Page 1 Introductory Exercise 7.1 1. In uniform circular motion the magnitude of acceleration = æ è ç ö ø ÷ v r 2 does not change while its direction (being always towards the centre of the circular path) changes. 2. If w 0 and w are in rad s 1 the value of a must be in rad s 2 . But, if w 0 and w are in degree s 1 the value of a must also be in degree s 2 . Thus, it is not necessary to express all angles in radian. One way change rad into degree using p rad = ° 180 . 3. During motion of an object along a curved path the speed and magnitude of its radial acceleration may remain constant. Due to change in direction of motion the velocity of the object will change even if its speed is constant. Further, the acceleration will also change even if the speed is constant. 4. (i) Radial acceleration ( ) a c = =  v r 2 1 2 2 1 ( ) cms cm = 4 cms 2 (ii) Tangential acceleration (a T ) = = dv dt d dt t ( ) 2 =2 cms 2 (iii) Magnitude of net acceleration = + ( ) ( ) a a c T 2 2 = +   ( ) ( ) 4 2 2 2 2 2 cms cms =2 5 cms 2 5.     v v 1 2 ® ® = =v (say) T r v = 2p     v PQ ® = ® av PQ t = = r T r r v 2 4 2 2 / / p = 2 2 p v \   v ® = av v 2 2 p 6. w t t = + 0 4 Centripetal acceleration = tangential acceleration r r t w a 2 = Þ w a t 2 = ( ) 4 4 2 t = t = 1 2 s Circular Motion 7 v 1 P r r O Q v 2 ® ® Page 2 Introductory Exercise 7.1 1. In uniform circular motion the magnitude of acceleration = æ è ç ö ø ÷ v r 2 does not change while its direction (being always towards the centre of the circular path) changes. 2. If w 0 and w are in rad s 1 the value of a must be in rad s 2 . But, if w 0 and w are in degree s 1 the value of a must also be in degree s 2 . Thus, it is not necessary to express all angles in radian. One way change rad into degree using p rad = ° 180 . 3. During motion of an object along a curved path the speed and magnitude of its radial acceleration may remain constant. Due to change in direction of motion the velocity of the object will change even if its speed is constant. Further, the acceleration will also change even if the speed is constant. 4. (i) Radial acceleration ( ) a c = =  v r 2 1 2 2 1 ( ) cms cm = 4 cms 2 (ii) Tangential acceleration (a T ) = = dv dt d dt t ( ) 2 =2 cms 2 (iii) Magnitude of net acceleration = + ( ) ( ) a a c T 2 2 = +   ( ) ( ) 4 2 2 2 2 2 cms cms =2 5 cms 2 5.     v v 1 2 ® ® = =v (say) T r v = 2p     v PQ ® = ® av PQ t = = r T r r v 2 4 2 2 / / p = 2 2 p v \   v ® = av v 2 2 p 6. w t t = + 0 4 Centripetal acceleration = tangential acceleration r r t w a 2 = Þ w a t 2 = ( ) 4 4 2 t = t = 1 2 s Circular Motion 7 v 1 P r r O Q v 2 ® ® Introductory Exercise 7.2 1. In uniform circular motion of a body the body is never in equilibrium as only one force (centripetal) acts on the body which forces the perform circular motion. 2. v gra h max ( ) = = ´ ´ 9.8 1.5/ 1.5 250 2 =35 ms 1 3. (a) T mg sinq = T mr cosq w = 2 \ tanq w = g r 2 Þ w q q q = = g r g l tan ( cos ) tan or 2p q T g l = sin Þ T l g = 2p q sin = æ è ç ö ø ÷ 2 2 1 2 p 9.8 \ f = = 1 1 2 9.8 p rve s 1 = ´ 9.8 2 60 p rev min 1 =29.9 rev min 1 (b) T mg mg = = sin q 2 = ´ ´ 2 5 9.8 =69.3N 4. (a) At rest : Required CPF =  = w N w 2 Þ mv r w 2 2 = …(i) At dip : Required CPF = ¢  N w mv r N w 2 = ¢  …(ii) Comparing Eqs. (i) and (ii), N w w ¢  = 2 \ N w ¢ = = ´ 3 2 3 2 16 kN =24 kN (b) At crest on increasing the speed (v), the value of N will decrease and for maximum value of v the of N will be just zero. Thus, mv r w max 2 0 =  Þ v wr m max = = gr (as w mg = ) = ´ 10 250 =50 ms 1 (c) At dip : N w mv r ¢ = + 2 = + w mg = = 2 32 w kN 5. Case I. If the driver turns the vehicle Circular Motion  141 w' N = w 2 w w r m mg l r q mg T sin q T l T cos q Page 3 Introductory Exercise 7.1 1. In uniform circular motion the magnitude of acceleration = æ è ç ö ø ÷ v r 2 does not change while its direction (being always towards the centre of the circular path) changes. 2. If w 0 and w are in rad s 1 the value of a must be in rad s 2 . But, if w 0 and w are in degree s 1 the value of a must also be in degree s 2 . Thus, it is not necessary to express all angles in radian. One way change rad into degree using p rad = ° 180 . 3. During motion of an object along a curved path the speed and magnitude of its radial acceleration may remain constant. Due to change in direction of motion the velocity of the object will change even if its speed is constant. Further, the acceleration will also change even if the speed is constant. 4. (i) Radial acceleration ( ) a c = =  v r 2 1 2 2 1 ( ) cms cm = 4 cms 2 (ii) Tangential acceleration (a T ) = = dv dt d dt t ( ) 2 =2 cms 2 (iii) Magnitude of net acceleration = + ( ) ( ) a a c T 2 2 = +   ( ) ( ) 4 2 2 2 2 2 cms cms =2 5 cms 2 5.     v v 1 2 ® ® = =v (say) T r v = 2p     v PQ ® = ® av PQ t = = r T r r v 2 4 2 2 / / p = 2 2 p v \   v ® = av v 2 2 p 6. w t t = + 0 4 Centripetal acceleration = tangential acceleration r r t w a 2 = Þ w a t 2 = ( ) 4 4 2 t = t = 1 2 s Circular Motion 7 v 1 P r r O Q v 2 ® ® Introductory Exercise 7.2 1. In uniform circular motion of a body the body is never in equilibrium as only one force (centripetal) acts on the body which forces the perform circular motion. 2. v gra h max ( ) = = ´ ´ 9.8 1.5/ 1.5 250 2 =35 ms 1 3. (a) T mg sinq = T mr cosq w = 2 \ tanq w = g r 2 Þ w q q q = = g r g l tan ( cos ) tan or 2p q T g l = sin Þ T l g = 2p q sin = æ è ç ö ø ÷ 2 2 1 2 p 9.8 \ f = = 1 1 2 9.8 p rve s 1 = ´ 9.8 2 60 p rev min 1 =29.9 rev min 1 (b) T mg mg = = sin q 2 = ´ ´ 2 5 9.8 =69.3N 4. (a) At rest : Required CPF =  = w N w 2 Þ mv r w 2 2 = …(i) At dip : Required CPF = ¢  N w mv r N w 2 = ¢  …(ii) Comparing Eqs. (i) and (ii), N w w ¢  = 2 \ N w ¢ = = ´ 3 2 3 2 16 kN =24 kN (b) At crest on increasing the speed (v), the value of N will decrease and for maximum value of v the of N will be just zero. Thus, mv r w max 2 0 =  Þ v wr m max = = gr (as w mg = ) = ´ 10 250 =50 ms 1 (c) At dip : N w mv r ¢ = + 2 = + w mg = = 2 32 w kN 5. Case I. If the driver turns the vehicle Circular Motion  141 w' N = w 2 w w r m mg l r q mg T sin q T l T cos q mv r mg 1 2 £ m [where v 1 = maximum speed of vehicle] Þ v gr 1 £ m Case II. If the driver tries to stop the vehicle by applying breaks. Maximum retardation = m g \ v g r 2 2 2 0 2 = + ( ) m Þ v gr 2 2 = m = 2 1 v As v v 2 1 > , driver should apply breaks to stop the vehicle rather than taking turn. 6. In the answer to question 3(a) if we replace q by f w = f g l sin If q is the angle made by the string with the vertical q + f = ° 90 i.e., f = °  90 q Þ sin cos f = q \ w q = g lcos Þ cosq w = g l 2 Introductory Exercise 7.3 1.   v 1 ® = u and   v 2 ® = v (say) Dv v v ® ® ® = +  2 1 ( ) =  ® ® v v 2 1 D D v u u v v u ® ® ® ® ® ® × =  ×  ( ) ( ) 2 1 2 1 = × + × ® ® ® ® u v u v 2 2 1 1   ® ® 2 2 1 v v     v v 2 2 1 2 2 2 ® ® + = + v u =  + u gL u 2 2 2   ( ) Dv ® =  2 2 2 u gL   ( ) Dv ® =  2 2 u gL 2. Ball motion from A to B : 0 2 2 2 2 = +  u g R min ( )( ) Þ u gR min 2 4 = Ball motion from C to A : v u g h 2 2 2 = +  min ( ) =  4 2 gR gh Þ v g R h =  2 2 ( ) 3. Decrease in KE of bob = Increase in PE of bob 142  Mechanics1 L v 2 – v 1 Di g v 1 ® ® ® ® h v C A 2R B PE of bob = K (say) KE of bob 2 = 1/2 mv 0 v 0 Page 4 Introductory Exercise 7.1 1. In uniform circular motion the magnitude of acceleration = æ è ç ö ø ÷ v r 2 does not change while its direction (being always towards the centre of the circular path) changes. 2. If w 0 and w are in rad s 1 the value of a must be in rad s 2 . But, if w 0 and w are in degree s 1 the value of a must also be in degree s 2 . Thus, it is not necessary to express all angles in radian. One way change rad into degree using p rad = ° 180 . 3. During motion of an object along a curved path the speed and magnitude of its radial acceleration may remain constant. Due to change in direction of motion the velocity of the object will change even if its speed is constant. Further, the acceleration will also change even if the speed is constant. 4. (i) Radial acceleration ( ) a c = =  v r 2 1 2 2 1 ( ) cms cm = 4 cms 2 (ii) Tangential acceleration (a T ) = = dv dt d dt t ( ) 2 =2 cms 2 (iii) Magnitude of net acceleration = + ( ) ( ) a a c T 2 2 = +   ( ) ( ) 4 2 2 2 2 2 cms cms =2 5 cms 2 5.     v v 1 2 ® ® = =v (say) T r v = 2p     v PQ ® = ® av PQ t = = r T r r v 2 4 2 2 / / p = 2 2 p v \   v ® = av v 2 2 p 6. w t t = + 0 4 Centripetal acceleration = tangential acceleration r r t w a 2 = Þ w a t 2 = ( ) 4 4 2 t = t = 1 2 s Circular Motion 7 v 1 P r r O Q v 2 ® ® Introductory Exercise 7.2 1. In uniform circular motion of a body the body is never in equilibrium as only one force (centripetal) acts on the body which forces the perform circular motion. 2. v gra h max ( ) = = ´ ´ 9.8 1.5/ 1.5 250 2 =35 ms 1 3. (a) T mg sinq = T mr cosq w = 2 \ tanq w = g r 2 Þ w q q q = = g r g l tan ( cos ) tan or 2p q T g l = sin Þ T l g = 2p q sin = æ è ç ö ø ÷ 2 2 1 2 p 9.8 \ f = = 1 1 2 9.8 p rve s 1 = ´ 9.8 2 60 p rev min 1 =29.9 rev min 1 (b) T mg mg = = sin q 2 = ´ ´ 2 5 9.8 =69.3N 4. (a) At rest : Required CPF =  = w N w 2 Þ mv r w 2 2 = …(i) At dip : Required CPF = ¢  N w mv r N w 2 = ¢  …(ii) Comparing Eqs. (i) and (ii), N w w ¢  = 2 \ N w ¢ = = ´ 3 2 3 2 16 kN =24 kN (b) At crest on increasing the speed (v), the value of N will decrease and for maximum value of v the of N will be just zero. Thus, mv r w max 2 0 =  Þ v wr m max = = gr (as w mg = ) = ´ 10 250 =50 ms 1 (c) At dip : N w mv r ¢ = + 2 = + w mg = = 2 32 w kN 5. Case I. If the driver turns the vehicle Circular Motion  141 w' N = w 2 w w r m mg l r q mg T sin q T l T cos q mv r mg 1 2 £ m [where v 1 = maximum speed of vehicle] Þ v gr 1 £ m Case II. If the driver tries to stop the vehicle by applying breaks. Maximum retardation = m g \ v g r 2 2 2 0 2 = + ( ) m Þ v gr 2 2 = m = 2 1 v As v v 2 1 > , driver should apply breaks to stop the vehicle rather than taking turn. 6. In the answer to question 3(a) if we replace q by f w = f g l sin If q is the angle made by the string with the vertical q + f = ° 90 i.e., f = °  90 q Þ sin cos f = q \ w q = g lcos Þ cosq w = g l 2 Introductory Exercise 7.3 1.   v 1 ® = u and   v 2 ® = v (say) Dv v v ® ® ® = +  2 1 ( ) =  ® ® v v 2 1 D D v u u v v u ® ® ® ® ® ® × =  ×  ( ) ( ) 2 1 2 1 = × + × ® ® ® ® u v u v 2 2 1 1   ® ® 2 2 1 v v     v v 2 2 1 2 2 2 ® ® + = + v u =  + u gL u 2 2 2   ( ) Dv ® =  2 2 2 u gL   ( ) Dv ® =  2 2 u gL 2. Ball motion from A to B : 0 2 2 2 2 = +  u g R min ( )( ) Þ u gR min 2 4 = Ball motion from C to A : v u g h 2 2 2 = +  min ( ) =  4 2 gR gh Þ v g R h =  2 2 ( ) 3. Decrease in KE of bob = Increase in PE of bob 142  Mechanics1 L v 2 – v 1 Di g v 1 ® ® ® ® h v C A 2R B PE of bob = K (say) KE of bob 2 = 1/2 mv 0 v 0 1 2 0 2 mv mgh = v gh 0 2 2 = v gl 0 2 1 =  ( cos ) q = ´ ´ ´  ° 2 5 1 60 9.8 ( cos ) =7 ms 1 AIEEE Corner Subjective Questions (Level 1) 1. v t = 4 2 \ dv dt t =8 i.e., a T = ´ = 8 3 24 ms 2 v = ´ = 4 3 2 36 ms 1 (at t = 3 s) \ a v c = = 2 2 4 36 54 ( ) = 24 ms 2 Angle between a ® net and a t ® q = = = °  ® ®  tan     tan 1 1 1 45 a a c T 2. v = 16 ms 1 and r = 50 m \ a v r c = = = 2 2 16 50 ( ) 5.12 ms 2 a a a c T net = + 2 2 = + ( ) 5.12 2 2 8 (given a T =  8 2 ms ) =9.5 ms 2 3. Speed ( ) v at the highest point ( ) P v u = cosq Now, a g c = \ v R g 2 = i.e., R v g u g = = 2 2 2 cos q 4. (a) a a c = ° = ´ cos30 25 3 2 = 21.65 ms 2 \ v R 2 =21.65 i.e., v = ´ 21.65 2.5 (Q R =2.5 m) = 7.36 ms 1 (b) a a T = ° = ´ sin 30 25 1 2 = 12.5 ms 2 5. (a) v t = = ´ 2.0 2.0 1 =2.0 cms 1 \ a v R c = = = 2 2 1 4 ( ) 2.0 cms 2 (b) v t =2.0 \ dv dt =2.0 i.e., a T =  2.0cms 2 (c) a a a C T net = + 2 2 = 4.47 cms 2 6. R uT u h g = = 2 \ u R g h = 2 i.e., u R g h 2 2 2 = Circular Motion  143 q u cos q u R a = g c v = u cos q Centre of curvature r P Range (R) = 10 m h h = 2.9 m u h PE of bob = K + mgh Page 5 Introductory Exercise 7.1 1. In uniform circular motion the magnitude of acceleration = æ è ç ö ø ÷ v r 2 does not change while its direction (being always towards the centre of the circular path) changes. 2. If w 0 and w are in rad s 1 the value of a must be in rad s 2 . But, if w 0 and w are in degree s 1 the value of a must also be in degree s 2 . Thus, it is not necessary to express all angles in radian. One way change rad into degree using p rad = ° 180 . 3. During motion of an object along a curved path the speed and magnitude of its radial acceleration may remain constant. Due to change in direction of motion the velocity of the object will change even if its speed is constant. Further, the acceleration will also change even if the speed is constant. 4. (i) Radial acceleration ( ) a c = =  v r 2 1 2 2 1 ( ) cms cm = 4 cms 2 (ii) Tangential acceleration (a T ) = = dv dt d dt t ( ) 2 =2 cms 2 (iii) Magnitude of net acceleration = + ( ) ( ) a a c T 2 2 = +   ( ) ( ) 4 2 2 2 2 2 cms cms =2 5 cms 2 5.     v v 1 2 ® ® = =v (say) T r v = 2p     v PQ ® = ® av PQ t = = r T r r v 2 4 2 2 / / p = 2 2 p v \   v ® = av v 2 2 p 6. w t t = + 0 4 Centripetal acceleration = tangential acceleration r r t w a 2 = Þ w a t 2 = ( ) 4 4 2 t = t = 1 2 s Circular Motion 7 v 1 P r r O Q v 2 ® ® Introductory Exercise 7.2 1. In uniform circular motion of a body the body is never in equilibrium as only one force (centripetal) acts on the body which forces the perform circular motion. 2. v gra h max ( ) = = ´ ´ 9.8 1.5/ 1.5 250 2 =35 ms 1 3. (a) T mg sinq = T mr cosq w = 2 \ tanq w = g r 2 Þ w q q q = = g r g l tan ( cos ) tan or 2p q T g l = sin Þ T l g = 2p q sin = æ è ç ö ø ÷ 2 2 1 2 p 9.8 \ f = = 1 1 2 9.8 p rve s 1 = ´ 9.8 2 60 p rev min 1 =29.9 rev min 1 (b) T mg mg = = sin q 2 = ´ ´ 2 5 9.8 =69.3N 4. (a) At rest : Required CPF =  = w N w 2 Þ mv r w 2 2 = …(i) At dip : Required CPF = ¢  N w mv r N w 2 = ¢  …(ii) Comparing Eqs. (i) and (ii), N w w ¢  = 2 \ N w ¢ = = ´ 3 2 3 2 16 kN =24 kN (b) At crest on increasing the speed (v), the value of N will decrease and for maximum value of v the of N will be just zero. Thus, mv r w max 2 0 =  Þ v wr m max = = gr (as w mg = ) = ´ 10 250 =50 ms 1 (c) At dip : N w mv r ¢ = + 2 = + w mg = = 2 32 w kN 5. Case I. If the driver turns the vehicle Circular Motion  141 w' N = w 2 w w r m mg l r q mg T sin q T l T cos q mv r mg 1 2 £ m [where v 1 = maximum speed of vehicle] Þ v gr 1 £ m Case II. If the driver tries to stop the vehicle by applying breaks. Maximum retardation = m g \ v g r 2 2 2 0 2 = + ( ) m Þ v gr 2 2 = m = 2 1 v As v v 2 1 > , driver should apply breaks to stop the vehicle rather than taking turn. 6. In the answer to question 3(a) if we replace q by f w = f g l sin If q is the angle made by the string with the vertical q + f = ° 90 i.e., f = °  90 q Þ sin cos f = q \ w q = g lcos Þ cosq w = g l 2 Introductory Exercise 7.3 1.   v 1 ® = u and   v 2 ® = v (say) Dv v v ® ® ® = +  2 1 ( ) =  ® ® v v 2 1 D D v u u v v u ® ® ® ® ® ® × =  ×  ( ) ( ) 2 1 2 1 = × + × ® ® ® ® u v u v 2 2 1 1   ® ® 2 2 1 v v     v v 2 2 1 2 2 2 ® ® + = + v u =  + u gL u 2 2 2   ( ) Dv ® =  2 2 2 u gL   ( ) Dv ® =  2 2 u gL 2. Ball motion from A to B : 0 2 2 2 2 = +  u g R min ( )( ) Þ u gR min 2 4 = Ball motion from C to A : v u g h 2 2 2 = +  min ( ) =  4 2 gR gh Þ v g R h =  2 2 ( ) 3. Decrease in KE of bob = Increase in PE of bob 142  Mechanics1 L v 2 – v 1 Di g v 1 ® ® ® ® h v C A 2R B PE of bob = K (say) KE of bob 2 = 1/2 mv 0 v 0 1 2 0 2 mv mgh = v gh 0 2 2 = v gl 0 2 1 =  ( cos ) q = ´ ´ ´  ° 2 5 1 60 9.8 ( cos ) =7 ms 1 AIEEE Corner Subjective Questions (Level 1) 1. v t = 4 2 \ dv dt t =8 i.e., a T = ´ = 8 3 24 ms 2 v = ´ = 4 3 2 36 ms 1 (at t = 3 s) \ a v c = = 2 2 4 36 54 ( ) = 24 ms 2 Angle between a ® net and a t ® q = = = °  ® ®  tan     tan 1 1 1 45 a a c T 2. v = 16 ms 1 and r = 50 m \ a v r c = = = 2 2 16 50 ( ) 5.12 ms 2 a a a c T net = + 2 2 = + ( ) 5.12 2 2 8 (given a T =  8 2 ms ) =9.5 ms 2 3. Speed ( ) v at the highest point ( ) P v u = cosq Now, a g c = \ v R g 2 = i.e., R v g u g = = 2 2 2 cos q 4. (a) a a c = ° = ´ cos30 25 3 2 = 21.65 ms 2 \ v R 2 =21.65 i.e., v = ´ 21.65 2.5 (Q R =2.5 m) = 7.36 ms 1 (b) a a T = ° = ´ sin 30 25 1 2 = 12.5 ms 2 5. (a) v t = = ´ 2.0 2.0 1 =2.0 cms 1 \ a v R c = = = 2 2 1 4 ( ) 2.0 cms 2 (b) v t =2.0 \ dv dt =2.0 i.e., a T =  2.0cms 2 (c) a a a C T net = + 2 2 = 4.47 cms 2 6. R uT u h g = = 2 \ u R g h = 2 i.e., u R g h 2 2 2 = Circular Motion  143 q u cos q u R a = g c v = u cos q Centre of curvature r P Range (R) = 10 m h h = 2.9 m u h PE of bob = K + mgh Thus, centripetal acceleration of the stone (at point P) while in circular motion = = × u r R g h r 2 2 2 1 = ´ ´ ´ 10 2 2 9.8 2.9 1.5 =112.6 ms 2 7. v =18 km/h 18.5 ms =  18 1 =  5 1 ms Angle of banking ( ) q =  tan 1 2 v rg = ´  tan ( ) 1 2 5 10 10 =  tan 1 1 40 8. mv r mg 2 = m i.e., m = = ´ = = v gr 2 2 5 10 10 1 4 ( ) 0.25 9. v rg = tanq = ´ ´ ° 50 10 30 tan =  17 1 ms 10. mr N w min 2 = …(i) and mg N = m …(ii) Solving Eqs. (i) and (ii), w m min = g r = ´ 10 3 0.15 = 4.7 rads 1 11. (a) ( ) cos T T mr 1 2 2 + = q w …(i) and ( )sin T T mg 1 2  = q …(ii) i.e., T mr T 2 2 1 cos cos q w q =  and T T mg 2 1 sin sin q q =  Þ cos cos sin q w q q =   mr T T mg 2 1 1 or T mg mr T 1 2 1 cos cos cos q q w q  =  or w q q =  2 1 T mg mr cos cos = ´ ´ æ è ç ö ø ÷  ´ ´ æ è ç ö ø ÷ ´ 2 200 3 5 4 10 3 4 4 3 =  8.37 rads 1 = ´ 8.37 rev min 1 2 1 60 p =39.94 rev min 1 . (b) From Eq. (ii), T T mg 2 1 =  sin q =  ´ æ è ç ö ø ÷ 200 4 100 4 5 =150 N 12. (a) For the block not to slip mL N mg w m m max 2 = = \ w m max = g L (b)If the angular speed is gradually increased, the block will also have translational acceleration ( ) =aL besides centripetal acceleration (=Lw 2 ). 144  Mechanics1 mN mg N T sin q 1 T sin q 2 mg 4m 4m T cos q 1 T cos q 2 T 2 T 1 aL At the time ( t) of just slipping At t = 0 2 Lw a net N mg L mNRead More
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