1. Physics

2. Session Particle Dynamics - 5

3. 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

4. Every day we see the sun rise and set. We see the sun moving round us. But ……. Session Opener Science says the earth moves around its axis and in a day the sun hardly moves. Have you asked yourself why our eyes observe a wrong phenomenon ?

5.   F v  F v  F v  F v  F v  F  F v  F v  F v Object A moves in a circular path radius r fixed : constrained motion  Directed towards center  Perpendicular to  Constant in magnitude  And change direction continuously. For uniform motion v is constant.  has a direction Circular Motion   F v  F v

6. Angular Variables (Constant Speed) v average =  avg. r Instantaneous angular velocity v = r r OO r  r  r  –  o = t Angular kinematical equation for constant .

7. Angular Variables (Variable v) v changes (a constant) v = v 0 + at  =  0 + t  : angular acceleration r r tt r tt r tt =  0 + t  is tangential.

8. Class Exercise

9. 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? = 200 radians/s. Solution :

10. Unit Vector Along Radius and Tangent [ox and oy : fixed reference frame]. r and  also define position Transfer origin from O to P ox || px’ oy || py’ P has acceleration  Reference frame non inertial O p x y r (t) y’ x’

11. Unit Vector Along Radius and Tangent (origin P) define a non inertial reference frame Define a unit vector perpendicular to (along )  y  x

12. Unit Vector Along Radius and Tangent  y  x Angular velocity vector is the rate of change of radial unit vector.

13. Radial and Tangential Acceleration v constant   constant can not change the magnitude of Particle moves in a circle  y  x v is perpendicular to arar

14. Radial and Tangential Acceleration When  changes = particle moves in the circle (r constant) = v changes  arar = tangential acceleration appears atat v a

15. Centripetal Force in Circular Motion Object in circular motion has a t and a r a t only changes magnitude of v. : non-uniform circular motion a r : necessity for circular motion. : no change in magnitude of v. r O arar As a r exists,an external force F cp must exists. F cp is a radial force, called centripetal force

16. Class Exercise

17. 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

18. Solution Hence answer is (b)

19. Solution N B is largest. From shape of the track,

20. 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.

21. Solution Pendulum moves along the arc of a vertical circle. Resolving the motion along T (X-axis) and perpendicular to T (Y-axis) mg sin ma (along y) (Tangential)

22. 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.

23. 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)

24. Solution = 0.05 m. (approx.)

25. Centripetal Force Source of Friction Object moves in circular track radius r Centripetal force is friction. fsfs

26. Banking of Curves Object moves along circular track. Track banked towards center O.  O y axis : N cos = mg x axis : N sin = F c =mv 2 /r mg N  Ncos=mg If velocity > v : object moves outward to increase r If velocity < v : object moves inward to decrease r.  < 90 0

27. Conical Pendulum P moves in horizontal circle at end of string OP fixed to rigid support O. Tension T supplies F cp Y axis : T cos = mg x axis : T sin = mv 2 /r (r=L sin ) Time t p to complete one revolution : L  mg T 

28. Class Exercise

29. Class Exercise - 1 Two similar cars, having masses of m 1 and m 2 move in circles of equal radii r. Car m 1 completes the circle in time T 1 and car m 2 completes the circle in time T 2. If the circular tracks are flat, and identical, then the ratio of T 1 to T 2 is

30. Solution Centripetal force As circles are identical and flat, friction supplies centripetal force in both cases. Hence answer is (d)

31. 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 .)

32. Solution In both cases, the friction force f supplies the braking force. F=Nmg. Deceleration = g. In applying the brakes car must stop within distance d. In taking a circular path, the maximum radius is d. So applying the brakes is the better option.

33. Class Exercise - 8 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/s 2 ) (Straight sections are of equal length)

34. Solution Motor cyclist B wins the race. In semicircular section: Length of track A = km

35. Solution Length of track B = 2 km Semicircular portion is negotiable in equal time. m.c.B is faster, so will complete straight parts faster.

36. 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/s 2 )

37. Solution The mass will not slide if 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.

38. Centrifugal Force in Circular Motion P moves in circle (radius r) with angular velocity  Reference frame centered at origin O : Reference frame is inertial Reference frame with P as origin Reference frame is non inertial. P P P P O r 

39. Centrifugal Force in Circular Motion P is at rest with respect to itself. A pseudo force F pseudo to be added as frame is non inertial. So in frame of P; F pseudo = m 2 r away from center. F pseudo is called centrifugal force. P O F cp F pseudo

40. Class Exercise

41. 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

42. 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 acceleration. Hence answer is (d)

43. 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.

44. Solution Apparent acceleration due to gravity: So except poles it increases everywhere. Hence answer is (d)

45. Thank you