Rotational Kinematics - Exercises Performed, Esercizi di Fisica

Scarica Rotational Kinematics - Exercises Performed e più Esercizi in PDF di Fisica solo su Docsity! Chapter 3 Rotational Kinematics Abstract Chapter 3 is devoted to rotational motion on horizontal and vertical planes and on loop-the-loop. 3.1 Basic Concepts and Formulae If s be the length of the arc and r the radius of the circle, the angle in radians is given by θ = s/r (3.1) If the angular displacement θ = θ2 − θ1 in the time interval t = t2 − t1, then the average angular velocity ω̄ = θ2 − θ1 t2 − t1 = θ t (3.2) The instantaneous angular velocity ω is defined as the limiting value of the ratio as t approaches zero. ω = dθ dt (3.3) In case the angular speed is not constant a particle would undergo an angular acceleration α. If ω1 and ω2 are the angular speeds at time t1 and t2, respectively, then the average acceleration of the particle is defined as ᾱ = ω2 − ω1 t2 − t1 = ω t (3.4) The instantaneous angular acceleration is the limiting value of the ratio as t approaches zero. α = dω dt (3.5) 103 104 3 Rotational Kinematics Rotation with Constant Angular Acceleration ω = ω0 + αt (3.6) ω2 = ω2 0 + 2αθ (3.7) θ = ω0t + 1 2 αt2 (3.8) θ = (ω0 + ω)t 2 (3.9) Linear and angular variables for circular motion, scalar form s = rθ (3.10) v = ωt (3.11) aT = αr (3.12) where aT is the tangential component of acceleration. The radial component of acceleration is aR = v2/R = ω2 R (3.13) Total acceleration a = √ a2 T + a2 R (3.14) Vector form: v = ω × r (3.15) a = α × r + ω × v (3.16) Motion in a Horizontal Plane Typical problems are as follows: (i) A coin is placed at a distance r from the centre of a gramophone record rotating with angular frequency ω = 2π f . Find the maximum frequency for which the coin will not slip if μ is the coefficient of friction. This problem is solved by equating the centripetal force to the frictional force. (ii) An object of mass m attached to a string is whirled around in a horizontal circle of radius r with a constant speed v. Find the tension in the string. The problem is solved by equating the centripetal force to the tension in the string. T = mv2/r 3.2 Problems 107 Eliminating v between the two equations and noting that cos θ = 1 − h/R we find h = R/3. (iv) A motorcyclist goes around in a vertical circle inside a spherical cage. Find the minimum speed at the top so that he may successfully complete the circular ride. Here we equate the reaction on the cage to the total weight of the rider plus motorcycle mg = mv2/R or v = √ gR (v) Loop-the-Loop is a track which consists of a frictionless slide connected to a vertical loop of radius R, Fig. 3.1. Let a particle start at a height h on the slide and acquire a velocity v at the bottom of the loop. If v < √ 2gR, the particle will not be able to climb up beyond the point B. It will oscillate in the lower semicircle about the point D. If √ 2gR < v < √ 5gR the particle will be able to climb up the arc BC and leave at some point E and describe a parabolic path. If v > √ 5gR, the particle will be able to execute a complete circle. This corresponds to a height h = 2.5R. Fig. 3.1 Loop-the-loop 3.2 Problems 3.2.1 Motion in a Horizontal Plane 3.1 Show that a particle with coordinates x = a cos t , y = a sin t and z = t traces a path in time which is a helix. [Adapted from Hyderabad Central University 1988] 108 3 Rotational Kinematics 3.2 A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a varies with time t as a = k2r t2, where k is a constant. Show that the power delivered to the particle by the forces acting on it is mk4r2t5/3 [Adapted from Indian Institute of Technology 1994] 3.3 A particle is moving in a plane with constant radial velocity of magnitude ṙ = 5 m/s and a constant angular velocity of magnitude θ̇ = 4 rad/s. Determine the magnitude of the velocity when the particle is 3 m from the origin. 3.4 A point moves along a circle of radius 40 cm with a constant tangential accel- eration of 10 cm/s2. What time is needed after the motion begins for the nor- mal acceleration of the point to be equal to the tangential acceleration? 3.5 A point moves along a circle of radius 4 cm. The distance x is related to time t by x = ct3, where c = 0.3 cm/s3. Find the normal and tangential acceleration of the point at the instant when its linear velocity is v = 0.4 m/s. 3.6 (a) Using the unit vectors î and ĵ write down an expression for the position vector in the polar form. (b) Show that the acceleration is directed towards the centre of the circular motion. 3.7 Find the angular acceleration of a wheel if the vector of the total acceleration of a point on the rim forms an angle 30◦ with the direction of linear velocity of the point in 1.0 s after uniformly accelerated motion begins. 3.8 A wheel rotates with a constant angular acceleration α = 3 rad/s2. At time t = 1.0 s after the motion begins the total acceleration of the wheel becomes a = 12 √ 10 cm/s2. Determine the radius of the wheel. 3.9 A car travels around a horizontal bend of radius R at constant speed V . (i) If the road surface has a coefficient of friction μs, what is the maximum speed, Vmax, at which the car can travel without sliding? (ii) Given μs = 0.85 and R = 150 m, what is Vmax? (iii) What is the magnitude and direction of the car’s acceleration at this speed? (iv) If μs = 0, at what angle would the bend need to be banked in order for the car to still be able to round it at the same maximum speed found in part (ii)? [University of Durham 2000] 3.10 The conical pendulum consists of a bob of mass m attached to the end of an inflexible light string tied to a fixed point O and swung around so that it describes a circle in a horizontal plane; while revolving the string generates a conical surface around the vertical axis ON, the height of the cone being ON = H , the projection of OP on the vertical axis (Fig. 3.2). Show that the angular velocity of the bob is given by ω = √ g/H , where g is the acceleration due to gravity. 3.2 Problems 109 Fig. 3.2 Conical pendulum 3.11 If the number of steady revolutions per minute of a conical pendulum is increased from 70 to 80, what would be the difference in the level of the bob? 3.12 A central wheel can rotate about its central axis, which is vertical. From a point on the rim hangs a simple pendulum. When the wheel is caused to rotate uniformly, the angle of inclination of the pendulum to the vertical is θ0. If the radius of the wheel is R cm and the length of the pendulum is 1 cm, obtain an expression for the number of rotations of the wheel per second. [University of Newcastle] 3.13 A coin is placed at a distance r from the centre of a gramophone record rotat- ing with angular frequency ω = 2π f . Find the maximum frequency for which the coin will not slip if μ is the coefficient of friction. 3.14 A particle of mass m is attached to a spring of initial length L0 and spring constant k and rotated in a horizontal plane with an angular velocity ω. What is the new length of the spring and the tension in the spring? 3.15 A hollow cylinder drum of radius r is placed with its axis vertical. It is rotated about an axis passing through its centre and perpendicular to the face and a coin is placed on the inside surface of the drum. If the coefficient of friction is μ, what is the frequency of rotation so that the coin does not fall down? 3.16 A bead B is threaded on a smooth circular wire frame of radius r , the radius vector r making an angle θ with the negative z-axis (see Fig. 3.3). If the frame is rotated with angular velocity ω about the z-axis then show that the bead will be in equilibrium if ω = √ g r cos θ . 112 3 Rotational Kinematics 3.2.2 Motion in a Vertical Plane 3.26 A particle is placed at the highest point of a smooth sphere of radius R and is given an infinitesimal displacement. At what point will it leave the sphere? [University of Cambridge] 3.27 A small sphere is attached to a fixed point by a string of length 30 cm and whirls round in a vertical circle under the action of gravity at such a speed that the tension in the string when the sphere is at the lowest point is three times the tension when the sphere is at its highest point. Find the speed of the sphere at the highest point. [University of Cambridge] 3.28 A light rigid rod of length L has a mass m attached to its end, forming a simple pendulum. The pendulum is put in the horizontal position and released from rest. Show that the tension in the suspension will be equal to the magnitude of weight at an angle θ = cos−1(1/3) with the vertical. 3.29 In a hollow sphere of diameter 20 m in a circus, a motorcyclist rides with sufficient speed in the vertical plane to prevent him from sliding down. If the coefficient of friction is 0.8, find the minimum speed of the motorcyclist. 3.30 The bob of a pendulum of mass m and length L is displaced 90◦ from the vertical and gently released. What should be the minimum strength of the string in order that it may not break upon passing through the lowest point? 3.31 The bob of a simple pendulum of length L is deflected through a small arc s from the equilibrium position and released. Show that when it passes through the equilibrium position its velocity will be s √ g/L , where g is the accelera- tion due to gravity. 3.32 A simple pendulum of length 1.0 metre with a bob of mass m swings with an angular amplitude of 60◦. What would be the tension in the string when its angular displacement is 45◦? 3.33 The bob of a pendulum is displaced through an angle θ with the vertical line and is gently released so that it begins to swing in a vertical circle. When it passes through the lowest point, the string experiences a tension equal to double the weight of the bob. Determine θ . 3.2.3 Loop-the-Loop 3.34 The bob of a simple pendulum of length 1.0 m has a velocity of 6 m/s when it is at the lowest point. At what height above the centre of the vertical circle will the bob leave the path? 3.2 Problems 113 3.35 A block of 2 g when released on an inclined plane describes a circle of radius 12 cm in the vertical plane on reaching the bottom. What is the minimum height of the incline? 3.36 A particle slides down an incline from rest and enters the loop-the-loop. If the particle starts from a point that is level with the highest point on the circular track then find the point where the particle leaves the circular groove above the lowest point. 3.37 A small block of mass m slides along the frictionless loop-the-loop track as in Fig. 3.7. If it starts at A at height h = 5R from the bottom of the track then show that the resultant force acting on the track at B at height R will be√ 65 mg. Fig. 3.7 3.38 In prob. (3.37), the block is released from a height h above the bottom of the loop such that the force it exerts against the track at the top of the loop is equal to its weight. Show that h = 3R. 3.39 A particle of mass m is moving in a vertical circle of radius R. When m is at the lowest position, its speed is 0.8944 √ 5gR. The particle will move up the track to some point p at which it will lose contact with the track and travel along a path shown by the dotted line (Fig. 3.8). Show that the angular position of θ will be 30◦. Fig. 3.8 114 3 Rotational Kinematics 3.40 A block is allowed to slide down a frictionless track freely under gravity. The track ends in a circular loop of radius R. Show that the minimum height from which the block must start is 2.5R so that it completes the circular track. 3.41 A nail is located at a certain distance vertically below the point of suspension of a simple pendulum. The pendulum bob is released from a position where the string makes an angle 60◦ with the vertical. Calculate the distance of the nail from the point of suspension such that the bob will just perform revolutions with the nail as centre. Assume the length of the pendulum to be 1 m. [Indian Institute of Technology 1975] 3.42 A test tube of mass 10 g closed with a cork of mass 1 g contains some ether. When the test tube is heated, the cork flies out under the pressure of the ether gas. The test tube is suspended by a weightless rigid bar of length 5 cm. What is the minimum velocity with which the cork would fly out of the test tube so that the test tube describes a full vertical circle about the point of suspension? Neglect the mass of ether. [Indian Institute of Technology 1969] 3.43 A car travels at a constant speed of 14.0 m/s round a level circular bend of radius 45 m. What is the minimum coefficient of static friction between the tyres and the road in order for the car to go round the bend without skidding? [University of Manchester 2008] 3.3 Solutions 3.3.1 Motion in a Horizontal Plane 3.1 x = a cos t (1) y = a sin t (2) z = t (3) Squaring (1) and (2) and adding x2 + y2 = a2(cos2 t + sin2 t) = a2 which is the equation of a circle. Since z = t , the circular path drifts along the z-axis so that the path is a helix. 3.2 a = dv dt = k2r t2 v = ∫ dv = k2r ∫ t2dt = k2r t3 3 Power, P = Fv = mav = mk2r t2 k2r t3 3 = mk4r2t5 3 3.3 Solutions 117 (ii) vmax = √ 0.85 × 9.8 × 150 = 35.35 m/s (iii) aN = v2 R = (35.35)2 150 = 8.33 m/s towards the centre of the circle (iv) tan θ = v2 gR = (35.35)2 9.8 × 150 = 0.85 ∴ θ = 40.36◦ 3.10 Equating the horizontal component of the tension to the centripetal force T sin α = mω2 R (1) Furthermore, the bob has no acceleration in the vertical direction. T cos α = mg (2) tan α = R H = ω2 R g ∴ ω = √ g H 3.11 Using the results of prob. (3.10), the difference in the level of the bob H = H1 − H2 = g [ 1 ω2 1 − 1 ω2 2 ] (1) ω1 = 2π f1 = 140 60 π (2) ω2 = 2π f2 = 160 60 π (3) Using (2) and (3) in (1) and g = 980 cm, H = 31.95 cm. 3.12 The centripetal force acting on the bob of the pendulum = mω2r , where r is the distance of the bob from the axis of rotation, Fig. 3.10. For equilibrium, the vertical component of the tension in the string of the pendulum must balance the weight of the bob ∴ T cos θ0 = mω2r (1) Further, the horizontal component of the tension in the string must be equal to the centripetal force. ∴ T sin θ0 = mω2r (2) Dividing (2) by (1) 118 3 Rotational Kinematics Fig. 3.10 tan θ0 = ω2r g = 4π2n2r g (3) where n = number of rotations per second. From the geometry of Fig. 3.10, r = R + L sin θ0 n = 1 2π √ g tan θ0 R + L sin θ0 3.13 The equilibrium condition requires that the centripetal force = the frictional force, mω2r = μ mg ∴ fmax = ω 2π = 1 2π √ μg r 3.14 Let the spring length be stretched by x . Equating the centripetal force to the spring force mω2(L0 + x) = kx ∴ x = mω2L0 k − mω2 Therefore, the new length L will be L = L0 + x = kL0 k − mω2 and the tension in the spring will be m ω2L = m ω2kL0 k − mω2 3.3 Solutions 119 3.15 As the drum rotates with angular velocity ω, the normal reaction on the coin acting horizontally would be equal to mω2r , (Fig. 3.11). As the coin tends to slip down under gravity a frictional force would act vertically up. If the coin is not to fall, the minimum frequency of rotation is given by the condition Frictional force = weight of the coin μmω2r = mg ∴ ω = 1 2π √ g μr Fig. 3.11 3.16 The bead is to be in equilibrium by the application of three forces, the weight mg acting down, the centrifugal force mω2 R acting horizontally and the nor- mal force acting radially along NO. Balancing the x- and z-components of forces (Fig. 3.12) N sin θ = mω2 R N cos θ = mg Fig. 3.12 122 3 Rotational Kinematics Fig. 3.14 centre of gravity produces a clockwise torque. The condition for the maximum speed vmax is given by equating these two torques: mv2 max r h = mga or vmax = √ gra h = √ 9.8 × 100 × 0.75 1.0 = 27.11 m/s 3.23 (a) When a vehicle takes a turn on a level road, the necessary centripetal force is provided by the friction between the tyres and the road. However, this results in a lot of wear and tear of tyres. Further, the frictional force may not be large enough to cause a sharp turn on a smooth road. If the road is constructed so that it is tilted from the horizontal, the road is said to be banked. Figure 3.15 shows the profile of a banked road at an angle θ with the horizontal. The necessary centripetal force is provided by the horizontal component of the normal reaction N and the horizontal component of frictional force. Three external forces act on the vehicle, and they are not balanced, the weight W , the normal reaction N , and the frictional force. Balancing the horizontal components mv2 r = μ mg cos2 θ + N sin θ or N sin θ = mv2 r − μ mg cos2 θ (1) Balancing the vertical components mg = N cos θ − μmg cos θ sin θ or N cos θ = mg + μmg cos θ sin θ (2) 3.3 Solutions 123 Fig. 3.15 Dividing (1) by (2) tan θ = v2/r − μ g cos2 θ g + μ g cos θ sin θ vmax = √ gr (μ + tan θ) (b) For θ = 30◦, μ = 0.25, g = 9.8 m/s2 and r = 100 m, vmax = 28.47 m/s. 3.24 At latitude λ the distance r of a point from the axis of rotation will be r = R cos λ where R is the radius of the earth. The angular velocity, however, is the same as for earth’s rotation ω = 2π T = 2π 86, 400 = 7.27 × 10−5 rad/s The linear velocity v = ωr = ωR cos λ = 7.27 × 10−5 × 6.4 × 106 × cos 60◦ = 232.64 m/s 3.25 The speed of the plane must be equal to the linear velocity of a point on the surface of the earth. Suppose the plane is flying close to the earth’s surface, ω = 7.27 × 10−5 rad/s (see prob. 3.24) v = ωR = 7.27 × 10−5 × 6.4 × 106 = 465.28 m/s = 1675 km/h. 3.3.2 Motion in a Vertical Plane 3.26 Let a particle of mass m be placed at A, the highest point on the sphere of radius R with the centre of O. Let it slide down from rest along the arc of the great circle and leave the surface at B, at depth h below A, Fig. 3.16. Let the radius OB make an angle θ with the vertical line OA. The centripetal 124 3 Rotational Kinematics Fig. 3.16 force experienced by the particle at B is mv2/R, where v is the velocity of the particle at this point. Now the weight mg of the particle acts vertically down so that its component along the radius BO is mg cos θ . So long as mg cos θ > mv2/R the particle will stick to the surface. The condition that the particle will leave the surface is mg cos θ = mv2 R (1) or cos θ = v2 gR (2) Now, in descending from A to B, the potential energy is converted into kinetic energy mgh = 1 2 mv2 (3) or v2 g = 2h (4) using (4) in (2) cos θ = 2h R (5) Drop a perpendicular BC on AO. Now cos θ = OC OB = R − h R (6) Combining (5) and (6), h = R 3 Thus the particle will leave the sphere at a point whose vertical distance below the highest point is R 3 . 3.3 Solutions 127 Measure potential energy with respect to A, the equilibrium position. At the point B, at height L the mechanical energy is entirely potential energy as the bob is at rest. As the vertical height is L , the potential energy will be mgL. When the bob is released, at the point A, the energy is entirely kinetic, poten- tial energy being zero, and is equal to 1 2 mv2 A. Conservation of mechanical energy requires that 1 2 mv2 A = mgL or v2 A = 2 gL (2) Using (2) in (1) TA = 2mg + mg = 3mg Thus the minimum strength of the string that it may not break upon passing through the lowest point is three times the weight of the bob. 3.31 Let the ball be deflected through a small angle θ from the equilibrium position A, Fig. 3.19. θ = s L (1) Fig. 3.19 where s is the corresponding arc. Drop a perpendicular BC on AO, so that the height through which the bob is raised is AC = h. Now, h = AC = OA − OC = L − L cos θ = L(1 − cos θ) = L [ 1 − 1 + θ2 2! + · · · ] ∴ h = Lθ2 2 = s2 2L (2) where we have used (1). 128 3 Rotational Kinematics From energy conservation, mgh = 1 2 mv2 ∴ v = √ 2gh = s √ g L 3.32 In coming down from angular displacement of 60◦ to 45◦, loss of potential energy is given by mg(h1 − h2) = mgL(1 − cos 60◦) − mgL(1 − cos 45◦) = 0.207 mgL Gain in kinetic energy = 1 2 mv2 ∴ 1 2 mv2 = 0.207 mgL = 0.207 mg (∵ L = 1 m) or v = 2.014 m/s The tension in the string would be T = mv2 L + mg cos 45◦ = mg [ 4.056 gL + 0.707 ] N = 1.12 mg N 3.33 When the bob is displaced through angle θ , the potential energy is mgL(1 − cos θ). At the lowest position the energy is entirely kinetic 1 2 mv2 = mg L (1 − cos θ) (1) The tension in the string will be T = mg + mv2 L = mg + 2mg (1 − cos θ) (2) where we have used (1) By problem T = 2mg (3) From (2) and (3) we find cos θ = 1 2 or θ = 60◦ 3.3 Solutions 129 3.3.3 Loop-the-Loop 3.34 If the bob of the pendulum has velocity u at B, the bottom of the vertical circle of radius r such that √ 2gr < u < √ 5gr (1) then the bob would leave some point P on the are DA (Fig. 3.20). Here √ 2gr = √ 2 × 9.8 × 1 = 4.427 m/s and √ 5gr = √ 5 × 9.8 × 1 = 7 m/s Fig. 3.20 Therefore (1) is satisfied for u = 6 m/s. Drop a perpendicular PE on the horizontal CD. Let PE = h and PO make an angle θ with OD. When the bob leaves the point P, the normal reaction must vanish. mv2 r − mg sin θ = 0 (2) Loss in kinetic energy = gain in potential energy 1 2 m u2 − 1 2 mv2 = mg(h + r) (3) sin θ = h r (4) Eliminating v2 between (2) and (3) and using (3), with u = 6 m/s and r = 1.0 m, h = 1 3 [ u2 g − 2r ] = 0.558 m 132 3 Rotational Kinematics From (1) and (2) h = 2.5R 3.41 Let the nail be located at D at distance x vertically below A, the point of suspension, Fig. 3.21. Initially the bob of the pendulum is positioned at B at height h above the equilibrium position C. h = L(1 − cos θ) = L(1 − cos 60◦) = 1 2 L (1) Fig. 3.21 where L = 1 m is the length of the pendulum. When the pendulum is released its velocity at C will be v = √ 2gh = √ gL (2) The velocity needed at C to make complete revolution in the vertical circle centred at the nail and radius r is v = √ 5gr (3) From (2) and (3) r = 1 5 L (4) Therefore x = AD = AC − DC = L − L 5 = 0.8 L = 0.8 × 1 m = 80 cm 3.42 If M and m are the mass of the test tube and cork, respectively, and their velocity V and υ respectively, momentum conservation gives 3.3 Solutions 133 MV = mv (1) or v = M m V = 10 1 V = 10 V (2) Condition for describing a full vertical circle is that the minimum velocity of the test tube should be V = √ 5gr = √ 5 × 980 × 5 = 156.5 cm/s Therefore the minimum velocity of the cork which flies out ought to be v = 10 V = 1565 cm/s = 15.65 m/s 3.43 Equating centripetal force to frictional force mv2 r = μ mg μ = v2 gr = (14)2 9.8 × 45 = 4 9