Partial preview of the text
Download rotational dynamics notes and more Study notes Physics in PDF only on Docsity!
A i 1. Rotational Dynamics 12th Science 12th - Physics Syllabus 4. Introduction 2. Characteristics of circular motion 2.1 Kinematics of circular motion 2.2 Dynamics of circular motion 3. Application of uniform circular motion 3.1 Vehicle along a horizontal circular track 3.2 Well (or wall) of death 3.3 Vehicle on a banked road 3.4 Conical pendulum ° 4. Vertical circular motion 4.4 Point mass undergoing vertical circular motion under gravity 4.2 Sphere of death 4.3 Vehicle at the top of a convex over bridge 5. Moment of inertia as an analogous quantity for mass 6. Radius of gyration 7. Theorem of parallel axes and theorem of perpendicular axes 7.4 Theorem of parallel axes 7.2 theorem of perpendicular axes 8. Angular momentum or moment of linear momentum 8.1 Expression for angular momentum in terms of moment of inertia g. Expression for torque in terms of moment of inertia 10. Conservation of angular momentum 1. Rolling motion 11.1 Linear acceleration and speed while pure rolling down an inclined plane { Theory Notes Q. Define circular motion and give its examples. Q. Define and explain radius vector. Q. Define and explain angular displacement. Circular Motion . . The motion of an object (particle) along the circumference of a circle is called as circular motion. Examples 1) Motion of moon around the earth. 2) Motion of every planet around the sun. Radius Vector or Position Vector (r) . 1) The vector drawn from the centre of the circle to the position of the particle performing circular motion 1S called as radius vector. . . ce 2) The radius vector is also known as position vector, if origin is at centre. 3) Magnitude of radius vector is always constant. 4) The direction of radius vector changes continuously. Explanation - . Consider a particle performing circular motion in anticlockwise direction along the circumference of a circle having radius r. Fig -Radius vector, Let the particle is at point A. The vector (ry) drawn from the centre Oto the point A ie. OAis radius vector at A. Similarly OBis the radius vector at point B. Angular displacement (30) Definition The angle traced at the centre of the circle by a radius vector in given time when the particle performing circular motion moves from one position to other, is called as angular displacement. Explanation 1) Consider a particle performing circular motion, moves from position A to B through distance s along the circle in time /. Let r be the radius of the circle. 2) During this time, the radius vector rotates through angle ZAOB=@. 3) Thus @ is the angular displacement of particle in time /. Y - y Fig - Angular Displacement Q. What is angular velocity? Q. Explain angular acceleration. Angular Velocity (@) . ; \) erhe rate of change of angular displacement with respect to time is called as angular velocity. ; , 2) If @) and @,, are angular displacements ofa particle at time h and 1 respectively then the angular velocity is given as - =A _ 30 h-t ot w= 3) If the angular displacement of the particle changes from @ to 0+60 in a very small time interval o/,then the instantaneous angular velocity is given as - Fig. Angular Velocity = Nm Br gpa @8 “O= dt Thus instantaneous angular velocity is the limiting rate of change of angular displacement. 4) SI unit of angular velocity is rad/sec. 5) Instantaneous angular velocity is vector quantity and it is in the direction of angular displacement given by right hand rule or right handed screw rule. 6) In UCM, the angular velocity is constant. Angular Acceleration (@ ) 1) The rate of change of angular velocity with respect to time is called as angular acceleration. 2) If the angular velocity of a particle performing circular motion is @, and @ at time f and f, then, Average angular acceleration(@ ) = a = oe. ath 3) Ifthe angular velocity of the particle changes from wto w + dw in a short time interval dt, then the instantaneous angular accel- eration is given by - Thus instantaneous angular acceleration is the limiting rate of change of angular velocity. 4) It is vector quantity. 5) Its SI unit is rad/ sec’ and dimensions are [’M°T™]. 6) Its direction is given by right hand rule or right handed screw rule. 7) If angular velocity of a particle performing circular motion increases, then angular acceleration is positive and it is in the direction of angular velocity. 8) If angular velocity of the particle decreases, then angular accel- eration is negative and it is in the opposite direction of angular velocity. decreasing increasing = Fig. Direction of angular acceleration Q. Prove that v = rw, Relation between Linear Velocity (v) and Angular Velocity (w) 1) When a particle performs circular motion, it has linear velocity (vy) which is along the tangent to the circle and angular velocity (w)along the axis of rotation. 2) Consider a particle performing circular motion along circumfer- ence of a circle having radius 7 with constant angular velocity @). 3) Let the particle displaces from point A to point B through small angle we in a very small time interval dy. The length of the arc AB is Os. Fig. Relation between v & w 4) We know that - = are angle = ca dius 09 = 8 “Os = 7.60 5) For infinitesimal quantity, ds will be a straight line. Hence linear velocity of the particle (v) can be given as - v= lms, _ |, 180 = limFy a) oo = (lim Gp anf tie SP = co} {linear velocity = radius X angular veloctiy] This is the relation between linear velocity and angular velocity in scalar form. Q. Give the relation between angular velocity and frequency. Q. Explain the angular velocity of particle in UCM. Q. What is radial acceleration? a2) an r velocity of the particle. spp 22 wn T Where T is period and w is angulal 3) Sl unit of 1 is sec” or hertz (Hz)i.e.r.p.s. or ©-p-S. Relation between Angular Velocity (wm) and Frequency (n) We have ~ on dn period of revolution (T)= o OT seen and linear velocity (V) = P.Q.ucssese (2) From equations (1) and (2) We also have 1 frequency (2) = (v= 2am) From equation (2) rw = 2zrn w= 2nn Angular Velocity of Particle in UCM . : 1) InUCM, the particle travels displacement of 2a in one revolution. .. angular displacement = 27 2) The time required for one revolution of particle is equal to the period of revolution (7). Angulalr displacement .. Angular velocity = tine taken 3) It is a vector quantity. 4) Its SI unit is rad/sec. Radial Acceleration or Acceleration in UCM In UCM, the magnitude of linear velocity remains constant but its direction changes continuously. Hence UCM is an accelerated motion. In vector form, radial acceleration is given as - > a,=-w'r G=-(Plinwufro=} Where / is unit vector in the direction of radius vector r, The negative sign shows that direction of acceleration is opposite to that of radius vector. We have This is the relation between radial acceleration and period. Direction of acceleration of a particle performing UCM is same as the direction Q. Define and explain centripetal force. rm QO of change in velocity,dv, when 6t — 0. If 66 is reduced, dv tends to be at right angle to v. ; . Hence acceleration is at right angle to the linear velocity. The accel- eration of the particle performing UCM is along the radius and towards the centre. Therefore it is called as radial acceleration or centripetal acceleration. _ In non-uniform circular motion, the magnitude of a, is not constant. The acceleration responsible for changing the magnitude of velocity and is called as tangential acceleration ar. In non-uniform circular motion, angular velocity w is also changing . dw at every instant. This is due to the angular acceleration @ = “dt Centripetal Force (CPF) The force acting on particle performing circular motion, which is directed along the radius towards the centre of the circle is called centripetal force. According to Newton’s second law of motion. force = mass acceleration Here, This is the equation for centripetal force (Fir). In this equation, Feris centripetal force m is mass of the particle performing circular motion. r is radius of the circle v is linear speed of the particle @ angular speed of the particle In vector form- —+ —my. —> = Fo =F or Foe =~ mw? This is the equation for centripetal force in vector form. In this equation, r is radius vector f is unit vector in the direction of Fig. Centripetal Force Centripetal force is real force necessary for maintaining the circular motion. It changes the direction of velocity. It does not perform any work because there is no displacement in the direction of force. (hare is Newton (NV) and CGS unit is dyne and dimensions are Q. Why banking of road is necessary? Q. Give the equation for the maximum safety speed with which a car can be safely driven along a curved hori- zontal road. Q. Derive the expression for the maximum safety speed with which vehicle should move along a curved hori- zontal road. State the signifi- cance of it. Q. Define banking of road and angle of banking. Q. What is banking of roads? Q. Obtain an expression for maximum safety speed with which vehicle can be safely driven along curved banked road. Q. Show that the angle of banking is independent of mass of vehicle. of Road oves along horizontal curved road, then n of vehicle is provided by force ‘he road surface. g with a speed v along a Necessity of Banking 1) When a vehicle mo i y yr circular moti necessary force fo ce eeo and t of friction between th! 2) Let m be the mass of the car movin horizontal curved road of radius '. Then for safe tat entripetal force = force of friction 2 «BE = yng r ure sb=fual Where yp is the coefficient of frictio: surface, ; ; bic the vehies i ives the maximum speed with whic icle 8 ean be catcly driven rizontal curved road. driven along hoi 1 ad. 4) This oe ctmerced is came for all vehicles, as it is independent of the mass of the vehicle. . 5) If we have to increase the speed limit of vehicle, then we have to increase the friction between road surface and tyre. 6) But increase in friction force causes the wear and tear of tyres. Therefore the friction force is not economical, alsoit is not reliable, 7) Therefore we need some another arrangement of curved road surface on which fast moving vehicle can go safely. Therefore the banking of road is necessary. in between tyres of car and road Banking of Road The process in which, outer edge of the road is raised above than the inner edge at a suitable angle, is called as banking of road. Angle of Banking The angle between surface of road and horizontal surface of road is known as angle of banking (8). Fig. Banking of road Expression for Maximum Safe Speed alon, Road (upper side limit) a Geewed Backed Consider a vehicle of mass m moving along a curved banked road. Nos 6 N 6 fos 0 Nsin 6 t) f mg fsin 0 Let 'r' be the radius of Let Vnn be the hago oy and @ be the angle of banking. m speed of the vehicle along banked road. Here the various forces acting on the vehicle are - a) Weight of the vehicle mg) acting vertically downwards, b) The Normal reaction N perpendicular to the road surface in upward direction. The normal reaction NV can be resolved into two mutually perpen- dicular components i.e. N.cos @ which is vertical component and N sin@ which is horizontal component. . As Nsin@ is less than the centrifugal force, the direction of force of static friction f between road and the tyres is directed along the inclination of the road downwards. . The friction force (f,) between tyres of vehicle and road surface can be resolved into- a) f.cos@ - along horizontal direction b) fsin@ - along vertically downward direction Here Ncos@ is balanced by the weight of the vehicle and component fsin@ of frictional force. «Neos @ = mg + fsin@ “mg = Ncos0—fsin6.......(1) Nsin@ and f,cos@ provide the centripetal force necessary for the circular motion. «. Nsin@ + fcos 0 = x Mili = Nsin6 + fcos 6.......(2) Divide equation (2) by (1) vi, _ Nsin@+fcos 6 rg Neos@—fsind The magnitude of frictional force depend on speed of vehicle for given road surface and tyres of vehicle. The frictional force produced will be - f= UN Where UL, is the coefficient of static friction. “re Neos@—y,Nsin@ Dividing numerator and denominator on LHS by Ncos@ , Nsin@ , “Nos @ . Yn — Neos@ — Ncos@ “rg Neos@ _ #,Nsin@ Ncos@ = Neos@ » Vos = tan + os “rg 1—yp,tan@ 2 Vmax = Vi r| Hs * tan 8 1—y,tan@ os Vinx = of sh Cases If 0 = 0°, If £, = cot A, Vinx = 00 Thus for = 45°, Vinx = 00 This shows that maximum safe speed of vehicle on a banked road is greater than that of curved horizontal road. Q. Derive the equation for the minimum speed of the vehicle driven in horizontal circle in a well (or wall) of death b) fsin@ - along vertically downward direction - Here weight of the vehicle (mg) is balanced by Ncos@ and fsin@ omg = fsin8 + N COS Oo... a) Nsin@ and f.cos@ are in opposite directions. Hence the required centripetal force will be me = Nsin @ — fic08 O......02) Divide equation (2) by (1) rg ~ The magnitude of frictional force depend on speed of vehicle for given road surface and tyres of vehicle. The frictional force produced will be - f= UN Where UL. is the coefficient of static friction Var _ Nsind ~ u.Ncos 8 ~ Neos@+yu.Nsind Dividing numerator and denominator on LHS by Ncos@ Nsin@ _ 4.Ncos6 — Neos@ Neos _ Neos@ , #.Nsin@ Neos@ " Ncos@ = tan d= n, ~ 1+y.tan@ | (tan-. \ "Vein =f 72 os Vinin jr 45) Cases If 4. = 0, Vain = Yrgtan@ If £4. 2 tan, Van = 0 Minimum speed of vehicle in a well (or wall) of death Consider a vehicle of mass m is driven in horizontal circle in a well of death having radius r. Let v is the speed of the vehicle Y mg Fig. Well of Death Here the various forces acting on the vehicle are - a) weight of the vehicle (mg) acting vertically downwards. b) Normal reaction N acting horizontally and towards the centre. c) Force of static friction f between vertical wall and the tyres acting vertically upwards. Weight of the vehicle is balanced by the force of static friction mg =f Normal reaction provides the necessary centripetal force- Ne p Force of static friction f. is always less than or equal to s6.N Q. Define conical pendulum. Obtain an expression for the angle made by string of conical pendulum with vertical. Hence deduce the expression for linear Speed of bob of the conical pendulum, © Vin = = Vis Ms This is the equation for the minimum required speed of the vehicle in a well (or wall) of death. ———— qounieal pengialue is a simple pendulum in which the bob describes the horizontal circle and string describes the cone. i dulum ular speed of conical pen consider bet oe ines 'm' is tied to a string of length / from a 8. . : ; 2) ite S revolving in horizontal circle of radius rwith velocity 'v'at the bottom of the string. s AX mg Fig. Conical Pendulum a 3) As the bob revolves, the string sweeps over a cone. Hence it is known as conical pendulum. 4) Let the string makes an angle @ with vertical. The forces acting on the bob are a) Weight mg acting vertically downward b) Tension T acting upwards along the string. 5) The tension T is resolved into two mutually perpendicular compo- nents i.e. Tcos@ vertically upward and Tsin@ horizontal and towards the centre of the circle, Tcos@ balances the weightmg 14 Q. What is vertical circular motion? Show that the motion of an object revolving in vertical circle is non-uni- form circular motion. Let the bob is revolving in horizontal circle of radius r with velocity el vat the bottom of the string. We have, vee = o/rgtan 0 } - a 1 2D | agg) From the figure sin® =F v.r= sind Hence equation (1) becomes - _ -2 isin @ i> Bi gtand ss = This is the equation for period of conical pendulum. If 2@ is small cos@ = 1 [Lv T= a as Thus time period is independent of @ if 20 is small, Vertical Circular Motion 1) Consider an object of mass m tied at the end of an inextensible string and rotated in vertical circle of radius r as shown in figure. The object will perform the vertical circular motion along the circle. 2 A M4, B y, B Fig. Vertical circular motion 3) Suppose Bis the lowest point while A is the highest point in the circle. When the body moves from B toA gravity opposes the motion. On the other hand as the body moves from A to B , the speed of the body increases because grav vity helps the motion. 6) Therefore speed of the body is minimum at the highest point and maximum at lowest point. 7 Thus vertical circular motion is not uniform. 4 » its speed decreases because 5) Q. Give the equations for minimum velocity and energy at different points in vertical circular motion. ©. Derive expressions for linear velocity at lowest point, midway and top posi- tion for a particle revolving ina vertical circle if it has to just complete circular motion without string slackening at top. Aly% Equation for Velocity and Energy at Different Points in Vertical Circular Motion ; 1) Consider a body of mass ‘m’ tied to an inextensible string and rotated in vertical circle of radius ‘r’. My Fig. Vertical circular motion : 2) Let V; be the velocity of body at highest point A, Vsbe the velocity at lowest point Band \. be velocity at midway point Cc. 3) The force acting on the body at highest position Aare a) Tension 7 acting in downward direction (towards the centre of circle) b) Weight mg acting in downward direction. To keep the body in circular motion, the necessary centripetal force is provided by the weight and the tension on the string. T+ mg = 1) 4) Similarly the force acting on the body at lowest point Bare a) Tension 7; acting vertically upwards (towards the centre of circle) b) Weight mgacting vertically downwards. But here two forces are acting in opposite directions, so . _ mv “omg =—* PE mg. (2) Velocity of the Body at Highest Point A (uppermost position) *~ 1) We have 2) There is certain velocity for the body below which the string becomes slack i.e. tension 7; vanishes (J) = 0) mvii 0 = M4 — mg vi tie ig “VA & é . . Vs = Te 3) This equation gives the minimum velocity at highest point, so that string doesn’t slack. 4) If the velocity is less than rg then the body will not be able to continue the circular motion. Q. Obtain expressions of Energies of Particle in vertical circular motion energy at different posi- Consider a body of mass 'm' tied to an inextensible string and rotated tions in the vertical circular | in vertical circle of radius '7". motion. Let v, be the velocity of body at highest point A, vs be the velocity }: f= yon B F=f ana —® Fig- Vertical circular motion A$-Ooint PB? at lowest point B and vc be velocity at midway point C. ™ - 1) At point A (highest) — Pp Don TE. = K.E.+ PE. clay TE. how) + 2mer SON £O But v= gr OTE = singe 2mer 2 =e ; => ~~ —(2) 0 TE. = MBP ccoreo() at yO s 2) At point B (lowest) Tne bib + Pek TE.=KE.+PE. TE.= dma +0 Neo: Vac As PE. = Oat lowest point and vs = Sgr a “TE= mgr oou(2) =z may —E) 3) At point C (midway) TE. = K.E.+ PE. “TE.= dmv + mgr But ve = y3gr “TE.= mgr + mer (3) “TR =+ OTE = ame 4) Thus from equation (1), (2) and (3) the T.E. at every point is same in 2 F . . . ie. = mgr. Hence total energy is conserved in vertical circular motion. 2. Derive an expression for difference in tensions at highest and lowest point for @ particle performing vertical | , Circular motion. r cL Q. Derive the equations for velocity at different points in | vertical circular motion ifa mass is tied to a rod. Difference in tensions at lowest and highest points The tension at lowest point B is - , he a + mg The tension at highest point A is ; ni T=" "~ mg Now - A mvi 1-1 = + me -| ; ~me| 2 2 Vv: my. TT = TA + mg “RT = 725 - vi) + 2m By law of conservation of energy- (P.E.+ KE. us =(PE.+ KE.) 20+ Smvi = mg2r+ dmv > o FS mv3— mv = (mg).2r 2. Fm} vi) = (mg).2r Waa V4 = 4gr.........(2) From equations (1) and (2) T-T =") (4gr) + Img - B-T = 6mg Vertical Circular Motion of a Mass Tied to a Rod Consider a mass m tied to a rigid rod and rotated in vertical circle of radius r. Velocity at Uppermost Point A Let v, be the velocity of the mass at the uppermost point A in vertical circular motion. As the rod is point A in vertical circular motion. As the rod is rigid no minimum velocity is required. Hence zero speed is possible at the uppermost point. VM =0 Velocity at Lowermost Point B 1) The potential energy of particle at the bottom is minimum. Therefore increase in P.E. of body when it moves from bottom to top of circular motion is mg(2r). -. PE. = 2mer 2) Increase in K.E. when particle moves from A to B is mvs ma From the law of conservation of energy, 2mer = mvs - Simi oe va = 0} = Simin “. Ve = 4gr “Va = /4gr = 2,/ er 20 
