PHYSICS - Rotational dynamics (MAHARASHTRA STATE BOARD)

PHYSICS
Chapter 1 - Rotational dynamics
Maharashtra state board
CLASS XII

1. PHYSICS CHAPTER – 1 ROTATIONAL DYNAMICS MAHARASHTRA STATE BOARD

2. CAN YOU RECALL? What is circular motion? What is the concept of centre of mass? What are kinematical equations of motion? Do you know real and pseudo forces, their origin and application?

3. Angular displacement (Circular) Denoted by ‘ ’ Angle through which an object moves on circular path. SI unit is Radian or Degree Displacement (Linear) Denoted by ‘ d’ or ‘s’ (Length) It is a minimum distance between initial and final position Displacement is sometime zero. SI unit is ‘m’

4. Angular Velocity Denoted by ‘ ’ Rate of change of the position angle of an object with respect to time.  = = SI unit is   Velocity Denoted by ‘ v’ Rate of change of displacement per unit time v = SI unit is ‘m/s’  

5. Angular Acceleration Denoted by ‘ ’ It is the time rate of change of angular velocity.  = = SI unit is  = = =   Acceleration Denoted by ‘ a’ Rate of change of velocity per unit time a = SI unit is ‘m/ ’ a = = = =  

6. CIRCULAR MOTION In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path E.g., – Artificial satellite

7. Characteristics of circular motion It is an accelerated motion It is periodic in nature

8. n Frequency of an object to rotate T Periodic time / Period of circular motion So, w = 2 n [ i.e. w = 2 ] ……….( n = ) Where, w Angular Velocity  

9. Relation between angular velocity and linear velocity Given, V = w.r ………..(magnitude) Where, v w r   Relation between angular velocity and acceleration Given, a = r ………..(magnitude) a = …………(w = ) a = vw …………(r = )  

10. Example: A fan is rotating at 90 rpm. It is then switched OFF. It stops after 21 revolutions. Calculate the time taken by it to stop assuming that the frictional torque is constant. Solution: = 3 The angle through which the blades of the fan move while stopping is θ = 2πN = 2π (21) = 42 π rad, ω = 0 (fan stops). Using equations analogous to kinematical equations of motion  

11. DYNAMICS OF CIRCULAR MOTION Centripetal force (CPF) Acceleration is responsible for circle motion, = - The force providing acceleration is centripetal or radial force F = ma = m = m ……..(magnitude) Resultant of all real force, = m = m ( - ) = - m = - m (v = wr or w = ) b) Centrifugal force Force away from centre Force equal in magnitude to real force but opposite It is pseudo force arising due to centripetal acceleration.  

12. Real force + Pseudo force It is non-real force, but not a imaginary force. Resultant force in frame of reference is ‘zero’ Hence, it is ‘net pseudo force’  

13. APPLICATION OF UCM Vehicle along a horizontal circular track: Theoretical proof: In this, force acting on car is Weight (mg) Normal reaction (N) Force of friction ( ) Here, mg = N …..(1) And = mr ( ) …….(v= wr ) = m …….[from equation (1)] …….. (m = ) =   Fig.: Vehicle on a horizontal road.

14. Now, ………..(According to given condition) N ……….( At maximum speed, = = v =   b) Well (or wall) of death: r And force acting on vehicle is, Normal reaction (N) Weight (mg) Force of static friction   Fig.: Well of death.

15. Here, friction force prevent the downward slipping So, its magnitude is equal to mg i.e., N = mr N = Now, frictional force always less than or equal to ) i.e.   Fig.: Well of death.

16. Vehicle on a Banked Road BANKED ROAD “Defined as the phenomenon in which the edge are raised for the curved roads above the inner edge to provide the necessary centripetal force to the vehical’s so that they take a safe turn.” Forces acting on the vehicle, weight mg, vertically downwards and normal reaction N Ncosθ Nsinθ Hence, N cos 𝜽 = mg …….( i ) and N sin 𝜽 = mr = …….(ii) Dividing ( i ) by (ii), tan   Vehicle on a banked road.

17. CASES Most safe speed: For a particular road, r and θ are fixed. So, tan tan V = Here, v b) Banking angle tan c) Upper speed limit For minimum possible speed: For maximum possible speed:   Fig: Banked road: lower speed limit. Fig: Banked road: upper speed limit.

18. UPPER SPEED LIMIT For minimum possible speed: Safest speed, , sin ……..(1) = N sin …….(2) Divide equation (1) by (2), For minimum possible speed, N For, For maximum possible speed: Speed, , - sin ……..(1) = N sin …….(2) Divide equation (1) by (2), For maximum possible speed, N  

19. c) Conical pendulum Consider, B Force acting on bob: Weight (mg) Now, Here, = m r ……..( i ) Divide equation ( i ) by (ii)  

20. = ………..(iii) Now, r T Time period of revolution of bob ………..(iv) In equation (iii) put value of ‘r’ w = T = 2  

21. VERTICAL CIRCULAR MOTION CASE 1: Mass tied to string Uppermost position (A) Both, weight (mg) and force due to tension are downward i.e., towards the center . There are only resultant centripetal force acts towards centre. So, mg + ………(a) But at the top: Minimum energy applied by string to mass So, So, mg =   Fig: Vertical circular motion.

22. VERTICAL CIRCULAR MOTION CASE 1: Mass tied to string Lowermost position (B) Force due to the tension, is vertically upwards, and opposite to mg. If is the speed at the lowermost point, So, ………(b) The verticle displacement is 2r = h Hence, Mg (2r) = - m ( ) = 4 g r We know that,   Fig: Vertical circular motion.

23. From equation (1) = 4 g r = 4 g r Also, subtracting equation (a) from (b) (4 g r) V …….(Topmost) V ………(Lowermost)  

24. VERTICAL CIRCULAR MOTION c) Position when string is at the horizontal Force due to the tension is the only force towards the centre as weight mg is perpendicular to the tension. This force due tension is the centripetal force. So, And Lowermost position   Fig: Vertical circular motion.

25. VERTICAL CIRCULAR MOTION CASE 2: Mass tied to rod Consider a bob tied to a rod and whirled along a vertical circle. Zero speed is possible at the uppermost point.   STRING ROD STRING ROD

26. Vehicle at the Top of a Convex Over-Bridge Forces acting on the vehicle are Weight mg and Normal reaction force N, M g – N = As the speed is increased, N goes on decreasing. Normal reaction is an indication of contact. Thus, for just maintaining contact, N = 0. This imposes an upper limit on the speed as   Fig.: Vehicle on a convex over-bridge. Sphere of Death

27. MOMENT OF INERTIA “A quantity expressing a body tendency to resist angular acceleration, which is sum of the product of the mass of each particle in the body with the square of its distance from the axis of rotation.” It depends upon Individual masses of objects (particles) Distribution of these masses about the given axis of rotation Where, m mass r distance from axis of rotation to the particle According to the definition, I = M  

28. Expression for moment of inertia (I) Consider be the masses of individual particle be the distance from axis of rotation to the individual particle According to the definition Moment of Inertia = = I = For single particle,  

29. Expression for KINETIC ENERGY Consider be individual masses of rigid body be the distance between axis of rotation and point masses Here, for individual mass of particle the K.E. will be K.E. = So, for all the point masses of the rotating body Rotational K.E. = …….+ Rotational K.E. = …….+ We know, v = r w Rotational K.E. = …….+ Rotational K.E. = …….+ Rotational K.E. = = …………..[I =  

30. Expression OF ‘I’ FOR UNIFORM RING Moment of Inertia of a ring. Said to be uniform if mass of any body practically situated uniformly on the circumference of circle It entire mass practically at equal distance from centre that is mass (m) Also distance R is same for the ring So, expression is, I = M ….[For ring]  

31. Expression OF ‘I’ FOR UNIFORM DISC Consider, Disc has negligible thickness It is uniform if its mass per unit area and it composition is same throughout. Where, M R FOR THE INNER SURFACE dm r 2 Area of ring = 2 . dr   Moment of Inertia of a disk.

32. Expression OF ‘I’ FOR UNIFORM DISC So, surface density ( ) = Here, r For whole, By integration the inertia: I = = = = dr = 2 dr = 2 = 2 =   Moment of Inertia of a disk. I = = …(For disc)  

33. RADIUS OF GYRATION (K) Value of ‘K’ is same as ‘R’ So, I = M Consider, Moment of inertia for ring Moment of inertia for disc Similarly, Conclusion,  

34. THEOREM OF PARALLEL AXIS Fig.: Theorem of parallel axes. Consider, ‘MOP’ is axis passing through point ‘O’ ‘ACB’ is axis passing through point ‘C’ H be the distance between point ‘O’ and ‘C’ be M.I of axis going from point ‘C’ be M.I of axis going from point ‘O’ D and N are individual point mass. Here, ……( i ) ……(ii)  

35. THEOREM OF PARALLEL AXIS Fig.: Theorem of parallel axes. + …..(iii) Here, …From ( i ) ….. Centre of mass From equation (iii)  

36. THEOREM OF perpendicular AXIS Fig.: Theorem of Perpendicular axes. Consider, and be the M.I of axis passing through point ‘M’ and ‘N’ P be the point individual mass PM = y , PN = x Here, Now, So, + dm  

37. Angular Momentum or Moment of Linear Momentum If is the instantaneous linear momentum is the Angular momentum where is the position vector from the axis of rotation. In magnitude, it is the product of linear momentum and its perpendicular distance from the axis of rotation. ∴ L = P × r sin θ where θ is the smaller angle between the directions of P and r  

38. Expression for Angular Momentum Consider, N number of particles of masses at respective perpendicular distances from the axis of rotation. Linear speeds, , , …….., Linear momentum of the first particle, Angular momentum, Similarly, , ,….., Thus, magnitude of angular momentum of the body is, L = + + ….. + L = ( + + ….. + L = I w Where, I = + + ….. +  

39. Expression for Torque Fig: Expression for torque. Consider, As the object rotates, all these particles perform circular motion with same angular acceleration α, but with different linear (tangential) accelerations Force experienced by the first particle is, Thus, the torque experienced by the first particle is of magnitude, Similarly, , ,……., Magnitude of the resultant torque is, + The relation is analogous to f = m a.  

40. Conservation of Angular Momentum Angular momentum or the moment of linear momentum of a system is given by, Differentiating with respect to time, Now, = and = Now, = 0 Thus, if = 0 or ……………(Principle of conservation of angular momentum)  

41. Rolling Motion Consider an object of moment of inertia I, rolling uniformly. v = Linear speed of the centre of mass R = Radius of the body = Angular speed of rotation of the body M = Mass of the body K = Radius of gyration of the body Total kinetic energy of rolling = Translational K.E. + Rotational K.E. + + (M ) (1 + )  

42. Linear Acceleration and Speed While Pure Rolling Down an Inclined Plane Gravitational P.E. is converted into K.E. of rolling. + (1 + ) Linear distance travelled along the plane is s = If a is the linear acceleration along the plane, 2as = - 0 For pure sliding, without friction, the acceleration is g sin θ and final velocity is .