1.Circular Motion (Marks 04/05)

Transcription

1 1.Cicula Motion (Maks 04/05) Intoduction :- Motion:- The displacement of paticle o body fom one place to anothe is called motion of body. Linea Motion:- The motion of the body alon a staiht line is called linea motion of that body. Cicula Motion:- The motion of the body alon a cicumfeence of a cicle is called cicula motion. It example of two dimensional motion Types of cicula motion :- 1)Unifom cicula motion ( U.C.M.) The motion of a body alon cicumfeence of a cicle with constant linea speed o anula speed (o anula velocity) is called unifom cicula motion. Exa:- Motion of tip of thee hands of a clock. )Non unifom cicula motion ( Non U.C.M. ):- The motion of a body alon cicumfeence of a cicle with vaiable speed is called Non-unifom cicula motion. Exa:- Vetical cicula motion is non-u.c.m. Radius Vecto:- A vecto dawn fom the cente of a cicle to the position of a paticle is called as adius ecto. Exa:- In the followin fi.(1) OA, OB and OC ae the adius vecto (Position vecto) The manitude of adius vecto of cicle is constant, but its diection chanes. Q.1.Define anula displacement, anula velocity and anula acceleation Anula displacement :- The anle thouh which adius vecto otates in iven time. OR Anula displacement is defined as a anle descibed by adius vecto in iven time at the cente of a cicle. It is vecto quantity when it has small value(infinitesimal). Conside, a paticle pefomin cicula motion, in anticlockwise diection(sense).let A be the initial position of paticle at a time t is equals to zeo. Suppose, in time t, it moves to the position B. Let θ is the anula displacement and s is the linea displacement in time t. Anula displacement (θ) = θ = S Ac AB adius S = θ If the paticle moves fom point A to B in shot time inteval δt and descibes the ac of lenth δs. Then, the anle taced by the adius vecto at cente of cicle is consideed as δθ δθ = δs δs = δθ Instantaneous anula displacement is a vecto quantity. SI unit :- Radian (ad) Dimension:- It has no dimension. (anula displacement is a dimensionless quantity) Diection :- Diection of anula displacement is iven by iht hand thumb ule o iht hand scew ule. Riht hand ule :- Imaine the axis of otation to be held in iht hand with the fiues culed ound the axis and the thumb stetched alon the axis. If the culed fines denote the sense of otation, then the thumb denotes the diection of the vecto. It is scala quantity when it has finite value.

2 If the anula velocity inceases then the anula acceleation is positive and is in the diection of anula velocity. Anula Velocity :- Rate of chane of anula displacement with espect to time is called as anula velocity. Anula Velocity = SI unit:- ad/s Anula Displacement Time δθ Anula Velocity (ω) = lim δt 0 δt ω = θ θ 1 t = dθ = θ t If the anula velocity deceases then the anula acceleation is neative and is in the diection opposite to the anula velocity. Dimension:- M L T 1 Anula velocity is always a vecto quantity. Diection :- The diection of anula velocity is iven by iht hand ule, and is in the diection of anula displacement. Anula acceleation (α) : - Anula acceleation of a paticle pefomin cicula motion is defined as the ate of chane of anula velocity with espect to time. Anula Acceleation(α) = SI unit:- ad/s Anula velocity Time δω Anula Acceleation α = lim δt 0 δt α = ω ω 1 t = dω = ω t Fo exa:- When an electic fan switched on then anula velocity of blades of a fan inceases with time i.e.ω inceases and α is positive. If the anula velocity deceases with time (fo ex. electic fan is switched off then anula velocity of blades of a fan deceases with time i.e. ω deceases and α is neative and anula acceleation is in the diection opposite to anula velocity. Q..Pove the elation,v = ω, whee symbols have thei usual meanins. OR Q.3.Deive the elation between the linea velocity and anula velocity. OR Dimension:- M L T Anula acceleation is a vecto quantity whose diection is iven by the iht hand thumb ule. Q.4.Define unifom cicula motion. Show that linea speed of paticle pefomin cicula motion is the poduct of adius of cicle and anula speed of paticle.

3 Relation between linea velocity (V) and anula velocity (ω) :- ( δθ) = lim δt 0 δt ν = ω In vecto fom v = ω This is the elation between linea velocity and anula velocity Relation between tanential acceleation (a T ) and anula acceleation (α) :- P = initial position of paticle Q = final position of paticle afte time δt δθ = anula displacement of paticle pefomin cicula motion. δs = ac lenth = adius of cicle. Conside a paticle pefomin cicula motion alon the cicumfeence of cicle adius and cente O Let, δt is the time taken by the paticle to move fom initial position P to final position Q. In the same time suppose δs is the distance coveed (ac PQ) which is equal to the staiht line path between P and Q by the paticle. δs The Linea Velcity v = lim = ds.(1) δt 0 δt But we know that, Anula displacement (δθ) = Ac PQ adius δθ = δs δs = δθ Equation (1) Becomes, Linea Velcity v = lim δt 0 ( δθ) δt Linea acceleation (a T ) = dv...(1) Anula acceleation α = dω But we know that v = ω equation (1) becomes a T = d ω a T =. dω.. a T =. α ( fom equation ()) Wheneve paticle pefomin cicula motion with the linea speed (v) of the paticle oes on chanin. The tanential acceleation (a T ) in manitude is iven by In vecto fom a T = dv a T = α This is the elation between tanential acceleation and anula acceleation Fo U.C.M. a T = 0 because linea speed is constant. Peiodic Motion :- Motion of the body which epeats itself afte equal inteval of time is called as peiodic motion.

4 The unifom cicula motion is an example of peiodic motion. Q.6. Define peiod and fequency of paticle pefomin unifom cicula motion. State thei S.I. units :- 1) Peiod of evolution (T) :- The time taken by a paticle (o a body) pefomin U.C.M. to complete one evolution is called as peiod of evolution. Conside a paticle pefomin U.C.M. which is situated at point A on the cicumfeence of the cicle of adius and cente O Velocity = v = π T distance coveed time taken n = 1 T = v π = ω π SI Unit :- hetz (Hz) Dimension :- M L T 1 Q. Define the tem adial acceleation(o centipetal acceleation). Obtain an expession fo the adial acceleation of a paticle pefomin U.C.M. OR Obtain an expession fo acceleation of an paticle pefomin U.C.M. Radial acceleation (o Centipetal acceleation):- The acceleation of paticle pefomin U.C.M. which is diected towads the cente and alon the adius of the cicula path is called as adical acceleation. T = π (i) v We know that, v = ω Equation becomes, T = π ω T = π ω (ii) T = 1 (iii) n Whee, n is fequency of evolution. Equation (i), (ii) and (iii) ae the expession fo peiod of evolution. SI unit :- Second (sec) Dimension :- M L T 1 T = 1 n = π v = π ω ) Fequency of evolution(n) :- The numbe of evolutions pefomed by the paticle pefomin U.C.M. in unit time (o one second) is called fequency of evolution. The fequency of evolution is the ecipocal of peiod of evolution. Deivation :- Conside a paticle pefomin U.C.M. with a linea velocity of manitude v and anula velocity of manitude ω alon the cicle of adius and cente O. Suppose, that, the paticle moves fom point A to point B in a vey small time inteval δt. Daw the vectos AA 1 and BB 1 to epesents the velocities of the paticle at point A and B also daw the vecto BC equal and paallel to the AA 1 and join the vecto CB 1. In BCB 1 Accodin to the tianle of addition of vecto.

5 BC + CB 1 = BB 1. CB 1 = BB 1. - BC CB 1. = BB 1 AA 1 ( AA 1 = BC ) a = ω...(iii) a = ω ω a = v ω...(iv) CB 1. = Velocity at point B Velocity at point A = chane in velocity. CB 1. = δv. 1 Fom fi, CBB 1 and AOB ae simila (S-A-S) In vecto fom, a = ω v Q. Deive an expession fo adial acceleation by usin calculus method. CB 1 AB = BB 1 OB δv δs = v δv = v δs Dividin the above equation on both side by δt we et, δv = v δs δt δt Takin limit as δt 0 on both sides of equation,we et, But, we know that dv ds ( δv) lim δt 0 δt = v lim (δs) δt 0 δt = v (linea velocity) a = v v a = v...(ii) a = 4π since,v=π T T But we know that, V = ω equation (ii) becomes a = ω dv = v ds = a (linea acceleation) A = Position of the paticle pefomin U.C.M. = adius vecto (position vecto) θ = ωt = anula displacement of a paticle in time t (x,y) = Co-odinates of point A x = x-co-odinate of point A y = y-co-odinate of point A v = tanential velocity of the paticle pefomin U.C.M. at point A Conside a paticle pefomin U.C.M. alon the cicumfeence of the cicle of adius and cente O as shown in the above fiue. In the above fi. < BOA = θ = ωt The adius vecto ( )is iven by Now fom fi. In BOA sinθ = AB OA = x + y = i x + j y...(1) cosθ = OB OA

6 sinθ = AB AB = sinθ AB = y = sinθ d d sinθ = OB cosωt = sinωt ω sinωt = cos ωt ω equation (1) becomes = i cosθ + j sin θ OB = cosθ OB = x = cosθ = i cosωt + j sinωt.() = d = i d v = d i cosωt + j sinωt cosωt + j d sinωt = i sinωt ω + j cosωt ω v = i ω sinωt + j ω cosωt...(3) Instantaneous linea acceleation (a) is iven by = d a = dv ( i ω sinωt + j ω cosωt) = i ω d sinωt + j ω d (cosωt) = i ω ( cosωt ω) + j ω( sinωt ω) = i ω cosωt j ω (sinωt) = ω ( i cosωt + j sinωt ) = ω This is an expession fo the adial acceleation. Q.5.Obtain the elation between the linea acceleation and anula acceleation in cicula motion. Relation between linea acceleation and anula acceleation We know that linea velocity of the paticle is iven by v = ω Takin deivative on both sides of the above equation, we et, dv = d( ω ) dv = ω d + dω a = ω v + α Resultant Acceleation ( a ) = a T + a c Whee, a T = tanential component of linea acceleation a c = Radial o centipetal component of linea acceleation a = Linea acceleation = Resultant Acceleation We know that the anula between a T and a c is Theefoe the manitude of the esultant acceleation is iven by a = a T + a In U.C.M., anula velocity is constant theefoe Resultant acceleation (a) = a c = v V Thus in U.C.M., the linea acceleation is centipetal acceleation. Q. Explain the concept of centipetal foce with it s example OR Define centipetal foce. Give it s any fou examples. Centipetal Foce:- The foce actin on a paticle pefomin U.C.M. which is diected towads the cente of the cicle. and alon the adius of the cicula path, is called as centipetal foce. If m is mass of an paticle pefomin U.C.M. then the manitude of centipetal foce is iven by Centipetal foce = mass centipetal accelaation.

7 F = m v In vecto fom :- = m a c...(1) F = m ω.. F = m v ω... (3) F = m ω F = mv Whee, = adius of cicle = Radius vecto = Unit vecto in the diection of The centipetal foce 1)is a eal foce actin on paticle pefomin cicula motion ) is a necessay foce fo maintainin cicula motion. 3) its diection is diffeent at diffeent points. 4) acts alon the adius of cicle and diected towads the cente of cicle. 5) does no wok SI unit :- Newton (N) Dimension :- M 1 L 1 T Examples :- CGS Unit :- dyne 1) A satellite evolves in a cicula obit aound the eath. In this case, the centipetal foce is due to the avitational foce of attaction between the satellite and the eath. ) In an atom, electon evolves aound the nucleus in a cicula obit. In this case, centipetal foce is due to the electostatic foce of attaction between positively chaed nucleus and neatively chaed electons. 3) If a ca movin alon the hoizontal cicula oad with unifom speed then, in this case the necessay centipetal foce is povided by the foce of fiction between tyes and the oad suface. 4) If an object tied to a one end of stin and evolved in a hoizontal cicle with constant speed then, in this case the centipetal foce is povided by tension in the stin. Q. Explain the concept of centifual foce with it s example OR Define centifual foce. Give it s any fou examples. Centifual Foce:- The foce actin on a paticle which is diected away fom the cente and alon the adius of the cicula path is called centifual foce. Centifual foce = F = m v Whee, m = mass of paticle...(1) F = m ω.. F = m v ω... (3) = adius of cicle ω= anula speed v = linea speed In vecto fom :- F = + m ω F = + mv Whee, = adius of cicle = Radius vecto = Unit vecto in the diection of The centifual foce 1)is a pseudo foce in U.C.M., is imained in ode to make Newton s laws of motion valid in non inetial fame of efeence. )is actin alon the adius, but diected away fom the cente of the cicle. Example:- 1) When a hoizontal mey-o-ound otates about it s vetical axis, all the chais ae pulled out due to centifual foce.

8 ) When a ca movin alon a hoizontal cuved oad, peson seatin in the ca expeiences a foce in the outwad diection. 3) The bulin of eath at equato and flattenin at the poles is due to the centifual foce. 4) A bucket full of wate otated in vetical cicle at paticula speed, wate does not fall because weiht of wate is balanced by the centifual foce actin on it. 5) Centifue is a device which is test tube used fo sepaatin heavie paticles and liht paticles and woks on the pinciple of centifual foce. 6) In washin machine, the centifual foce is used fo dyin clothes in a die. 7) A coin kept on a otatin tun table moves away fom the axis of otation due to the centifual foce actin on it. Q. Distinuish between Centipetal and Centifual foce. Centipetal foce Centifual foce 1.It is diected towads the cente and alon the adius of cicula path. 1.It is diected away fom the cente and alon the adius of cicula path..it aises in inetial.it aises in non inetial fame of efeence. fame of efeence /acceleated fame of efeence. 3. It is a eal foce. 3.It is an imainay foce called as pseudo foce. 4. A satellite evolves aound the eath due to centipetal foce. 4.Chai in a mey-oound ae pulled out due to centifual foce. 5. In vecto fom F = mω 5.In vecto fom f = + mω Q. Explain why centifual foce is called as pseudo foce. The foce whose oiin is not due to known inteaction such as avitational foce, electostatic foce o nuclea foce is called pseudo foce. These foces ae pesent in only in the acceleated /non-inetial fame of efeence and they vanish if the fame of efeence is made zeo. The centifual foce is not poduced due to the know inteaction in a natue. Theefoe, the centifual foce is called as pseudo foce. Q. Obtain an expession fo the maximum speed with which a vehicle (ca) can be safely diven alon a hoizontal cuved aod. OR Deive an expession fo the optimum speed of the vehicle movin alon the cuved oad which is not banked. The ca movin alon a hoizontal cuved oad is pefomin cicula motion, fo which a centipetal foce must act on the ca. In such a case, the centipetal foce is povided by the fiction between the tyes of wheels of the ca and the suface of oad. Conside, a ca of mass m movin with a speed v alon the hoizontal cuved oad of adius. If μ is coefficient of fiction between the tyes and the oad suface. Maximum foce of fiction = μm...(1) But we know that, Centipetal foce = mv Fom equation (1) and (), we et, Centipetal foce = foce of fiction mv = μm v = μ v = μ.() This as an expession fo the maximum speed of the ca with which it can be safely diven alon the hoizontal cuved oad. If the speed of the ca is eate than that calculated by above expession fo the iven cuved oad, then the ca will be in dane of bein thown out of the cuved oad. Q. What is bankin of oad? and define the anle of bankin. Bankin of Roads:-

9 In ode to potect a vehicle movin alon a cuved oad at hih speed, the suface of oad is kept inclined to the hoizontal in such a way that the oute ede of the oad is at hihe level than that of inne ede. This aanement is called as bankin of oad. Anle of bankin :- The anle between the inclined Suface of oad and the hoizontal is called anle of bankin. Q. Why it is necessay to make a cuved oad banked? OR Why is the necessity of bankin of oad? ( maks) 1)When a vehicle moves alon a hoizontal cuved oad, the necessay centipetal foce is supplied by the foce of fiction between the wheels of vehicle and the suface of the oad. )The fictional foce is not enouh and eliable (unbelievable) evey time as it chanes when the oad becomes oily o wet in the ainy season 3)To incease the centipetal foce, the oad should be made ouh but it will cause the damae of the tyes of the wheels. 4)Thus, due to the lack of centipetal foce, the vehicle will be in dane of the thown off the oad. 5)When the oad is banked, the hoizontal component of nomal eaction (N sin θ) povides the necessay centipetal foce equied fo cicula motion of the vehicle. 6)Thus, to povide necessay centipetal foce at the cuved oad, the bankin of oad is necessay. Q. Obtain an expession fo a maximum speed (optimum speed) with which the vehicle can be safely diven alon the banked oad. State the factos on which it depends. OR Deive an expession fo the anle of bankin. Show that it is independent of mass of the vehicle. Let be the adius of the cuved oad and θ be the anle of bankin. Suppose that, a ca of mass m is movin with a speed v alon the banked oad.let f be the fictional foce between the tyes of the vehicle and the oad suface as shown in below fiue. Thee ae two foces actin on the ca which ae iven below. 1)Weiht of the vehicle (m) actin in the vetically downwad diection. ) The nomal eaction (N) actin at iht anle to the suface of the oad. The nomal eaction (N) can be esolved into two components. 1)N cos θ actin in the vetically upwad diection. ) N sin θ actin alon the hoizontal towads the cente of the cuved path. The fictional foce f between the tyes and the oad suface can be esolved into two ectanula components as shown in the above fiue. 1) f cos θ alon the hoizontal diection. ) f sin θ alon the vetically downwad diection. The component N cos θ balances the weiht (m) of the vehicle and the component f sin θ of fictional foce. N cos θ = m + f sin θ N cos θ f sin θ = m (1) The hoizontal component N sin θ and + f cos θ of fictional foce povides necessay centipetal foce. N sin θ+ f cos θ = mv () Divide equation () by equation (1), we et, N sin θ+f cos θ = mv N cos θ f sin θ m

10 The manitude of the fictional foce depend on the speed of vehicle fo iven oad suface and tyes of vehicle. Let v max be the maximum speed of the vehicle, the fictional foce poduced at this speed should be f = μn N sin θ+f cos θ = v max N cos θ f sin θ (3).(4) Fom equations (3) and (4), we et, N sin θ + μn cos θ N cos θ μn sin θ = v max Divide numeato and denominato by Ncos θ of L.H.S. of the above equation we et, we et, Theefoe we et, Theefoe N sin θ + μn cos θ Ncosθ N cos θ μn sin θ Ncosθ tanθ + μ 1 μ tanθ = v max = v max v max = μ + tanθ 1 μ tanθ v max = μ+tanθ 1 μ tanθ (5) Fo the hoizontal cuved oad, the anle θ = 0 and hence tan θ = 0, theefoe above equation becomes, we et, v max = μ (6) Fom the equations (5) and (6), we conclude that the maximum safe speed on the banked oad is eate than that of cuved hoizontal cuved oad. If the fictional foce is consideed to be zeo that is μ = 0 and let v be the maximum speed of the vehicle movin alon the banked oad then v = 0 + tanθ 1 0 tanθ v = tanθ.(7) This is an expession fo the maximum speed with which a vehicle (ca) can be safely diven alon the banked oad. In this expession, m is absent, theefoe we can say that velocity of the vehicle is independent of the mass of the vehicle. The velocity of the vehicle depends upon the followin factos:- 1)adius of the cuved oad () ) Acceleation due to avity () 3) anle of bankin (θ) Fom equation (7), we can obtain, tan θ = v 1 v θ = tan This is an expession fo the anle of bankin. In this expession, m is absent, theefoe we can say that the anle of bankin is independent of the mass of the vehicle. The anle of bankin depends upon the followin factos:- 1)maximum velocity of the vehicle (v) ) Acceleation due to avity () 3) adius of the cuved path () Q. What is conical pendulum? Obtain an expession fo the peiod of conical pendulum. OR Define peiod of conical pendulum and obtain an expession fo its peiod. OR Define conical pendulum. Obtain an expession fo anle made by the stin of conical pendulum with vetical. Hence, deduce an expession fo linea speed of bob of conical pendulum.

11 Ans:- Conical pendulum:- Conical pendulum is a simple pendulum which has iven such a motion that the bob descibes a hoizontal cicle and the stin descibes a cone. Peiod of conical pendulum:- The hoizontal component (T sin θ) povides the necessay centipetal foce fo the cicula motion of the bob. T sin θ = mv () Divide equation () by equation (1), we et, T sin θ = mv T cos θ m tan θ = v..(3) But fom fi. In SOB, tan θ = We know that, v =. ω v = ω by the bob. Let, m = mass of the bob l = lenth of the pendulum v = speed of the bob = adius of the hoizontal cicle descibed θ = semi-vetical anle of the cone descibed by the stin. Thee ae two foces actin on the bob which ae iven below. 1)Weiht (m) of the bob actin in the vetically downwad diection. ) The tension (T ) actin alon the diection BS. The tension (T ) can be esolved into two components 1)T cos θ actin in vetically upwad diection. ) T sin θ actin alon hoizontal and towads the cente of the hoizontal cicle. The vetical component (T cos θ) balances the weiht (m) of the bob. T cos θ = m (1) eq n (3) becomes = ω 1 = ω ω = ω = But we know that that T = π Now, fom fi, T = ω T = π equation (4) becomes, π Cos θ = l h = l cosθ h.. (4) T = π l cosθ...(5)

12 T l and T The equations (4) and (5) epesents peiod of conical pendulum. In above equation if θ = 0 i.e. vey small the cosθ 1, we et, T = π l...(5) Thus peiod of conical pendulum is the same as that of simple pendulum of lenth h, whee h is the axial heiht of the cone. The peiod of conical pendulum depends upon 1 Fom equations tan θ = v Theefoe, and tan θ = v = h T = m + m h T = m 1 + h T = m 1 + h, we et, 1) Lenth of pendulum ) Anle of inclination to veitical 3) Acceleation due to avity at a iven place 4) Independent of the mass of the bob 5) Independent of θ (if θ is vey small) Fequency of conical penduim :- We know that, n = π n = 1 π OR 1 h n = 1 T OR = π 1 n = 1 π l cos θ l cosθ This is the equation fo fequency of conical pendulum. Vetical Cicula Motion:- Q.1)Deive an expession fo linea velocity at lowest point, midway and top position fo a paticle evolvin in a vetical cicle. It has to just complete cicula motion without slackenin stin at the top. Q.) What is a vetical cicula motion? show that motion of an object evolvin in a vetical motion is a non unifom cicula motion. Q.3) Obtain expession fo tensions at hihest position, midway position and bottom position fo an object evolvin in a vetical cicle. Q.4) Obtain expession fo tensions at hihest position, midway position and bottom position fo an object evolvin in a vetical cicle. Q.5) Obtain expession fo eney at diffeent positions in vetical cicula motion. Q 6) show that the total eney of an object pefomin vetical cicula motion is conseved. Expession fo the Tension in the of conical pendulum Squain and addin equations (1) and (), we et, T cos θ + T sin θ = m + mv T = m + mv

13 the vetical cicula motion of the body is not unifom i.e. it s velocity chanes at evey point. Let, V L as the velocity at lowest in the stin at lowest end, hihest point espectively. a) Velocity at hihest point (B) :- At hihest point (B), (T h + m) povides necessay centipetal foce fo vetical cicula motion of the body. Conside, a spheical body of mass m is tied to one end of a stin and it is whiled in a vetical cicle of adius as shown in the above fi. Duin the vetical cicula motion of the spheical body, thee ae two foces actin on it. These ae:- 1) It s weiht m actin in vetically downwad diection. ) The tension T in the stin which is diected towads the cente o of the vetical cicle. When the body is at point C then it s weiht m can be esolved into two components. 1) m cosθ alon diection OC. ) m sinθ actin at iht anle to OC. When the body moves fom the bottom to top i.e. alon ADB then m sinθ opposes its upwad motion. So that velocity of the body oes on deceasin while it s upwad motion and it is minimum at the top (at a point B).But when the body moves fom top to bottom i.e. alon BEA then m sinθ is in the diection so that velocity of body oes on inceasin while it s downwad motion and it is maximum at the lowest point (point A) Thus, it is concluded that the velocity of the body oes on chanin duin it s vetical cicula motion. Thus, vetical cicula motion is an example of non unifom cicula motion. Equations fo velocity at diffeent position of vetical cicula motion:- Conside, a body of mass m is tied to one end of the stin and whiled in a vetical cicle of adius and cente O as shown in the above fi. we know that, T h + m = mv h...(1) If tension T H at hihest point is zeo, the entie centipetal foce is supplied by the weiht m of the body. m = mv h = v h v h =... ( T H = 0).() This as an expession fo the minimum velocity v h of the body at hihest point B,So that the stin will not slack. If v h is less than then the stin will become slack and body will not complete veticle cicle. B) Velocity at lowest point (A) :- Let, the potential eney at hoizontal level passin thouh the point A is zeo (P.E. at point A = 0) When the body moves fom point A to B then it is aised thouh the vetical distance AOB and it is equals to i.e. AOB = h = Now, accodin to the law of consevation of eney, Total eney at point A = Total Eney at point B P.E. at point A + K. E. at point A = P. E. at point B + K. E. at point B mv L = mh + 1 mv h 1 mv L = m + 1 m 1 mv L = m ( + 1 ) 1 v L = 4+

14 v L = 5 v L = 5...(3) This is an expession fo the minimum velocity at the lowest point A so that it can safely pefom vetical cicula motion. If velocity v L is less than 5 then the body will not pefom veticle cicula motion and the stin will be slaken at the hihest point B C) Velocity at middle point (Hoizontal level of cente of vetical cicle) D :- Let the point v m is the velocity at point D and whee the heiht of point D fom the point A is AO =h d = Accodin to the law of consevation of eney, cos θ = OF OF = cos θ h c = (1 cosθ) Accodin to the law of convesion of eney Total Eney at point C = Total Eney at point A P. E. at point C + K. E. at point C = P. E. at point A + K. E. at point A mh c + 1 mv = mv L mh c + mv = mv L m (h c + v ) = mv L Total eney at point D = Total eney at point A P.E. at point D + K. E.at point D = P.E. at point A + K. E. at point A. mh d + 1 mv m = mv L m + 1 v m = 1 m 5 V = v = v L h c v = v L (1-cosθ) v = 5 (1 cosθ) 5 (1 cosθ) OR + 1 v m +v m v m = 5 = 5 = 3 v m = 3...(4) This is an expession fo minimum velocity (v m ) of the body at point D (Hoizontal position) So that, the stin will not be slacken at hihest point B D ) Velocity at any point C :- Let, v is the velocity at point C and hee the heiht of point C is calculated in the followin manne. But in OFC OA = OF + FA = OF + h c h c = OF cos θ = OF OC V = 5l l(1 cosθ)...(5) This is an expession fo minimum velocity v of the body at any point C of the vetical cicle so that the stin will not be slacken at hihest point. Eneies at diffeent points (consevation of eney) A) Eney at lowest point (A) :- (T.E.) at point A = (P.E.) at point A + (K.E.) at point A = mv c = 1 m 5 = 1 5m = 5 m (6) B) Eney at hihest point (B) :- (T.E.) at point B = (P.E.) at point B + (K.E.) at point B = mh b + 1 mv c

15 = m + 1 m = m + 1 m = 4m + m = 5 m. 1 C) Eney at middle point (D) :- (T.E.) at point D = (P.E) at point D + (K.E.) at point D = mh d + 1 mv m = m + 1 m. 3 = m + 3 m = m + 3m T m = mv m = m3 d) Tension in the stin at any point :- T mcosθ = mv T m = 3m T = mv + mcosθ Special Case: Diffeence in tension at lowest and hihest points. At lowest point B, T L = mv L At hihest point A, T H = mv H Now T L -T H = + m - m mv L = mv L = m + m mv H m mv H v L v H + m + m = 5 m (8) = m 5 + m Similally, we can detemine the eney of body, at any point C on the veticle cicle and it will be 5 m. We conclude that, the eney is conseved (constant) at evey point on the vetical cicle. Tension in the stin at diffeent points :- A)Tension in the stin at hihest point (B) :- T h + m = mv h T h = mv h - m T h = m - m= m m T h = 0 B) Tension in the stin at lowest point (A) :- T L - m = mv L T L = mv L + m = m5 T L = 6m + m C) Tension in the stin at middle Point (D) :- = m 4 + m = 4m + m T L -T H = 6m In non unifom cicula motion, the adial component of acceleation (a = v ), which is pependicula to instantaneous velocity and diected towads cente of cicle and is not constant, since the speed has diffeent values at diffeent points. In non unifom cicula motion, thee is component of acceleation alon tanent to a point on cicumfeence of cicle and is not equal to the ate of chane of speed. This tanential component of acceleation is alon v, if paticle is speedin up and opposite to v if is slowin down. a ad = v and a tanential = d v = α Kinetical Equations fo cicula motion in analoy with linea motion :- The physical quantities fo cicula motion in analoy with linea motion ae iven :-

16 Linea motion Linea displacement (s) Linea velocity (v) Linea acceleation (a) Linea momentum (p) 1) v= u + at Whee v=final velocity u=initial velocity ) v = u +as 3)s=ut + 1 at Cicula motion Anula displacement(θ) Anula velocity (ω) Anula acceleation (α) Anula momentum (L) 1) ω = ω 1 + α t Whee ω = Final velocity ω 1 = initial velocity ) ω = ω 1 + αθ 3) θ = ω 1 t + 1 α t 9) Centipetal o centifual Foce (F c ) = mv = mω = mvω 10) Equations fo cicula motion in analoy with the linea motion Whee i) ω = ω 1 + α t ω = Final velocity ω 1 = initial velocity ii) ω = ω 1 + αθ Impotant Fomulae Type I & II 1)Anula displacement(θ) =ωt = π T t = πn t iii) θ = ω 1 t + 1 α t Type III 1)The maximum speed of the vehicle movin alon the hoizontal cuved oad is iven by )Anula velocity(ω) = v = π T = πn 3)Anula acceleation(α) = ω ω 1 t Whee, ω = final anula velocity ω 1 = initial anula velocity n = final fequency = π n n 1 t v = μ Whee,μ =coefficient of fiction between the oad suface and the tyes = adius of cuved oad = acceleation due to avity )The maximum speed of the vehicle movin alon the bandked oad n 1 = initial fequency 4)Linea velocity (v) =ω = π T = πn v max = μ + tanθ 1 μ tanθ 5)Linea acceleation (a) = α 6) Peiod of evolution (T) = 1 = π n 7)Fequency of eolution(n) = 1 T = v v = π ω π = ω π Whee,θ = anle of bankin And if the fictional foce is nelected o coefficient of fiction is consideed to be zeo then maximum speed of the vehicle is iven by v = tanθ 8) Centipetal accelation (a c ) = v = ω = vω The anle of bankin is iven by

17 v 1 θ = tan 3)A cyclist movin vey fast alon a hoizontal cuved oad has to lean inwad in ode to keep his balance, the maximum speed of the cycle is iven by v = tanθ Wheee, = adius of the cuved oad = acceleation due to avity θ = anle of inclination o anle made by the cylist with the vetical Hee also 1 v θ = tan Type IV 1)Conical Pendulum :- Peiod of conical pendulum Whee, = l = lenth of the stin = adius of vetical cicle ) Velocity at lowest point v L = 5 = 5l Whee, = l = lenth of the stin = adius of vetical cicle 3) Velocity at midpoint(hoizontal level) v L = 3 = 3l Whee, = l = lenth of the stin = adius of vetical cicle 4) Velocity at any point v L = v 1 (1 cosθ) Whee, = l = lenth of the stin = adius of vetical cicle T = π l cos θ = π = π l 5)Eney at points of vetical cicle Fequency of conical pendulum n = 1 T = 1 π = 1 l cos θ π = 1 π l S No. Position of the object P.E.(mh) K.E.( 1 mv ) T.E. Whee, h= distance between cente of hoizontal cicle and the iid suppot (hiht of conical pendulum) l = lenth of conical pendulum θ = semivetical anle of conical pendulum Type V Vetical Cicula Motion 1) Velocity at hihest point 1 At Lowest Point At Middle Point 3 At Hihest Point 0 5 m 5 m m 3 m 5 m m 1 m 5 m v h = = l 6)Tension in the stin at diffeent points of vetical

18 cicle :- 1) Tension at any point T = mv + mcosθ ) Tension at hihest point T = mv H - m 0 3) Tension at Lowest point T = mv L + m 6m 3) Tension at middle point T = mv m 3m Poblems Type I & II 1. What is the anula speed of a paticle which, movin unifomly in a cicle, pefoms 40 evolutions pe minute? (Ans 5.1 ad/s). If the speed of evolutions of an object chanes fom Hz to 4 Hz in 5 seconds, calculate its anula acceleation. (Ans.51 ad/s ) 3. The fequency of a paticle pefomin unifom cicula motion chanes fom 60.p.m to 180.p.m in 0 seconds. Calculate the anula acceleation. (0.68 ad/s ) 4. Detemine the anula acceleation of a otatin body which slows down fom 500.p.m to est in 10 seconds. (Ans ad /s ) 5. The speed of evolutions of an object, pefomin cicula motion is inceased fom 30 ev/min to 10 ev/min in 5 seconds. Calculate the anula acceleation, assumin it to be unifom. If the adius of the cicle is 1.5 m calculate the initial and final linea speeds of the paticle. (Ans 0.38 ad/s, 4.71 m/s, m/s) 6. The minute hand of a clock is 10cm lon. Calculate the linea speed of the tip of the minute hand. (Ans x 10-4 m/s) 7. Calculate the anula speed of otation of the eath about its axis. If the adius of the eath is 6400 km, what is the linea speed of an object situated on the equato? (Ans 7.6 x 10-5 ad/s, m/s) 8.An aeoplane movin with a speed of 300m/s takes a tun of adius 1500 m in a hoizontal plane Calculate its anula speed and the time it takes to complete a semicicle. (Ans 0. ad/s,15.7s) 9. A paticle pefoms cicula motion with a constant anula acceleation of 4 ad /s. If the adius of the cicula path is 0cm and the initial anula speed of the paticle is ad /s, find the (i) anula speed of the paticle afte 0.5 second (ii) anula displacement of the paticle in 0.5 second (iii) tanential acceleation of the paticle. (Ans (i) 4 ad /s (ii) 1.5 ad (iii)0.8 m / s ) 10. To simulate the acceleation of lae ockets astonauts ae spun at the end of a lon otatin beam of adius 9.8 m What anula velocity is equied to eneate a centipetal acceleation, 8 times the acceleation due to avity? (Ans.88 ad/s) 11. A 0.5 k mass is evolved in a hoizontal cicle of adius 0cm. Calculate the centipetal foce actin upon it, if its anula speed of evolution is 0.6 ad/s. (Ans N) 1. A moto ca of mass 100 k is bein diven with a unifom speed of 75.6 km/h oun a hoizotal cuve of adius 50m. Find the centipetal foce actin on the the ca. (Ans x 10 4 N) 13. An object of mass 0.5 k is tied to a stin and whiled on a hoizontal cicle of adius 60 cm with a speed of 5 m/s What is the foce exeted on the object by the stin. (Ans 10.4N) 14. An object of mass 50 ams is tied to one end of a stin and evolved unifomly in a hoizontal cicle of adius 10 cm. If the maximum tension the stin can bea is 100 N, find the maximum speed with which the object can evolve. Also

19 Find the maximum numbe of evolutions the object can pefom in one minute. (Ans 1.91 m/s p.m.) 15. One end of a stin, 1 m lon, is fixed and a body of mass 500 is tied to the othe end. If the beakin tension is 98 N, find the maximum anula velocity of the body that the stin can withstand when otated in a hoizontal cicle. (Ans 14 ad/s ) 16. A piece of stone of mass 00ams is held in a slin and whiled with constant speed in a hoizontal cicle of adius 1.5 m. If it pefoms 90 evolutions pe minute calculate its (i) peiod,(ii) anula velocity (iii) linea velocity (iv) centopetal acceleation and (v) centipetal foce. (Ans (i) 0.67 s,(ii)9.4 ad/s (iii)14.13m (iv)133.1 m/s (v)6.6n) 17. Two objects of equal masses ae taken. One object is tied to one end of a stin and kept on a smooth table. The stin is passed thouh a small hole in the table and the othe object is tied to its fee end. How many evolutions pe minute should be pefomed by the object on the table in a cicle of adius 5 cm so that it balances the suspended object. Nelect the effect of fiction. (Ans 59.8.p.m ) 18.A mey-o-ound (diamete = 10 m)otates with a fequency of 6.p.m. Find the centifual foce expeienced by a boy (mass = 50k) in the mey-o-ound. (Ans 98.6N) 19.Wheel otates at constant fequency 600 evolutions pe minute. Find the anle thouh which it otates in one second. (Ans 00π adian) 0.Two wheels A and B ae evolvin with same linea velocities. If adius of the wheel A is twice that B Compee thei anula velocities (Ans ω A : ω B = 1:) Type -III 1. A cuve on a hihway is in the fom of an ac of adius 100 m The oad suface is 13 m wide and its oute ede is 5 m above the inne ede. Find the maximum speed with which vehicles can be safely diven alon the cuve. (Ans 0.m/s). Find the anle of bankin of a ailway tuck of adius of cuvatue 1600 m if the optimum velocity of the tain is 03m/s. Also find the elevation of the oute tack above the inne tack if the distance between the tacks is 1.8 m. (Ans cm) 3. The cicumfeence of a tack is 1.56 km Find the anle of bankin of the tack if the maximum speed at which a ca can be diven safely alon it is 5 m/ s. (Ans ) 4. A vehicle is movin alon a cicula oad which is inclined to the hoizontal at 10 0 The maximum velocity with which it can move safely is 36 km / h Calclulate the lenth of the cicula oad [Hint:- Lenth of the cicula oad = π, whee is the adius of the path] (Ans m) 5. Find the anle which a bicycle and its ide must make with the vetical when tavellin at 18 km/ h aound a hoizontal cuve of adius 10m. (Ans ) 6.The speed at which a ca is neotiatin a cicula tun of adius 40m on an unbanked level oad is 36km/h. If µ between the oad suface and the tyes is 0. will the divin be safe? =9.8m/s (No) 7.A coin is kept on hoizontal disc at the distance 0.5 m fom cente of the disc The disc is otatin about vetical axis passin thouh its cente. If coefficient of fiction between disc and coin is 0.3 calculate maximum anula velocity without slippin the coin. (Ans.45ad/s) 8.Moto ca is oin alon tun of adius 70m with velocity 50km/h. If oad is banked at anle 30 0 will the ca safe? (Ans Vmax = 19.9m/s and ca will safe) 9. Find the maximum speed of a ca which can be safely diven alon a cuve of adius 100m if coefficient of fiction between the tyes and the oad is 0.. (Ans 14m/s) 30. A ca can be safely diven with a maximum speed of 5. km/h alon a hoizontal cuve of adius 0m Find the coefficient of fiction between the ca tyes and the oad suface (Ans 0.5) 31. A amophone disc pefoms 78 evolutions pe minute. If the coefficient of fiction between the coin and the disc is

20 0.5, find the maximum distance fom the cente of the disc at which the coin can be kept without slippin (Ans 7.35 cm ) Type -IV 3. A conical pendulum has a lenth 150 cm and the adius of the cicle is 50cm Find the peiod of the pendulum. (Ans.387 s) 33. Fo a conical pendulum, the time equied fo 30 evolutions is 45 seconds. If the lenth of the pendulum is 1 m what is the adius of the cicula path?. (Ans 0.83 m ) 34. A bob of mass 00 is suspended by a liht stin of 80 cm. The adius of the cicle taced by the bob is 40 cm.find the velocity of the bob and the peiod of pendulum. (Ans 1.5 m/s 1.67s) 35. A bob of mass 50 hans by a liht stin fom a fixed point and otates in a hoizontal cicle. If the lenth of the stin is 50cm and the bob makes 10 evolutions in 14 s find the adius of the obit and the tension in the stin. (11.3cm;0.503N) Type V 36. A 500 mass tied to a stin is whiled in a vetical cicle of adius 1m Find the minimum velocity at the top so that the stin does not slacken. (Ans 3.13 m/s) 37. A moto cyclist ides in vetical cicles in a hollow sphee of adius 5m Find the minimum speed equied so that he does not lose contact with the sphee at the hihest point. (Ans 7m/s) 38. The vetical section of a flyove bide is in the fom of an ac of a cicle of adius 100m Find the maximum speed with which a tuck can coss the bide without losin contact with it at its hihest point. The cente of avity of the tuck is 1 m above the oad suface. (Ans 31.46m/s) 39. A small body is tied to a weihtless stin and is otated in a vetical cicle of 5 cm adius. What is the minimum evolutions of otation so that the stin does not slacken at the top. (Ans 1 ev/s) 40. A 50 mass is tied to a stin and is whiled in a vetical cicle of adilus 1m if the tension in the stin at the hihest point of the cicula pathe is zeo find the tension intje stin when the stin makes an anle of 60 0 with the vetical. (Ans.05 N) 41. A moto cyclist ides in vetical cicles in a hollow sphee of adius 5 m. Find the minimum anula speed equied so that he does not lose contact with the sphee at the hihest point. (Ans 1.4 ad / s) 4. A bucket containin wate is tied to one end of a ope 8 m lon and evloved about the othe end in a vetical cicle. Find the minimum numbe of evolutions pe minute in ode that the wate in the bucket does not spill.(ans p.m) 43.A body of mass 5k tied with a stin of lenth 1m is otated in vetical cicle. Find kinetic eney, potential eney and total eney of the stone at hihest position and lowest position. (Ans. Ep=98J (hihest), Ep=0 (lowest), Ek=4.5J(lowest), Ek=1.5J (hihest), E=1.5J (hihest), E=1.5 J (lowest)) BOARD BOOK PROBLEMS 44.A tuntable otates at 100 ev/min. Calculate its anula speed in ad/s and in deees/s. (Ans:- ω = 600 deee/s) 45.An poject of mass 100 move aound cicumfeence of cicle of adius m with constant anula speed 7.5 ad/s. Compute a) Linea speed b) foce diected towads cente. (Ans:- v = 15 m/s,f=11.5 N) 46. A spheical bob of diamete 3 cm havin a mass 100 is attached at the end of stin of lenth 48.5 cm. Find anula velocity and tension in the stin, if bob is otated now at 600.p.m. in a hoizontal cicle (Nelect the weiht of stin and bob). If the same bob is now whiled in a vetical cicle of same bob is now whiled in a vetical cicle of same adius, what will be the diffeence in tension at lowest and hihest point? (Ans:- ω = 6.84 ad/s,t=197.4 N,T -T 1 =5.88 N) 47. A flat cuve on hihway has a adius of cuvatue 400 m. A ca ounds the cuve at a speed of 3 m/s. What is the minimum value of coefficient of fiction that

21 will pevent ca fom slidin? ( = 9.8 m/s )(Ans:- μ = 0. 61) 48. A mete aue tain is movin at 60 km/h alon a cuved oad of adius of cuvatue 500 m at a cetain place. Find the elevation of oute ail above inne ail, so that thee is no side pessue on ial. ( = 9.8 m/s ) (Ans θ =3 15, h =0.0567cm) 49. An aicaft in level fliht completes a cicula tun in 100 second. i) What is the adius of cicula tun? ii) What is the anle of bankin, if the velocity of aicaft is 40 m/s? (Ans = m, θ =14 5 ) 50. In a cicus, moto cyclist havin mass of 50 k move in a spheical cae of adius 3m. Calculate the least velocity with which he must pass hihest point without loosin. Also calculate his anula speed at hihest point. (Ans:- v= 5.41 m/s,ω = ad/s) 51. A coin is kept at a distance 10 cm fom the cente of tun table of adius 1 m just beins to slip, when the tun table otates at speed of 90.p.m. Calculate the coefficient of static fiction between coin and the tuntable. ( = 9.8 m/s ) (Ans:-μ = ) 5. A stone is whiled in a vetical cicle at the end of a ope of lenth 0.5 m. Find the velocity of stone at (i) lowest position. (ii) midway position. (iii) top position to just complete the cicle ( = 9.8 m/s ) ( Ans:-V l =4.949 m/s,v m =3.834 m/s,v h =.13 m/s) 53. A stone of mass 1 k is whiled in a hoizontal cicle attached at the end of 1 m lon stin. If the stin makes an anle of 30 with vetical calculatei) peiod ii) Centipetal foce (=9.8 m/s ) (Ans:T=1.868 s, F=5.658 N) 54. A wheel has adius 30 cm. The wheel stats fo m est and attains a speed of 300.p.m. in 3 minutes. What is its anula acceleation? What is anula displacement within this time? (Ans:α = ad/s, θ=85 ad) 55. A oto has a diamete of 4.0 m. The oto is otated about cental vetical axis. The occupant emains pinned aainst wall, when the linea velocity of dum is 8 m/s. Compute coefficient of static fiction between the wall of oto and clothin of occupant (fi. 1.13) Also calculate anula velocity of dum? How many evolutions will it make in a minute? (Ans:-μ = , ω =4 ad/s,n=38.19 ev/min) Unsolved PROBLEMS FOR PRACTICE (Boad Book) 56. Calculate the anula velocity and linea velocity of a tip of minute hand of lenth 10 cm. (Ans : x 10-3 ad/s, x 10-4 m/s ) 57. Popelle blades in aeoplane ae m lon? a) When popelle is otatin at 1800 ev/min, compute the tanential velocity of tip of the blade. b) What is the tanential velocity at a point on blade midway between tip and aixs? (Ans : m/s, m/s) 58. A ca of mass 000 k ounds a cuve of adius 50 m at 90 km/h. compute its, a) Anula speed b) Centipetal acceleation c) Centipetal foce (Ans:0.1adian/s,.5 m/s,5000 N) 59. A bucket containin wate is whiled in a vetical cicle at ams lenth. Find the minimum speed at top to ensue that no wate spills out. Also find coespondin anula speed. (Assume = 0.75 m) ( Ans :.711 m/s, ad/s) 60. A moto cyclist at a speed of 5 m/s is descibin a cicle of adius 5 m. Find his inclination with vetical. What is the value of coefficient of fiction between tye and ound? ( Ans : 5 51, 0.100) 61. A stone weihtin 1 k is whiled in a vetical cicle attached at the end of a ope of lenth 0.5 m. Find tension at i) Lowest position ii) Mid position iii) Hihest position ( Ans : 58.8N, 0 N, 9.4 N) 6. An object of mass 0.5 k attached to a od of lenth 0.5 m is whiled in a cicle at constant anula speed. If the maximum tension in the stin is 5 k wt, calculate i) speed of stone ii) maximum numbe of evolutions it can complete in a minute. ( Ans : m/s, 16.7.p.m.) 63. A moto van weihtin 4400 k ounds a level cuve of adius 00 m on unbanked oad at 60 km/h. What should be minimum value of coefficient of fiction to pevent skiddin? At what anle the od should be banked fo this velocity? ( Ans: ) 64. A stin of lenth 0.5 m caies a bob with a peiod πs. Calculate anle of inclination of stin with vetical and tension in the stin. ( Ans : 9 5, N) 65. A pilot of mass 50 k in a jet aicaft while executin a loop-the loop with constant speed of 50 m/s. If the adius of cicle is 5 km, compute the foce exeted by seat on the pilot. a) at the top of loop. b) at the bottom of loop.

22 ( Ans : N, N) 66. A ball is eleased fom heiht alon the slope and move alon a cicula tack of adius R without fallin vetically downwads Show that h = 5 R 67.A block of mass 1 k is eleased fom P on a fictionless tack ends in quate cicula tack of adius m at the bottom as shown in the fiue below. What is the manitude of adial acceleation and total acceleation of the block when it aive at Q? ( Ans : 7.71 m/s ) 68.A cicula ace couse tack has a adius of 500 m and is banked to 10. If the coefficient of fiction between tyes of vechile and the oad suface is 0.5 compute: a) the maximum speed to avoid slippin. b) the optimum speed to avoid wea and tea of tyes. ( = 9.8 m/s ) (Ans : m/s, 9.39 m/s) 69. The lenth of hou hand of a wist watch is 1.5 cm. Find manitude of a) anula velocity b) linea velocity c) anula acceleation d) adial acceleation e) tanential acceleation f) linea acceleation of a paticle on tip of hou hand. ( Ans: (a) x 10-4 ad/s, b).18 x 10-6 m/s, c) 0 ad/s d) x m/s, e) 0, f) x m/s ) Poblems fo Home wok 70. If the fequency of evolution of an object chanes fom Hz to 4Hz in seconds. Calculate its anula acceleation. (Ans 6.8 ad/s ) 71. An aicaft takes a tun alon a cicula path of adius 1500m. If the linea speed if the aicaft is 300 m / s, find (i) its anula speed (ii) the time taken by it to complete 1/5 th of the cicula path. (Ans. (i) 0. ad / s (ii) 6.8 s) 7.Assumin that the eath evolves ound the sun in a cicula obit of adius 1.5 x 10 8 km. calculate its (i) anula speed, (ii) linea speed and (iii) centipetal acceleation.( (i) x 10-7 ad/s (ii).987 x 10 4 m/s(iii) x 10-3 m/s ) 73. An object of mass 0.4 k is whiled in a hozontal cicle of adius m If it pefoms 60 ev/min calculate the centipetal foce actin on it (Ans N) 74. An object of mass 0. k evolves unifomly alon a hoizontal cicle of adius 0.5 m If the centipetal foce actin on the object is 0.1 N. Calculate the anula velocity of the object (Ans 1 ad/s ) 75. A body of mass k is tied to the end of a stin m lon and evolved in a hoizontal cicle. If the beakin tension of the stin is 400 N calculate the maximum velocity of the body. (Ans 0 m/s) 76. A cetain stin beaks unde a tension of 45k wt, A mass of 100 is attached to this stin of lenth 500cm and whiled in a hoizontal cicle. Find the maximum numbe of evolutions pe second without beakin the stin. (Ans 4.73 ev/s ) 77. An object of mass 400 ams is whiled in a hoizontal cicle of adius m. If it pefoms 60.p.m. calculate the centipetal foce actin on it. (Ans N) 78. A body of mass k is tied to the fee end of a stin of lenth 1.5 m and evolved alon a hoizontal cicle with the

23 othe end fxed. The body makes 300.p.m Calculate the linea velocity, centipetal acceleation and foce actin on it. (Ans 47.1 m/s 1479 m/s 958 N) 79. Calculate the linea speed of the tip of the second hand of a clock if the second hand is 4 cm lon. ( x10-3 m /s) 80. The minute hand of a clock is 7 cm lon. Calculate the linea speed of the tip if the minute hand (1.1 x 10-4 m / s) 81. A paticle pefoms unifom cicula motion in a cicle of adius m. If its fequency if evolution is 60.p.m., find the (i)peiod of evolution (ii) linea speed (iii) centipetal acceleation of the paticle. Ans (i) 1 s (ii) 1.56 m / s (iii) m / s 8. A body of mass 1k tied at the end of a stin 0.5 m lon is whiled in a hoizontal cicle the othe end of the a stin bein fixed. The beakin tension in the stin is 50 N Find the eatest speed that can be iven to the body (5 m/ s ) 83. A body of mass 0.4 k is tied to a stin 1 m lon and whiled in a hoizontal cicle with a speed of m/ s with the othe end of the stin fixed. Find the (i) tension in the stin (ii) maximum speed with which the body can be whiled if the beakin stenth of the stin is 10 k wt. (Ans (i) 1.6 N (ii) m / s) 84. A stin beaks unde a tension of 10k wt. If the stin is used to evolve a body of mass 1 ams in a hoizontal cicle of adius 50 cm, find (i)the maximum speed with which the body can be evolved (ii) the coespondin peiod of evolution.[ans(i)63.90 m/ s (ii) x 10 - s] 85. An electon (mass = 9.1 x k) evolves aoumd a poton in a stable obit of adius x m. If the electo static foce of attaction between the electon and the poton is x N, find the velocity of the electon. (Ans.18 x 10 6 m / s) 86. Find the anula speed of the spinnin motion of the eath fo which a body on the equato will have zeo appaent weiht. What will be the lenth of the day at that time? (Radius of the eath = 6400km) (Ans ω= 1.37 x 10-3 ad / s, T = 5077 second) 87. A tuntable has a constant anula speed of 45.p.m. Expess this in adians pe second and deees pe second. If the adius of the tuntabe is 0.5 m what is the linea speed of a point on its im? (4.71 ad/s 70 0 /s.355 m/s) 88. A body of mass 1 k is tied to the end of stin 1m and evolved alon hoizontal cicle with the othe end fixed and makes 40 pm Detemine i) anula velocity ii) Linea velocity iii) centipetal acceleation iv) centipetal foce v) centifual foce (Ans.5.1ad/s,5.1m/s m/s, N, N ) 89.A lon beam is otated at a constant anula speed of 1.75 ad/s in a hoizontal plane about a vetical axis situated at one end. How fa fom the axis must a man be standin on the beam. if the expeiences a centifual acceleation of 9.8 m/s? (Ans. 3. m) 90.An electon moves alon a cicula path havin a adius of cuvatue of 10cm. If it has a centipetal acceleation of 4 x m/s find the linea speed. (Ans. x 10 5 m/s) 91.How fast should the eath otate about its axis so that the appaent weiht of a body at the equato be zeo? How lon would a day be then? Take adius of eath = 6400km 9. The beakin tension of a stin is 80 k wt A body of mass 1 k is tied to the stin and evolved in a hoizontal cicle of adius m. Find the maximum numbe of evolutions pe second the body can make.(3.153.p.s) 93. A body of mass 5k is tied at the end of a stin 1. m lon and evolved in a hoizontal cicle. If the beakin tension in the stin is 300 N find the maximum numbe of evolutions pe minute the body can make.(ans 67.6.p.m ) 94. A body of mass 1 k is tied to a stin and evolved in a hoizontal cicle of adius 1m. Calculate the maximum numbe of evolutipns pe minute so that the stin does not beak. The beakin tension of the stin is 9.86 N. (Ans 30.p.m. ) 95. A body of mass 100 ams is tied to one end of a stin of lenth 5m and whiled in a hoizontal cicle with the othe end of the stin fxed. Find the maximum fequency with which the body cna be whiled if the stin beaks unde a tension of 45 k wt. (Ans 4.73 Hz) 96. What is the anula velocity of the minute hand of a clock? What is its anula displacement in 0 minutes? If the