg, v= 36km/hr = 10m/sec, r = 30m Horizontal force needed is r mv 2 = 30 ) 10 ( 150 2 ? = 30 100 150 ? = 500N 5. in the diagram R cos ? = mg ..(i) R sin ? = r mv 2 ..(ii) Dividing equation (i) with equation (ii) Tan ? = rmg mv 2 = rg v 2 v = 36km/hr = 10m/sec, r = 30m Tan ? = rg v 2 = 10 30 100 ? = (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 v 2 ? ? = tan –1 rg v 2 = tan –1 100 25 = tan –1 (1/4) R mv 2 /R mg Page 2 7.1 SOLUTIONS TO CONCEPTS circular motion;; CHAPTER 7 1. Distance between Earth & Moon r = 3.85 × 10 5 km = 3.85 × 10 8 m T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10 6 sec v = T r 2 ? = 6 8 10 36 . 2 10 85 . 3 14 . 3 2 ? ? ? ? = 1025.42m/sec a = r v 2 = 8 2 10 85 . 3 ) 42 . 1025 ( ? = 0.00273m/sec 2 = 2.73 × 10 –3 m/sec 2 2. Diameter of earth = 12800km Radius R = 6400km = 64 × 10 5 m V = T R 2 ? = 3600 24 10 64 14 . 3 2 5 ? ? ? ? m/sec = 465.185 a = R V 2 = 5 2 10 64 ) 5185 . 46 ( ? = 0.0338m/sec 2 3. V = 2t, r = 1cm a) Radial acceleration at t = 1 sec. a = r v 2 = 1 2 2 = 4cm/sec 2 b) Tangential acceleration at t = 1sec. a = dt dv = ) t 2 ( dt d = 2cm/sec 2 c) Magnitude of acceleration at t = 1sec a = 2 2 2 4 ? = 20 cm/sec 2 4. Given that m = 150kg, v= 36km/hr = 10m/sec, r = 30m Horizontal force needed is r mv 2 = 30 ) 10 ( 150 2 ? = 30 100 150 ? = 500N 5. in the diagram R cos ? = mg ..(i) R sin ? = r mv 2 ..(ii) Dividing equation (i) with equation (ii) Tan ? = rmg mv 2 = rg v 2 v = 36km/hr = 10m/sec, r = 30m Tan ? = rg v 2 = 10 30 100 ? = (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 v 2 ? ? = tan –1 rg v 2 = tan –1 100 25 = tan –1 (1/4) R mv 2 /R mg Chapter 7 7.2 7. The road is horizontal (no banking) R mv 2 = ?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 v 2 ? tan 30° = rg v 2 ? 3 1 = rg v 2 ? v 2 = 3 rg = 3 10 50 ? ? 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 –11 m m = mass of electron = 9.1 × 10 –3 kg. charge of electron = 1.6 × 10 –19 c. r mv 2 = 2 2 r q k ? v 2 = rm kq 2 = 31 11 38 9 10 1 . 9 10 3 . 5 10 6 . 1 6 . 1 10 9 ? ? ? ? ? ? ? ? ? ? = 13 10 23 . 48 04 . 23 ? ? v 2 = 0.477 × 10 13 = 4.7 × 10 12 ? v = 12 10 7 . 4 ? = 2.2 × 10 6 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 = 60 3 100 ? rps ?? = 2 ? n = 2 ? × 180 100 = 9 10 ? rad/sec r = 10cm =0.1m, g = 10m/sec 2 ?mg ? mr ? 2 ? ? = g r 2 ? ? 10 9 10 1 . 0 2 ? ? ? ? ? ? ? ? ? ? ? 81 2 ? ? R mg ??g mv 2 /R Page 3 7.1 SOLUTIONS TO CONCEPTS circular motion;; CHAPTER 7 1. Distance between Earth & Moon r = 3.85 × 10 5 km = 3.85 × 10 8 m T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10 6 sec v = T r 2 ? = 6 8 10 36 . 2 10 85 . 3 14 . 3 2 ? ? ? ? = 1025.42m/sec a = r v 2 = 8 2 10 85 . 3 ) 42 . 1025 ( ? = 0.00273m/sec 2 = 2.73 × 10 –3 m/sec 2 2. Diameter of earth = 12800km Radius R = 6400km = 64 × 10 5 m V = T R 2 ? = 3600 24 10 64 14 . 3 2 5 ? ? ? ? m/sec = 465.185 a = R V 2 = 5 2 10 64 ) 5185 . 46 ( ? = 0.0338m/sec 2 3. V = 2t, r = 1cm a) Radial acceleration at t = 1 sec. a = r v 2 = 1 2 2 = 4cm/sec 2 b) Tangential acceleration at t = 1sec. a = dt dv = ) t 2 ( dt d = 2cm/sec 2 c) Magnitude of acceleration at t = 1sec a = 2 2 2 4 ? = 20 cm/sec 2 4. Given that m = 150kg, v= 36km/hr = 10m/sec, r = 30m Horizontal force needed is r mv 2 = 30 ) 10 ( 150 2 ? = 30 100 150 ? = 500N 5. in the diagram R cos ? = mg ..(i) R sin ? = r mv 2 ..(ii) Dividing equation (i) with equation (ii) Tan ? = rmg mv 2 = rg v 2 v = 36km/hr = 10m/sec, r = 30m Tan ? = rg v 2 = 10 30 100 ? = (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 v 2 ? ? = tan –1 rg v 2 = tan –1 100 25 = tan –1 (1/4) R mv 2 /R mg Chapter 7 7.2 7. The road is horizontal (no banking) R mv 2 = ?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 v 2 ? tan 30° = rg v 2 ? 3 1 = rg v 2 ? v 2 = 3 rg = 3 10 50 ? ? 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 –11 m m = mass of electron = 9.1 × 10 –3 kg. charge of electron = 1.6 × 10 –19 c. r mv 2 = 2 2 r q k ? v 2 = rm kq 2 = 31 11 38 9 10 1 . 9 10 3 . 5 10 6 . 1 6 . 1 10 9 ? ? ? ? ? ? ? ? ? ? = 13 10 23 . 48 04 . 23 ? ? v 2 = 0.477 × 10 13 = 4.7 × 10 12 ? v = 12 10 7 . 4 ? = 2.2 × 10 6 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 = 60 3 100 ? rps ?? = 2 ? n = 2 ? × 180 100 = 9 10 ? rad/sec r = 10cm =0.1m, g = 10m/sec 2 ?mg ? mr ? 2 ? ? = g r 2 ? ? 10 9 10 1 . 0 2 ? ? ? ? ? ? ? ? ? ? ? 81 2 ? ? R mg ??g mv 2 /R 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/sec 2 From the figure T sin ? = r mv 2 ..(i) T cos ? = mg ..(ii) ? ? ? cos sin = rmg mv 2 ? tan ? = rg v 2 ? ? = tan –1 ? ? ? ? ? ? ? ? rg v 2 = 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 mv 2 = 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 mv 2 + mg cos ? ? T = ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 1 8 . 9 ) 1 . 0 ( 1 ) 4 . 1 ( 1 . 0 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 ) r m mg ( mg 2 ? ? ? = g 2 ? = 10 10 6400 3600 24 2 3 2 ? ? ? ? ? ? ? ? ? ? = 3.5 × 10 –3 b) When the balance reading is half the true weight, ? mg ) r m mg ( mg 2 ? ? ? = 1/2 ? 2 r = g/2 ? ???? r 2 g ??? 3 10 6400 2 10 ? ? rad/sec ? Duration of the day is T = ? ? 2 = 8 . 9 10 6400 2 2 3 ? ? ? ? sec = 49 10 64 2 6 ? ? ? sec = 3600 7 8000 2 ? ? ? hr = 2hr mv 2 /R mg ? ? mg ? mv 2 /r ? T mg sin ? ? mg cos ? ? T mg sin ? ? mg cos ? ? T m ? 2 /R mg R Page 4 7.1 SOLUTIONS TO CONCEPTS circular motion;; CHAPTER 7 1. Distance between Earth & Moon r = 3.85 × 10 5 km = 3.85 × 10 8 m T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10 6 sec v = T r 2 ? = 6 8 10 36 . 2 10 85 . 3 14 . 3 2 ? ? ? ? = 1025.42m/sec a = r v 2 = 8 2 10 85 . 3 ) 42 . 1025 ( ? = 0.00273m/sec 2 = 2.73 × 10 –3 m/sec 2 2. Diameter of earth = 12800km Radius R = 6400km = 64 × 10 5 m V = T R 2 ? = 3600 24 10 64 14 . 3 2 5 ? ? ? ? m/sec = 465.185 a = R V 2 = 5 2 10 64 ) 5185 . 46 ( ? = 0.0338m/sec 2 3. V = 2t, r = 1cm a) Radial acceleration at t = 1 sec. a = r v 2 = 1 2 2 = 4cm/sec 2 b) Tangential acceleration at t = 1sec. a = dt dv = ) t 2 ( dt d = 2cm/sec 2 c) Magnitude of acceleration at t = 1sec a = 2 2 2 4 ? = 20 cm/sec 2 4. Given that m = 150kg, v= 36km/hr = 10m/sec, r = 30m Horizontal force needed is r mv 2 = 30 ) 10 ( 150 2 ? = 30 100 150 ? = 500N 5. in the diagram R cos ? = mg ..(i) R sin ? = r mv 2 ..(ii) Dividing equation (i) with equation (ii) Tan ? = rmg mv 2 = rg v 2 v = 36km/hr = 10m/sec, r = 30m Tan ? = rg v 2 = 10 30 100 ? = (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 v 2 ? ? = tan –1 rg v 2 = tan –1 100 25 = tan –1 (1/4) R mv 2 /R mg Chapter 7 7.2 7. The road is horizontal (no banking) R mv 2 = ?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 v 2 ? tan 30° = rg v 2 ? 3 1 = rg v 2 ? v 2 = 3 rg = 3 10 50 ? ? 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 –11 m m = mass of electron = 9.1 × 10 –3 kg. charge of electron = 1.6 × 10 –19 c. r mv 2 = 2 2 r q k ? v 2 = rm kq 2 = 31 11 38 9 10 1 . 9 10 3 . 5 10 6 . 1 6 . 1 10 9 ? ? ? ? ? ? ? ? ? ? = 13 10 23 . 48 04 . 23 ? ? v 2 = 0.477 × 10 13 = 4.7 × 10 12 ? v = 12 10 7 . 4 ? = 2.2 × 10 6 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 = 60 3 100 ? rps ?? = 2 ? n = 2 ? × 180 100 = 9 10 ? rad/sec r = 10cm =0.1m, g = 10m/sec 2 ?mg ? mr ? 2 ? ? = g r 2 ? ? 10 9 10 1 . 0 2 ? ? ? ? ? ? ? ? ? ? ? 81 2 ? ? R mg ??g mv 2 /R 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/sec 2 From the figure T sin ? = r mv 2 ..(i) T cos ? = mg ..(ii) ? ? ? cos sin = rmg mv 2 ? tan ? = rg v 2 ? ? = tan –1 ? ? ? ? ? ? ? ? rg v 2 = 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 mv 2 = 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 mv 2 + mg cos ? ? T = ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 1 8 . 9 ) 1 . 0 ( 1 ) 4 . 1 ( 1 . 0 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 ) r m mg ( mg 2 ? ? ? = g 2 ? = 10 10 6400 3600 24 2 3 2 ? ? ? ? ? ? ? ? ? ? = 3.5 × 10 –3 b) When the balance reading is half the true weight, ? mg ) r m mg ( mg 2 ? ? ? = 1/2 ? 2 r = g/2 ? ???? r 2 g ??? 3 10 6400 2 10 ? ? rad/sec ? Duration of the day is T = ? ? 2 = 8 . 9 10 6400 2 2 3 ? ? ? ? sec = 49 10 64 2 6 ? ? ? sec = 3600 7 8000 2 ? ? ? hr = 2hr mv 2 /R mg ? ? mg ? mv 2 /r ? T mg sin ? ? mg cos ? ? T mg sin ? ? mg cos ? ? T m ? 2 /R mg R 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 v 2 = tan –1 ? ? ? ? ? ? ?10 20 100 = tan –1 ? ? ? ? ? ? 2 1 or tan ? = 0.5 When the car travels at max. speed so that it slips upward, ?R 1 acts downward as shown in Fig.1 So, R 1 – mg cos ? – r mv 2 1 sin ? = 0 ..(i) And ?R 1 + mg sin ? – r mv 2 1 cos ? = 0 ..(ii) Solving the equation we get, V 1 = ? ? ? ? ? ? tan 1 tan rg = 2 . 1 1 . 0 10 20 ? ? = 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 ? = R 2 L . So, R mv 2 = mg cos ? ? v = ? ? ? ? ? ? R 2 L cos gR ? 20. Since the motion is nonuniform, the acceleration has both radial & tangential component a r = r v 2 a t = 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 g 2 = 2 4 a 2 r v ? ? ? ? ? ? ? ? ? ? v 4 = ( ? 2 g 2 – a 2 ) r 2 ? v = [( ? 2 g 2 – a 2 ) r 2 ] 1/4 ? ? ?R 1 ? ? mv 1 2 /r mg R 1 ? ? mv 2 2 /r ?R 2 ? ? mg R 2 ? ? 2 ? ? mv 2 /R mg 2 ?= L/R ? ? ? mv 2 /R mg 2 ? ? 2 ?= L/R ? ? ? mv 2 /R 2 ? ? 2 ?= L/R ? mv 2 /R ? mg m dv/dt m N mv 2 /R m Page 5 7.1 SOLUTIONS TO CONCEPTS circular motion;; CHAPTER 7 1. Distance between Earth & Moon r = 3.85 × 10 5 km = 3.85 × 10 8 m T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10 6 sec v = T r 2 ? = 6 8 10 36 . 2 10 85 . 3 14 . 3 2 ? ? ? ? = 1025.42m/sec a = r v 2 = 8 2 10 85 . 3 ) 42 . 1025 ( ? = 0.00273m/sec 2 = 2.73 × 10 –3 m/sec 2 2. Diameter of earth = 12800km Radius R = 6400km = 64 × 10 5 m V = T R 2 ? = 3600 24 10 64 14 . 3 2 5 ? ? ? ? m/sec = 465.185 a = R V 2 = 5 2 10 64 ) 5185 . 46 ( ? = 0.0338m/sec 2 3. V = 2t, r = 1cm a) Radial acceleration at t = 1 sec. a = r v 2 = 1 2 2 = 4cm/sec 2 b) Tangential acceleration at t = 1sec. a = dt dv = ) t 2 ( dt d = 2cm/sec 2 c) Magnitude of acceleration at t = 1sec a = 2 2 2 4 ? = 20 cm/sec 2 4. Given that m = 150kg, v= 36km/hr = 10m/sec, r = 30m Horizontal force needed is r mv 2 = 30 ) 10 ( 150 2 ? = 30 100 150 ? = 500N 5. in the diagram R cos ? = mg ..(i) R sin ? = r mv 2 ..(ii) Dividing equation (i) with equation (ii) Tan ? = rmg mv 2 = rg v 2 v = 36km/hr = 10m/sec, r = 30m Tan ? = rg v 2 = 10 30 100 ? = (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 v 2 ? ? = tan –1 rg v 2 = tan –1 100 25 = tan –1 (1/4) R mv 2 /R mg Chapter 7 7.2 7. The road is horizontal (no banking) R mv 2 = ?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 v 2 ? tan 30° = rg v 2 ? 3 1 = rg v 2 ? v 2 = 3 rg = 3 10 50 ? ? 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 –11 m m = mass of electron = 9.1 × 10 –3 kg. charge of electron = 1.6 × 10 –19 c. r mv 2 = 2 2 r q k ? v 2 = rm kq 2 = 31 11 38 9 10 1 . 9 10 3 . 5 10 6 . 1 6 . 1 10 9 ? ? ? ? ? ? ? ? ? ? = 13 10 23 . 48 04 . 23 ? ? v 2 = 0.477 × 10 13 = 4.7 × 10 12 ? v = 12 10 7 . 4 ? = 2.2 × 10 6 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 = 60 3 100 ? rps ?? = 2 ? n = 2 ? × 180 100 = 9 10 ? rad/sec r = 10cm =0.1m, g = 10m/sec 2 ?mg ? mr ? 2 ? ? = g r 2 ? ? 10 9 10 1 . 0 2 ? ? ? ? ? ? ? ? ? ? ? 81 2 ? ? R mg ??g mv 2 /R 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/sec 2 From the figure T sin ? = r mv 2 ..(i) T cos ? = mg ..(ii) ? ? ? cos sin = rmg mv 2 ? tan ? = rg v 2 ? ? = tan –1 ? ? ? ? ? ? ? ? rg v 2 = 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 mv 2 = 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 mv 2 + mg cos ? ? T = ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 1 8 . 9 ) 1 . 0 ( 1 ) 4 . 1 ( 1 . 0 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 ) r m mg ( mg 2 ? ? ? = g 2 ? = 10 10 6400 3600 24 2 3 2 ? ? ? ? ? ? ? ? ? ? = 3.5 × 10 –3 b) When the balance reading is half the true weight, ? mg ) r m mg ( mg 2 ? ? ? = 1/2 ? 2 r = g/2 ? ???? r 2 g ??? 3 10 6400 2 10 ? ? rad/sec ? Duration of the day is T = ? ? 2 = 8 . 9 10 6400 2 2 3 ? ? ? ? sec = 49 10 64 2 6 ? ? ? sec = 3600 7 8000 2 ? ? ? hr = 2hr mv 2 /R mg ? ? mg ? mv 2 /r ? T mg sin ? ? mg cos ? ? T mg sin ? ? mg cos ? ? T m ? 2 /R mg R 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 v 2 = tan –1 ? ? ? ? ? ? ?10 20 100 = tan –1 ? ? ? ? ? ? 2 1 or tan ? = 0.5 When the car travels at max. speed so that it slips upward, ?R 1 acts downward as shown in Fig.1 So, R 1 – mg cos ? – r mv 2 1 sin ? = 0 ..(i) And ?R 1 + mg sin ? – r mv 2 1 cos ? = 0 ..(ii) Solving the equation we get, V 1 = ? ? ? ? ? ? tan 1 tan rg = 2 . 1 1 . 0 10 20 ? ? = 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 ? = R 2 L . So, R mv 2 = mg cos ? ? v = ? ? ? ? ? ? R 2 L cos gR ? 20. Since the motion is nonuniform, the acceleration has both radial & tangential component a r = r v 2 a t = 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 g 2 = 2 4 a 2 r v ? ? ? ? ? ? ? ? ? ? v 4 = ( ? 2 g 2 – a 2 ) r 2 ? v = [( ? 2 g 2 – a 2 ) r 2 ] 1/4 ? ? ?R 1 ? ? mv 1 2 /r mg R 1 ? ? mv 2 2 /r ?R 2 ? ? mg R 2 ? ? 2 ? ? mv 2 /R mg 2 ?= L/R ? ? ? mv 2 /R mg 2 ? ? 2 ?= L/R ? ? ? mv 2 /R 2 ? ? 2 ?= L/R ? mv 2 /R ? mg m dv/dt m N mv 2 /R m Chapter 7 7.5 21. a) When the ruler makes uniform circular motion in the horizontal plane, (fig–a) ? mg = m ? ? ? L ? ? ? ?? L g ? ? b) When the ruler makes uniformly accelerated circular motion,(fig–b) ? mg = 2 2 2 2 ) mL ( ) L m ( ? ? ? ? ? 2 4 + ? 2 = 2 2 2 L g ? ? ? 2 = 4 / 1 2 2 L g ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (When viewed from top) 22. Radius of the curves = 100m Weight = 100kg Velocity = 18km/hr = 5m/sec a) at B mg – R mv 2 = N ? N = (100 × 10) – 100 25 100 ? = 1000 – 25 = 975N At d, N = mg + R mv 2 = 1000 + 25 = 1025 N b) At B & D the cycle has no tendency to slide. So at B & D, frictional force is zero. At ‘C’, mg sin ? = F ? F = 1000 × 2 1 = 707N c) (i) Before ‘C’ mg cos ? – N = R mv 2 ? N = mg cos ? – R mv 2 = 707 – 25 = 683N (ii) N – mg cos ? = R mv 2 ? N = R mv 2 + mg cos ? = 25 + 707 = 732N d) To find out the minimum desired coeff. of friction, we have to consider a point just before C. (where N is minimum) Now, ? N = mg sin ? ? ? × 682 = 707 So, ? = 1.037 ? 23. d = 3m ? R = 1.5m R = distance from the centre to one of the kids N = 20 rev per min = 20/60 = 1/3 rev per sec ??= 2 ?r = 2 ?/3 m = 15kg ? Frictional force F = mr ? 2 = 15 × (1.5) × 9 ) 2 ( 2 ? = 5 × (0.5) × 4 ? 2 = 10 ? 2 ? Frictional force on one of the kids is 10 ? 2 ? 24. If the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force acts downward. Here, r = R sin ? From FBD –1 R 1 – mg cos ? – m ? ? 2 (R sin ?) sin ? = 0 ..(i) [because r = R sin ?] and ?R 1 mg sin ? – m ? 1 2 (R sin ?) cos ? = 0 ..(ii) Substituting the value of R 1 from Eq (i) in Eq(ii), it can be found out that ? 1 = 2 / 1 ) sin (cos sin R ) cos (sin g ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Again, for minimum speed, the frictional force ?R 2 acts upward. From FBD–2, it can be proved that, ??? 1 2 L ? mg mg ? ? ? ? R L (Fig–a) (Fig–b) m ? 2 2 L ? mg mL ? ? ?R 1 ? ? m ? 1 2 r R 1 (FBD – 1) ?R 2 ? ? m ? 2 2 r R 2 (FBD – 2) ? E D C B A mv 2 /R mg N B 15kg mg m ? 2 r ? F 15kgRead More

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- Chapter 8 : Work and Energy - HC Verma Solution, Physics
- Chapter 9 : Centre of Mass, Linear Momentum, Collision - HC Verma Solution, Physics
- Chapter 10 : Rotational Mechanics - HC Verma Solution, Physics
- Chapter 11 : Gravitation - HC Verma Solution, Physics
- Chapter 12 : Simple Harmonics Motion - HC Verma Solution, Physics
- Chapter 13 : Fluid Mechanics - HC Verma Solution, Physics