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Physics Session Particle Dynamics - 5 Session Objective 1. Circular motion 2. Angular variables 3. Unit vector along radius and tangent 4. Radial and tangential acceleration 5. Dynamics of circular motion 6. Centripetal force in circular motion 7. Circular hoops 8. Centrifugal force in circular motion Session Opener Have you asked yourself why our eyes observe a wrong Every Science day says we the see earth the sun moves risephenomenon ? and around set. its Weaxis seeand the in sun a day moving the round sun hardly us. But moves. ……. Circular Motion A moves in atocircular Object F Perpendicular v path radius r fixed : constrained motion F Directed towards center F Constant in magnitude F And v change direction continuously. s r has a direction For uniform motion v is constant. Lim. 0 ds rd F v Angular Variables (Constant Speed) average s angular velocity : avg. vaverage r t t t vaverage= avg.r d Lim t 0 Instantaneous t dt angular velocity v = r r vector relation v tan rt cons tan t v :cons rad t :axial vector. SIunit d dt sec r – o = t r O r Angular kinematical equation for constant . Angular Variables (Variable v) 1 v changes (t) v(t) (a v :constant) along t angent r acceleration along t angent v = v0 + + t at := 0 + t = 0 : angular acceleration 1 22 1 s v0tt at t 0 2 2 2 20 2 r is tangential. r t Class Exercise Class Exercise - 5 A particle moves with a constant linear speed of 10 m/s in a circular path of radius 5 cm. What is its angular velocity? Solution : v 10 m/s r 5 102 m = 200 radians/s. Unit Vector Along Radius and Tangent r i r cos j r sin [ox and oy : fixed reference frame]. r and also define position y y’ j p O (t) r (t) i x’ x Transfer origin from O to P ox || px’ oy || py’ P has acceleration Reference frame non inertial Unit Vector Along Radius and Tangent r can be defined as er i cos jsin r rer r rer (er : unit vector along r) Define a unit vector e perpendicular toer (along v ) e i sin jcos e er j y i x er ,e (origin P) define a non inertial reference frame Unit Vector Along Radius and Tangent er icos jsin der d d isin jcos e dt dt dt Angular velocity vector is the rate of change of radial unit vector. 2 d er dt2 e j y d d icos jsin dt dt 2 er er i x Radial and Tangential Acceleration Particle moves in a circle r rer r cons tan t constant v constant 2 d dr r ar 2rer 2v re v : tangential dt dt v r ar : radial v r : opposite er 2 v ar 2r v r e er v j y ar is perpendicular to v ar ar can not change the magnitude of v i x Radial and Tangential Acceleration When changes = particle moves in the circle (r constant) = v changes = tangential acceleration at appears v r rer dr 1 v re dt 2 2 2 2 a r t r dv d r e 2 er dt dtat t Tan 2 2 r T ear r rer a a at ar Centripetal Force in Circular Motion Object in circular motion has at and ar art only : necessity changes formagnitude circular motion. of v. : no change circular in magnitude of v. : non-uniform motion v2 ar v 2r r Fcp 2 v m2r m mv r Fcp ar O r As ar exists,an external force Fcp must exists. F cp is a radial force, called centripetal force Class Exercise Class Exercise - 3 A vehicle moves with constant speed along the track ABC. The normal reaction by the road on the vehicle at A, B and C are respectively. Then (a) NA NB NC (b) NB NA NC (c) NC NA NB (d) NB NA NC C A B Solution mv 2 mg NA rA mv 2 NA mg rA NB mv2 rB NA mv2 rB mv2 NB = mg+ r B NB –mg= B mg (B) A mg m v2 rA (A ) Hence answer is (b) Solution mv 2 mg NC rC mv 2 NC mg rC NB is largest. From shape of the track, mv 2 mv 2 rA rC .So NA NC rA rC N C C m g (C ) m v2 rC Class Exercise - 7 A pendulum, constructed by attaching a tiny mass m at the end of a light string of length L, is oscillating in a vertical plane. When the pendulum makes an angle with vertical, its speed is equal to v. Find the tension in the string and the tangential acceleration at that instant. Solution Pendulum moves along the arc of a vertical circle. Resolving the motion along T (X-axis) and perpendicular to T (Y-axis) mg sinma (along y) (Tangential) a g sin mv2 T mg cos (along x) (radial) L T m(g cos m v2 L y L 2 v ) L T ma mg x Class Exercise - 10 A mass m of 50 kg is set moving in a horizontal circular path around a fixed centre O to which it is connected by a spring of unstretched length of 1 m and a spring constant of 905 N/m. Find out the amount by which the spring will stretch if the speed of the mass is 1 m/s and the spring is light. 1m M F ix e d c e n tre 1m /s Solution As the mass moves, it tends to slip outwards, providing a stretching force. Spring provides the reaction (restoring force), which is the cause of centripetal force (|F| = kx) mv 2 kx (as spring has stretched by x) (r x) (1+ x ) mv2 mv2 2 (r x)x x rx 0 O k k M F F = re s to rin g fo rc e Solution k r 1 m, v 1 m/s, m 50 kg, 905 m 2 1 2 4mv x r r 2 k x 1 4 50 1 1 2 905 = 0.05 m. (approx.) Centripetal Force Source of F cp Friction Centripetal force is friction. fs Object moves in circular track radius r Fcp fmax v2 mg m r v2max g (cons tan t) r Banking of Curves Object moves along circular track. Track banked towards center O. N Ncos=mg y axis : N 2cos = 2mg v v Tan g tan x axis : Nrgsin = rFc=mv2/r mv2 Nsin r O mg If If velocity velocity < > vv:: object object moves moves outward inward toto decrease increase rr. < 900 Conical Pendulum P moves in horizontal circle at end of string OP fixed to rigid support O. Tension T supplies Fcp 2 Y axis :T cos = mg v2 sin v2 rg tan 2gL x axis (r=L sin ) rg : T sin = mv /r cos Time tp to complete one revolution : 2r L cos tp 2 v g 1 2 2 m2v2 2 Note T m g 2 r L T mg Class Exercise Class Exercise - 1 Two similar cars, having masses of m1 and m2 move in circles of equal radii r. Car m1 completes the circle in time T1 and car m2 completes the circle in time T2. If the circular tracks are flat, and identical, then the ratio of T1 to T2 is (a) m1 m2 m2 (b) m1 (c) m2 m1 (d) 1 Solution Centripetal force F1 (car m1) m112r centripetal force F2 (car m2 ) 42 m1r T12 42 m2r T22 As circles are identical and flat, friction supplies centripetal force in both cases. m1g m2g m1r 42 T12 m2r 42 T22 g 42r g T12 ............(1) 42r T22 Hence answer is (d) Dividing (2) & (1) we get, .................(2) T12 T22 1 Class Exercise - 6 The driver of a car, moving at a speed of v, suddenly finds a wall across the road at a distance d. Should he apply the brakes or turn in a circle of radius d to avoid a collision with the wall? (Coefficient of kinetic friction between the road and the tyre of the car is .) Solution In both cases, the friction force f supplies the braking force. F=Nmg. Deceleration = g. In applying the brakes car must stop within distance d. 2 v v 2f vi2 2ax 0 v 2 2( g) d d 2 g In taking a circular path, the maximum radius is d. mv 2 mv 2 v2 f mg d d d g So applying the brakes is the better option. f Class Exercise - 8 ra A 0.1 B 0.2 rb rA 1km rB 2 km A B Two motor cyclists start a race along a flat race track. Each track has two straight sections connected by a semicircular section, whose radii for track A and track B are 1 km and 2 km respectively. Friction coefficients of A and B are 0.1 and 0.2 respectively. The rules of the race requires that each of the motor cyclist must travel at constant speed without skidding. Which car wins the race? (g = 10 m/s2) (Straight sections are of equal length) Solution Motor cyclist B wins the race. v2 Skidding occurs for g r Maximun speed v rg v A 0.1 1 10 103 10 10 m/ s vB 0.2 2 10 103 20 10 m/ s In semicircular section: Length of track A = km T1 Time taken by m.c.A 102 S T1 10 Solution Length of track B = 2km T2 Time taken by m.c.B 102 s T1 10 Semicircular portion is negotiable in equal time. m.c.B is faster, so will complete straight parts faster. Class Exercise - 9 Figure shows a centrifuge, consisting of a cylinder of radius 0.1 m, which spins around its central axis at the rate of 10 revolutions per second. A mass of 500 g lies against the wall of the centrifuge as it spins. What is the minimum value of the coefficient of static friction between the mass and the wall so that the mass does not slide? (g = 10 m/s2) m Solution The mass will not slide if mg f The mass will press the wall at a force equal and opposite to the centripetal force supplied by the wall which is the reaction force. fmin m2r mg 2 10 rad / s,r 0.1,g 10 m / s2 g 1m 2r f 10 2 100 0.1 1 0.025 40 m mg Centrifugal Force in Circular Motion P moves in circle (radius r) with angular velocity Reference frame centered at origin O : Reference frame is inertial Fcp m2r. (along PO) Y' ' X' ' X' ' P r Reference frame with P as origin Y' ' Reference frame is non inertial. X' ' P O X' ' P P Y' ' Y' ' Centrifugal Force in Circular Motion P is at rest with respect to itself. A pseudo force Fpseudo to be added as frame is non inertial. So in frame of P; F Fcp Fpseudo 0. Fcp P O Fpseudo Fpseudo = m2r away from center. Fpseudo is called centrifugal force. Class Exercise Class Exercise - 2 A particle of mass m moves in a circular path of radius r with a uniform angular speed of in the xy plane. When viewed from a reference frame rotating around the z-axis with radius a and angular speed , the centrifugal force on the particle is equal to (a) m (2 02 ) a (b) m2a (b) m 0a (d) m 02a Solution Centrifugal force is a pseudo force equal to –m × (Acceleration of the frame of iron inertial frame) Non inertial frame in this case has radial 2 acceleration 0 a . centrifugal force m02 a Hence answer is (d) Class Exercise - 4 If the earth stops rotating, the apparent value of g on its surface will (assuming the earth to be a sphere) (a) decrease everywhere (b) increase everywhere (c) increase at pole and remain same everywhere (d) increase everywhere but remain the same at poles. Solution Apparent acceleration due to gravity: g g2 2R sin2 (2g 2R) g At Equater 90 g g 2R (Least) At Poll 0 g g (Largest) When 0 g g So except poles it increases everywhere. Hence answer is (d) Equator Thank you