PHYSICS CHAPTER 8 CHAPTER 8: Rotational of rigid body (8 Hours) 1 PHYSICS CHAPTER 8 WHAT IS ROTATIONAL OF RIGID BODY ? The study of the motions of a rigid body under the influence of forces and torques. WHAT IS RIGID BODY ? an object or system of particles in which the distances between particles are fixed and remain constant. 2 PHYSICS CHAPTER 8 Learning Outcome: 8.1 Rotational Kinematics (2 hour) At the end of this chapter, students should be able to: Define and use: angular displacement () average angular velocity (av) instantaneous angular velocity () average angular acceleration (av) instantaneous angular acceleration (). Relate parameters in rotational motion with their corresponding quantities in linear motion. Write and use, 2 v s r , v r , at r , ac r 2 r 3 PHYSICS CHAPTER 8 Learning Outcome: Use equations for rotational motion with constant angular acceleration, o t 1 2 ot t and 2 2 o 2 2 4 PHYSICS CHAPTER 8 8.1.1 Parameters in rotational motion 1) Angular displacement, is defined as an angle through which a point or line has been rotated in a specified direction about a specified axis. The S.I. unit of the angular displacement is radian (rad). Figure 8.1 shows a point P on a rotating compact disc (CD) moves through an arc length s on a circular path of radius r about a fixed axis through point O. Figure 8.1 5 PHYSICS CHAPTER 8 From Figure 8.1, thus s θ r where OR s rθ θ : angle (angular displacement) in radian s : arc length r : radius of the circle Others unit for angular displacement is degree () and revolution (rev). Conversion factor : 1 rev 2π rad 360 Sign convention of angular displacement : Positive – if the rotational motion is anticlockwise. Negative – if the rotational motion is clockwise. 6 PHYSICS CHAPTER 8 2) Angular velocity Average angular velocity, av is defined as the rate of change of angular displacement. Equation : θ 2 θ1 θ ωav t 2 t1 t where θ2 : final angular displacement in radian θ1 : initial angular displacement in radian t : time interval Instantaneous angular velocity, is defined as the instantaneous rate of change of angular displacement. Equation : θ dθ limit t 0 t dt 7 PHYSICS CHAPTER 8 It is a vector quantity. The unit of angular velocity is radian per second (rad s-1) Others unit is revolution per minute (rev min1 or rpm) Conversion factor: 2 1 1 rpm rad s rad s 1 60 30 Note : Every part of a rotating rigid body has the same angular velocity. Direction of the angular velocity Its direction can be determine by using right hand grip rule where Thumb Curl fingers : direction of angular velocity : direction of rotation 8 PHYSICS CHAPTER 8 Figures 8.2 and 8.3 show the right hand grip rule for determining the direction of the angular velocity. Figure 8.2 Figure 8.3 9 PHYSICS CHAPTER 8 Example 8.1 : The angular displacement, of the wheel is given by θ 5t 2 t where in radians and t in seconds. The diameter of the wheel is 0.56 m. Determine a. the angle, in degree, at time 2.2 s and 4.8 s, b. the distance that a particle on the rim moves during that time interval, c. the average angular velocity, in rad s1 and in rev min1 (rpm), between 2.2 s and 4.8 s, d. the instantaneous angular velocity at time 3.0 s. 10 PHYSICS Solution : CHAPTER 8 d 0.56 r 0.28 m 2 2 a. At time, t1 =2.2 s : At time, t2 =4.8 s : 11 PHYSICS Solution : CHAPTER 8 d 0.56 r 0.28 m 2 2 b. By applying the equation of arc length, c. The average angular velocity in rad s1 is given by 12 PHYSICS CHAPTER 8 Solution : c. and the average angular velocity in rev min1 is d. The instantaneous angular velocity as a function of time is At time, t =3.0 s : 13 PHYSICS CHAPTER 8 3) Angular acceleration Average angular acceleration, av is defined as the rate of change of angular velocity. Equation : ω2 ω1 ω av t2 t1 t where ω2 : finalangular velocity ω1 : initialangular velocity t : time interval Instantaneous angular acceleration, is defined as the instantaneous rate of change of angular velocity. Equation : ω dω α limit t 0 t dt 14 PHYSICS CHAPTER 8 It is a vector quantity. The unit of angular acceleration is rad s2. Note: If the angular acceleration, is positive, then the angular velocity, is increasing. If the angular acceleration, is negative, then the angular velocity, is decreasing. Direction of the angular acceleration If the rotation is speeding up, and in the same direction as shown in Figure 8.4. α Figure 8.4 15 PHYSICS CHAPTER 8 If the rotation is slowing down, and have the opposite direction as shown in Figure 8.5. α Figure 8.5 Example 8.3 : The instantaneous angular velocity, of the flywheel is given by 3 2 ω 8t t where in radian per second and t in seconds. Determine a. the average angular acceleration between 2.2 s and 4.8 s, b. the instantaneous angular acceleration at time, 3.0 s. 16 PHYSICS CHAPTER 8 Solution : 17 PHYSICS CHAPTER 8 Solution : b. The instantaneous angular acceleration as a function of time is At time, t =3.0 s : 18 PHYSICS CHAPTER 8 Exercise 8.1A : 1. If a disc 30 cm in diameter rolls 65 m along a straight line without slipping, calculate a. the number of revolutions would it makes in the process, b. the angular displacement would be through by a speck of gum on its rim. ANS. : 69 rev; 138 rad 2. During a certain period of time, the angular displacement of a swinging door is described by θ 5.00 10.0t 2.00t 2 where is in radians and t is in seconds. Determine the angular displacement, angular speed and angular acceleration a. at time, t =0, b. at time, t =3.00 s. ANS. : 5.00 rad, 10.0 rad s1, 4.00 rad s2; 53.0 rad, 22.0 rad s1, 4.00 rad s2 19 PHYSICS CHAPTER 8 8.1.2 Relationship between linear and rotational motion Relationship between linear velocity, v and angular velocity, When a rigid body is rotates about rotation axis O , every particle in the body moves in a circle as shown in the Figure 8.6. y v P r O s x Figure 8.6 20 PHYSICS CHAPTER 8 Point P moves in a circle of radius r with the tangential velocity v where its magnitude is given by ds and s rθ v dt d vr dt v r The direction of the linear (tangential) velocity always tangent to the circular path. Every particle on the rigid body has the same angular speed (magnitude of angular velocity) but the tangential speed is not the same because the radius of the circle, r is changing depend on the position of the particle. 21 PHYSICS CHAPTER 8 Relationship between tangential acceleration, at and angular acceleration, If the rigid body is gaining the angular speed then the tangential velocity of a particle also increasing thus two component of acceleration are occurred as shown in Figure 8.7. y O at P a ac x Figure 8.7 22 PHYSICS CHAPTER 8 The components are tangential acceleration, at and centripetal acceleration, ac given by dv at dt but and v rω d at r at r dt v2 2 ac r v r The vector sum of centripetal and tangential acceleration of a particle in a rotating body is resultant (linear) acceleration, a given by a at ac Vector form and its magnitude, 2 2 a at ac 23 PHYSICS CHAPTER 8 8.1.3 Rotational motion with uniform angular acceleration Linear motion s u v a t Table 8.1 shows the symbols used in linear and rotational kinematics. Quantity Displacement Initial velocity Rotational motion θ ω0 Acceleration ω α Time t Final velocity Quantity Angular displacement Angular velocity (initial) Angular velocity (final) Angular acceleration time Table 8.1 24 PHYSICS CHAPTER 8 Table 8.2 shows the comparison of linear and rotational motion with constant acceleration. Linear motion Rotational motion a constant α constant v u at ω ω0 αt 1 2 s ut at 2 1 2 θ ω0t αt 2 v 2 u 2 2as ω2 ω02 2αθ 1 s v u t 2 where in radian. 1 θ ω ω0 t 2 Table 8.2 25 PHYSICS CHAPTER 8 Example 8.3 : A car is travelling with a velocity of 17.0 m s1 on a straight horizontal highway. The wheels of the car has a radius of 48.0 cm. If the car then speeds up with an acceleration of 2.00 m s2 for 5.00 s, calculate a. the number of revolutions of the wheels during this period, b. the angular speed of the wheels after 5.00 s. Solution : u 17.0 m s 1 , r 0.48 m, a 2.00 m s 2 , t a. The initial angular velocity is 5.00 s and the angular acceleration of the wheels is given by 26 PHYSICS CHAPTER 8 Solution : u 17.0 m s 1 , r 0.48 m, a 2.00 m s 2 , t 5.00 s a. By applying the equation of rotational motion with constant angular acceleration, thus b. The angular speed of the wheels after 5.00 s is 27 PHYSICS CHAPTER 8 Example 8.5 : The wheels of a bicycle make 30 revolutions as the bicycle reduces its speed uniformly from 50.0 km h-1 to 35.0 km h-1. The wheels have a diameter of 70 cm. a. Calculate the angular acceleration. b. If the bicycle continues to decelerate at this rate, determine the time taken for the bicycle to stop. Solution : 28 PHYSICS CHAPTER 8 Solution : a. The initial angular speed of the wheels is and the final angular speed of the wheels is therefore b. The car stops thus Hence 29 PHYSICS CHAPTER 8 Example 8.5 : A blade of a ceiling fan has a radius of 0.400 m is rotating about a fixed axis with an initial angular velocity of 0.150 rev s-1. The angular acceleration of the blade is 0.750 rev s-2. Determine a. the angular velocity after 4.00 s, b. the number of revolutions for the blade turns in this time interval, c. the tangential speed of a point on the tip of the blade at time, t =4.00 s, d. the magnitude of the resultant acceleration of a point on the tip of the blade at t =4.00 s. Solution : r 0.400 m, ω0 0.150 2π 0.300π rad s 1 , α 0.750 2π 1.50π rad s 2 a. Given t =4.00 s, thus 30 PHYSICS CHAPTER 8 Solution : b. The number of revolutions of the blade is c. The tangential speed of a point is given by 31 PHYSICS CHAPTER 8 Solution : d. The magnitude of the resultant acceleration is 32 PHYSICS CHAPTER 8 Example 8.6 : A coin with a diameter of 2.40 cm is dropped on edge on a horizontal surface. The coin starts out with an initial angular speed of 18 rad s1 and rolls in a straight line without slipping. If the rotation slows down with an angular acceleration of magnitude 1.90 rad s2, calculate the distance travelled by the coin before coming to rest. Solution : 1 ω 18 rad s 1 ω 0 rad s 0 2 d 2.40 10 m α 1.90 rad s 2 s The radius of the coin is d r 1.20 10 2 m 2 33 PHYSICS CHAPTER 8 Solution : The initial speed of the point at the edge the coin is and the final speed is The linear acceleration of the point at the edge the coin is given by Therefore the distance travelled by the coin is 34 PHYSICS CHAPTER 8 Exercise 8.1B : 1. A disk 8.00 cm in radius rotates at a constant rate of 1200 rev min-1 about its central axis. Determine a. its angular speed, b. the tangential speed at a point 3.00 cm from its centre, c. the radial acceleration of a point on the rim, d. the total distance a point on the rim moves in 2.00 s. ANS. : 126 rad s1; 3.77 m s1; 1.26 103 m s2; 20.1 m 2. A 0.35 m diameter grinding wheel rotates at 2500 rpm. Calculate a. its angular velocity in rad s1, b. the linear speed and the radial acceleration of a point on the edge of the grinding wheel. ANS. : 262 rad s1; 46 m s1, 1.2 104 m s2 35 PHYSICS CHAPTER 8 3. A rotating wheel required 3.00 s to rotate through 37.0 revolution. Its angular speed at the end of the 3.00 s interval is 98.0 rad s-1. Calculate the constant angular acceleration of the wheel. ANS. : 13.6 rad s2 4. A wheel rotates with a constant angular acceleration of 3.50 rad s2. a. If the angular speed of the wheel is 2.00 rad s1 at t =0, through what angular displacement does the wheel rotate in 2.00 s. b. Through how many revolutions has the wheel turned during this time interval? c. What is the angular speed of the wheel at t = 2.00 s? ANS. : 11.0 rad; 1.75 rev; 9.00 rad s1 36 PHYSICS CHAPTER 8 Learning Outcome: 8.2 Equilibrium of a uniform rigid body (2 hours) At the end of this chapter, students should be able to: Define and use torque, State and use conditions for equilibrium of rigid body, τ F 0 and 0 37 PHYSICS 8.2.1 CHAPTER 8 Torque (moment of a force), The magnitude of the torque is defined as the product of a force and its perpendicular distance from the line of action of the force to the point (rotation axis). OR τ Fd where Because of τ : magnitude of the torque F : magnitude of the force d : perpendicular distance (moment arm) d r sin where r : distance between the pivot point (rotation axis) and the point of application of force. Thus rF sin OR r F where : angle between F and r 38 PHYSICS CHAPTER 8 It is a vector quantity. The dimension of torque is F d ML T 2 2 The unit of torque is N m (newton metre), a vector product unlike the joule (unit of work), also equal to a newton metre, which is scalar product. Torque is occurred because of turning (twisting) effects of the forces on a body. Sign convention of torque: Positive - turning tendency of the force is anticlockwise. Negative - turning tendency of the force is clockwise. The value of torque depends on the rotation axis and the magnitude of applied force. 39 PHYSICS CHAPTER 8 Case 1 : Consider a force is applied to a metre rule which is pivoted at one end as shown in Figures 8.12a and 8.12b. F τ Fd (anticlockwise) d Figure 8.12a Line of action of a force Pivot point (rotation axis) d r sin θ r θ Figure 8.12b F Point of action of a force τ Fd Fr sin θ (anticlockwise) 40 PHYSICS CHAPTER 8 Case 2 : Consider three forces are applied to the metre rule which is pivoted at one end (point O) as shown in Figures 8.13. F3 d1 r1 sin θ1 r2 O d 2 r2 sin θ2 θ2 θ1 r1 F2 Figure 8.13 F1 τ1 F1d1 F1r1 sin θ1 τ 2 F2 d 2 F2 r2 sin θ2 τ 3 F3 d 3 F3 r3 sin θ3 0 Therefore the resultant (nett) torque is τ τ O τ1 τ 2 τ 3 O F1d1 F2 d 2 Caution : If the line of action of a force is through the rotation axis then τ Fr sin θ andθ 0 τ 0 41 PHYSICS CHAPTER 8 8.2.2 Equilibrium of a rigid body Rigid body is defined as a body with definite shape that doesn’t change, so that the particles that compose it stay in fixed position relative to one another even though a force is exerted on it. If the rigid body is in equilibrium, means the body is translational and rotational equilibrium. There are two conditions for the equilibrium of forces acting on a rigid body. The vector sum of all forces acting on a rigid body must be zero. F F nett 0 OR F x 0, F y 0, F z 0 42 PHYSICS CHAPTER 8 The vector sum of all external torques acting on a rigid body must be zero about any rotation axis. τ nett 0 This ensures rotational equilibrium. This is equivalent to the three independent scalar equations along the direction of the coordinate axes, τ any axis 0 Centre of gravity, CG is defined as the point at which the whole weight of a body may be considered to act. A force that exerts on the centre of gravity of an object will cause a translational motion. 43 PHYSICS CHAPTER 8 Figures 8.14 and 8.15 show the centre of gravity for uniform (symmetric) object i.e. rod and sphere rod – refer to the midway point between its end. l CG l 2 l 2 Figure 8.14 sphere – refer to geometric centre. CG Figure 8.15 44 PHYSICS CHAPTER 8 Problem solving strategies for equilibrium of a rigid body The following procedure is recommended when dealing with problems involving the equilibrium of a rigid body: Sketch a simple diagram of the system to help conceptualize the problem. Sketch a separate free body diagram for each body. Choose a convenient coordinate axes for each body and construct a table to resolve the forces into their components and to determine the torque by each force. Apply the condition for equilibrium of a rigid body : F x 0; F y 0 and τ any axis 0 Solve the equations for the unknowns. 45 PHYSICS CHAPTER 8 Example 8.7 : A 35 cm O 75 cm B W1 W2 Figure 8.9 A hanging flower basket having weight, W2 =23 N is hung out over the edge of a balcony railing on a uniform horizontal beam AB of length 110 cm that rests on the balcony railing. The basket is counterbalanced by a body of weight, W1 as shown in Figure 8.9. If the mass of the beam is 3.0 kg, calculate a. the weight, W1 needed, b. the force exerted on the beam at point O. (Given g =9.81 m s2) 46 PHYSICS CHAPTER 8 N Solution : m 3 kg; W2 23 N The free body diagram of the beam : A Let point O as the rotation axis. Force W1 W2 mg N W2 y-comp. (N) 0.35 m 0.20 m O 0.55 m CG mg 0.55 m W1 Torque (N m), o=Fd=Frsin W1 W1 0.75 0.75W1 23 230.35 8.05 39.81 29.4 N B 0.75 m 29.40.20 5.88 0 47 PHYSICS CHAPTER 8 Solution : Since the beam remains at rest thus the system in equilibrium. a. Hence b. 48 PHYSICS CHAPTER 8 Example 8.9 : A uniform ladder AB of length 10 m and mass 5.0 kg leans against a smooth wall as shown in Figure 8.10. The height of the A end A of the ladder is 8.0 m from the rough floor. a. Determine the horizontal and vertical forces the floor exerts on the end B of the ladder when a firefighter of mass 60 kg is 3.0 m from B. b. If the ladder is just on the verge of slipping when the firefighter is 7.0 m up the ladder , Calculate the coefficient of static friction between ladder and floor. (Given g =9.81 m s2) smooth wall B rough floor Figure 8.10 PHYSICS CHAPTER 8 Solution : ml 5.0 kg; m f 60 kg a. The free body diagram of the ladder : Let point B as the rotation axis. Force ml g mf g N1 N2 fs x-comp. (N) y-comp. (N) Torque (N m), B=Fd=Frsin A α N1 β 8.0 m CG 8 sin α 0.8 10 6 sin β 0.6 10 10 m ml g β mf g β 5.0 m α fs 3.0 m N2 B 6.0 m 50 PHYSICS CHAPTER 8 Solution : Since the ladder in equilibrium thus 51 PHYSICS CHAPTER 8 sin α 0.8; sin β 0.6 Solution : b. The free body diagram of the ladder : Let point B as the rotation axis. Torque (N m), x-comp. y-comp. Force (N) (N) B=Fd=Frsin ml g mf g N1 N2 fs A α N1 β mf g β 10 m 7.0 m 8.0 m ml g β N2 5.0 m α B fs 6.0 m 52 PHYSICS CHAPTER 8 Solution : Consider the ladder stills in equilibrium thus 53 PHYSICS CHAPTER 8 Example 8.9 : A floodlight of mass 20.0 kg in a park is supported at the end of a 10.0 kg uniform horizontal beam that is hinged to a pole as shown in Figure 8.11. A cable at an angle 30 with the beam helps to support the light. a. Sketch a free body diagram of the beam. b. Determine i. the tension in the cable, ii. the force exerted on the beam by the pole. (Given g =9.81 m s2) Figure 8.11 54 PHYSICS CHAPTER 8 Solution : m f 20.0 kg; mb 10.0 kg a. The free body diagram of the beam : T 30 S O 0.5l CG l mb g mf g b. Let point O as the rotation axis. Force x-comp. (N) y-comp. (N) mf g mb g T S Torque (N m), o=Fd=Frsin 55 PHYSICS CHAPTER 8 Solution : b. The floodlight and beam remain at rest thus i. 56 PHYSICS CHAPTER 8 Solution : b. ii. Therefore the magnitude of the force is and its direction is given by 57 PHYSICS CHAPTER 8 Exercise 8.2 : Use gravitational acceleration, g = 9.81 m s2 1. F1 B a A F2 b D C γ F3 Figure 8.12 Figure 8.12 shows the forces, F1 =10 N, F2= 50 N and F3= 60 N are applied to a rectangle with side lengths, a = 4.0 cm and b = 5.0 cm. The angle is 30. Calculate the resultant torque about point D. ANS. : -3.7 N m 58 PHYSICS CHAPTER 8 2. Figure 8.13 A see-saw consists of a uniform board of mass 10 kg and length 3.50 m supports a father and daughter with masses 60 kg and 45 kg, respectively as shown in Figure 8.13. The fulcrum is under the centre of gravity of the board. Determine a. the magnitude of the force exerted by the fulcrum on the board, b. where the father should sit from the fulcrum to balance the system. ANS. : 1128 N; 1.31 m 59 PHYSICS CHAPTER 8 3. Figure 8.14 A traffic light hangs from a structure as show in Figure 8.14. The uniform aluminum pole AB is 7.5 m long has a mass of 8.0 kg. The mass of the traffic light is 12.0 kg. Determine a. the tension in the horizontal massless cable CD, b. the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole. ANS. : 248 N; 197 N, 248 N 60 PHYSICS CHAPTER 8 Exercise 5.2 : 4. 30.0 cm 15.0 cm 50.0 F Figure 8.15 A uniform 10.0 N picture frame is supported by two light string as shown in Figure 8.15. The horizontal force, F is applied for holding the frame in the position shown. a. Sketch the free body diagram of the picture frame. b. Calculate i. the tension in the ropes, ii. the magnitude of the horizontal force, F . ANS. : 1.42 N, 11.2 N; 7.20 N 61 PHYSICS CHAPTER 8 Learning Outcome: 8.3 Rotational dynamics (1 hours) At the end of this chapter, students should be able to: Define the moment of inertia of a rigid body about an axis, I mi ri 2 State and use torque, I 62 PHYSICS CHAPTER 8 8.3.1 Moment of inertia, I Figure 8.16 shows a rigid body about a fixed axis O with angular velocity . m1 mn rn r1 Or 3 r2 m2 m3 Figure 8.16 is defined as the sum of the products of the mass of each particle and the square of its respective distance from the rotation axis. 63 PHYSICS OR CHAPTER 8 I 2 m1r1 2 m2 r2 2 m3 r3 2 ...mn rn n m r 2 i i i 1 where I : moment of inertia of a rigid body about rotation axis m : mass of particle r : distance from the particle to the rotation axis It is a scalar quantity. Moment of inertia, I in the rotational kinematics is analogous to the mass, m in linear kinematics. The S.I. unit of moment of inertia is kg m2. The factors which affect the moment of inertia, I of a rigid body: a. the mass of the body, b. the shape of the body, c. the position of the rotation axis. 64 PHYSICS CHAPTER 8 Moments of inertia of various bodies Table below shows the moments of inertia for a number of objects about axes through the centre of mass. Shape Diagram Equation Hoop or ring or thin cylindrical shell CM I CM MR Solid cylinder or disk CM I CM 2 1 MR 2 2 65 PHYSICS Shape Uniform rod or long thin rod with rotation axis through the centre of mass. CHAPTER 8 Diagram CM Solid Sphere CM Equation I CM 1 ML2 12 I CM 2 MR 2 5 66 PHYSICS Shape Hollow Sphere or thin spherical shell CHAPTER 8 Diagram CM Equation I CM 2 MR 2 3 67 PHYSICS CHAPTER 8 Example 8.10 : Four spheres are arranged in a rectangular shape of sides 250 cm and 120 cm as shown in Figure. 2 kg 3 kg 60 cm A O 5 kg 250 cm B 60 cm 4 kg The spheres are connected by light rods . Determine the moment of inertia of the system about an axis a. through point O, b. along the line AB. 68 PHYSICS CHAPTER 8 Solution : m1 2 kg; m2 3 kg; m3 a. rotation axis about point O, 4 kg; m4 5 kg m2 m1 r2 r1 O r4 m4 0.6 m 1.25 m r3 m3 Since r1= r2= r3= r4= r thus and the connecting rods are light therefore 69 PHYSICS CHAPTER 8 Solution : m1 2 kg; m2 3 kg; m3 b. rotation axis along the line AB, m1 4 kg; m4 5 kg m2 r2 r1 B A r4 m4 r1= r2= r3= r4= r=0.6 m therefore r3 m3 70 PHYSICS CHAPTER 8 8.3.2 Torque, Relationship between torque, and angular acceleration, Consider a force, F acts on a rigid body freely pivoted on an axis through point O as shown in Figure. a1 m1 mn r1 an rn O r2 F a2 m2 The body rotates in the anticlockwise direction and a nett torque is produced. 71 PHYSICS CHAPTER 8 A particle of mass, m1 of distance r1 from the rotation axis O will experience a nett force F1 . The nett force on this particle is F1 m1a1 and a1 r1α F1 m1r1α The torque on the mass m1 is The total (nett) torque on the rigid body is given by 1 r1F1 sin 90 2 1 m1r1 2 2 2 m r m r ... m r 11 2 2 n n n n 2 2 and m r m r i i I i i i 1 i 1 I 72 PHYSICS CHAPTER 8 From the equation, the nett torque acting on the rigid body is proportional to the body’s angular acceleration. Note : Nett torqu e , I is analogous to the Nett force, F ma 73 PHYSICS CHAPTER 8 Example 8.11 : Forces, F1 = 5.60 N and F2 = 10.3 N are applied tangentially to a F2 disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure. O F1 30.0 cm Calculate, a. the nett torque on the disc. b. the magnitude of angular acceleration by the disc. 1 influence 2 ( Use the moment of inertia, I CM MR ) 2 74 PHYSICS CHAPTER 8 Solution : R 0.30 m; M 5.00 kg a. The nett torque on the disc is b. By applying the relationship between torque and angular acceleration, 75 PHYSICS CHAPTER 8 Example 8.12 : A wheel of radius 0.20 m is mounted on a frictionless horizontal axis. The moment of inertia of the wheel about the axis is 0.050 kg m2. A light string wrapped around the wheel is attached to a 2.0 kg block that slides on a horizontal frictionless surface. A horizontal force of magnitude P = 3.0 N is applied to the block as shown in Figure. Assume the string does not slip on the wheel. a. Sketch a free body diagram of the wheel and the block. b. Calculate the magnitude of the angular acceleration of the wheel. 76 PHYSICS CHAPTER 8 Solution : R 0.20 m; I a. Free body diagram : for wheel, 0.050 kg m 2 ; P 3.0 N; m 2.0 kg for block, 77 PHYSICS Solution : R 0.20 m; I b. For wheel, CHAPTER 8 0.050 kg m 2 ; P 3.0 N; m 2.0 kg For block, By substituting eq. (1) into eq. (2), thus 78 PHYSICS CHAPTER 8 Example 8.13 : An object of mass 1.50 kg is suspended from a rough pulley of radius 20.0 cm by light string as shown in Figure. The pulley has a moment of inertia 0.020 kg m2 about the axis of the pulley. The object is released from rest and the pulley rotates without encountering frictional force. Assume that the string does not slip on the pulley. After 0.3 s, determine a. the linear acceleration of the object, b. the angular acceleration of the pulley, c. the tension in the string, d. the liner velocity of the object, e. the distance travelled by the object. (Given g = 9.81 m s-2) R 1.50 kg 79 PHYSICS CHAPTER 8 Solution : a. Free body diagram : for pulley, for block, 80 PHYSICS Solution : CHAPTER 8 R 0.20 m; I 0.020 kg m 2 ; m 1.50 kg; u 0; t 0.3 s a. By substituting eq. (1) into eq. (2), thus b. By using the relationship between a and , hence 81 PHYSICS Solution : CHAPTER 8 R 0.20 m; I 0.020 kg m 2 ; m 1.50 kg; u 0; t 0.3 s c. From eq. (1), thus d. By applying the equation of liner motion, thus e. The distance travelled by the object in 0.3 s is 82 PHYSICS CHAPTER 8 Exercise 8.3 : Use gravitational acceleration, g = 9.81 m s2 1. Figure 7.16 shows four masses that are held at the corners of a square by a very light 70 g 40 cm frame. Calculate the moment of inertia 80 cm of the system about an axis perpendicular B to the plane A 150 g a. through point A, and 150 g b. through point B. 80 cm ANS. : 0.141 kg m2; 0.211 kg m2 70 g 83 PHYSICS CHAPTER 8 2. A 5.00 kg object placed on a frictionless horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object as in Figure 8.17. The pulley has a radius of 0.250 m and moment of inertia I. The block on the table is moving with a constant acceleration of 2.00 m s2. a. Sketch free body diagrams of both objects and pulley. b. Calculate T1 and T2 the tensions in the string. 2.00 m s 2 T1 T2 Figure 8.17 c. Determine I. ANS. : 10.0 N, 70.3 N; 1.88 kg m2 84 PHYSICS CHAPTER 8 Learning Outcome: 8.4 Work and energy of rotational motion(2 hours) At the end of this chapter, students should be able to: Solve problems related to: 1 2 Rotational kinetic energy, K r I Work, W Power, P 2 85 PHYSICS CHAPTER 8 8.4 Work and energy of rotational motion. 8.4.1 Rotational kinetic energy, Kr Consider a rigid body rotating about the axis OZ as shown in Figure. Z v1 m1 mn r1 vn rn O r3 r2 m3 v2 m2 v3 Every particle in the body is in the circular motion about point O. 86 PHYSICS CHAPTER 8 The rigid body has a rotational kinetic energy which is the total of kinetic energy of all the particles in the body is given by 1 1 1 1 2 2 2 K r m1v1 m2 v2 m3v3 ... mn vn2 2 2 2 2 1 1 1 1 2 2 2 2 2 2 K r m1r1 ω m2 r2 ω m3 r3 ω ... mn rn2 ω2 2 2 2 2 1 2 K r ω m1r12 m2 r22 m3 r32 ... mn rn2 2 n 1 2 n 2 2 K r ω mi ri and mi ri I 2 i 1 i 1 1 2 K r Iω 2 87 PHYSICS CHAPTER 8 From the formula for translational kinetic energy, Ktr 1 2 K tr mv 2 After comparing both equations thus is analogous to v I is analogous to m For rolling body without slipping, the total kinetic energy of the body, K is given by K K tr K r where K tr : translati onal kinetic energy K r : rotational kinetic energy 88 PHYSICS CHAPTER 8 Example 8.14 : A solid sphere of radius 15.0 cm and mass 10.0 kg rolls down an inclined plane make an angle 25 to the horizontal. If the sphere rolls without slipping from rest to the distance of 75.0 cm and the inclined surface is smooth, calculate a. the total kinetic energy of the sphere, b. the linear speed of the sphere, c. the angular speed about the centre of mass. 2 (Given the moment of inertia of solid sphere is I CM mR 2and 5 the gravitational acceleration, g = 9.81 m s2) 89 PHYSICS CHAPTER 8 Solution : R 0.15 m; m 10.0 kg s 0.75 m R h s sin 25 v CM 25 a. From the principle of conservation of energy, 90 PHYSICS CHAPTER 8 Solution : R 0.15 m; m 10.0 kg b. The linear speed of the sphere is given by c. By using the relationship between v and , thus 91 PHYSICS CHAPTER 8 8.4.2 Work, W Consider a tangential force, F acts on the solid disc of radius R freely pivoted on an axis through O as shown in Figure 8.19. ds d R R O F Figure 8.19 The work done by the tangential force is given by dW Fds and ds Rdθ dW FRdθ dW θ2 θ1 τdθ W θ2 θ1 τdθ 92 PHYSICS CHAPTER 8 If the torque is constant thus 2 W d 1 W 2 1 W is analogous to the W Fs where τ : torque Δθ : change in angular displacement W : work done Work-rotational kinetic energy theorem states W K r K r f K r i 1 2 1 2 W Iω Iω0 2 2 93 PHYSICS CHAPTER 8 8.4.3 Power, P From the definition of instantaneous power, dW and dW τdθ P dt τdθ and dθ ω P dt dt P τω is analogous to the P Fv Caution : The unit of kinetic energy, work and power in the rotational kinematics is same as their unit in translational kinematics. 94 PHYSICS CHAPTER 8 Example 8.15 : A horizontal merry-go-round has a radius of 2.40 m and a moment of inertia 2100 kg m2 about a vertical axle through its centre. A tangential force of magnitude 18.0 N is applied to the edge of the merry-go- round for 15.0 s. If the merry-go-round is initially at rest and ignore the frictional torque, determine a. the rotational kinetic energy of the merry-go-round, b. the work done by the force on the merry-go-round, c. the average power supplied by the force. (Given g = 9.81 m s2) Solution : R 2.40 m F 95 PHYSICS CHAPTER 8 Solution : R 2.40 m; I 2100 kg m 2 ; F 18.0 N; t 15.0 s; ω0 0 a. By applying the relationship between nett torque and angular acceleration, thus Use the equation of rotational motion with uniform angular acceleration, Therefore the rotational kinetic energy for 15.0 s is 96 PHYSICS CHAPTER 8 Solution : R 2.40 m; I 2100 kg m 2 ; F 18.0 N; t 15.0 s; ω0 0 b. The angular displacement, for 15.0 s is given by By applying the formulae of work done in rotational motion, thus c. The average power supplied by the force is 97 PHYSICS CHAPTER 8 Learning Outcome: 8.5 Conservation of angular momentum (1 hour) At the end of this chapter, students should be able to: Define and use the formulae of angular momentum, L Iω State and use the principle of conservation of angular momentum 98 PHYSICS CHAPTER 8 8.5 Conservation of angular momentum 8.5.1 Angular momentum, L is defined as the product of the angular velocity of a body and its moment of inertia about the rotation axis. OR is analogous to the p mv L I where L : angular momentum I : moment of inertia of a body ω : angular velocity of a body It is a vector quantity. Its dimension is M L2 T1 The S.I. unit of the angular momentum is kg m2 s1. 99 PHYSICS CHAPTER 8 The relationship between angular momentum, L with linear momentum, p is given by vector notation : L r p r mv magnitude form : L rp sin θ mvr sin θ where r : distance from the particle to the rotation axis θ : the angle between r with v Newton’s second law of motion in term of linear momentum is dp F Fnett dt hence we can write the Newton’s second law in angular form as dL τ τ nett dt and states that a vector sum of all the torques acting on a rigid body is proportional to the rate of change of angular momentum. 100 PHYSICS CHAPTER 8 8.5.2 Principle of conservation of angular momentum states that a total angular momentum of a system about an rotation axis is constant if no external torque acts on the system. OR I constant Therefore If the τ 0 dL τ 0 dt dL 0 and dL L f - Li Li L f 101 PHYSICS CHAPTER 8 Example 8.16 : A 200 kg wooden disc of radius 3.00 m is rotating with angular speed 4.0 rad s-1 about the rotation axis as shown in Figure. A 50 kg bag of sand falls onto the disc at the edge of the wooden disc. R ω0 R Before After Calculate, a. the angular speed of the system after the bag of sand falling onto the disc. (treat the bag of sand as a particle) b. the initial and final rotational kinetic energy of the system. Why the rotational kinetic energy is not the same? 1 (Use the moment of inertia of disc is MR2 ) 2 102 PHYSICS CHAPTER 8 Solution : R 3.00 m; ω0 4.0 rad s 1 ; mw a. The moment of inertia of the disc, 200 kg; mb 50 kg The moment of inertia of the bag of sand, By applying the principle of conservation of angular momentum, 103 PHYSICS CHAPTER 8 Solution : R 3.00 m; ω0 4.0 rad s 1 ; mw b. The initial rotational kinetic energy, 200 kg; mb 50 kg The final rotational kinetic energy, thus It is because 104 PHYSICS CHAPTER 8 Example 8.17 : A student on a stool rotates freely with an angular speed of 2.95 rev s1. The student holds a 1.25 kg mass in each outstretched arm that is 0.759 m from the rotation axis. The moment of inertia for the system of student-stool without the masses is 5.43 kg m2. When the student pulls his arms inward, the angular speed increases to 3.54 rev s1. a. Determine the new distance of each mass from the rotation axis. b. Calculate the initial and the final rotational kinetic energy of the system. Solution : 105 PHYSICS CHAPTER 8 Solution : m 1.25 kg; ω0 18.5 rad s 1; I ss 5.43 kg m 2 ; rb 0.759 m; ω 22.2 rad s 1; 0 rb rb m m ra ra Before After 106 PHYSICS Solution : CHAPTER 8 m 1.25 kg; ω0 18.5 rad s 1; I ss 5.43 kg m 2 ; rb 0.759 m; ω 22.2 rad s 1; a. The moment of inertia of the system initially is The moment of inertia of the system finally is By using the principle of conservation of angular momentum, thus 107 PHYSICS CHAPTER 8 18.5 rad s 1; I ss 5.43 kg m 2 ; rb 0.759 m; ω 22.2 rad s 1; Solution : m 1.25 kg; ω0 b. The initial rotational kinetic energy is given by and the final rotational kinetic energy is 108 PHYSICS CHAPTER 8 Example 8.18 : The pulley in the Figure has a radius of 0.120 m and a moment of inertia 0.055 g cm2. The rope does not slip on the pulley rim. Calculate the speed of the 5.00 kg block just before it strikes the floor. (Given g = 9.81 m s2) 5.00 kg 7.00 m 2.00 kg 109 PHYSICS CHAPTER 8 Solution : m1 5.00 kg; m2 2.00 kg; R 0.120 m; h 7.00 m The moment of inertia of the pulley, 3 4 2 10 kg 10 m 2 9 2 I 0.055 g 1 cm 5.5 10 kg m 1 cm2 1 g m2 v m1 7.00 m m2 Initial v m1 7.00 m Final 110 PHYSICS CHAPTER 8 5.00 kg; m2 2.00 kg; R 0.120 m; h 7.00 m; I 5.5 10 9 kg m 2 Solution : m1 By using the principle of conservation of energy, thus 111 PHYSICS CHAPTER 8 Exercise 8.5: Use gravitational acceleration, g = 9.81 m s2 1. A woman of mass 60 kg stands at the rim of a horizontal turntable having a moment of inertia of 500 kg m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about the frictionless vertical axle through its centre. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m s1 relative to the Earth. a. In the what direction and with what value of angular speed does the turntable rotate? b. How much work does the woman do to set herself and the turntable into motion? ANS. : 0.360 rad s1 ,U think; 99.9 J 112 PHYSICS CHAPTER 8 2. Determine the angular momentum of the Earth a. about its rotation axis (assume the Earth is a uniform solid sphere), and b. about its orbit around the Sun (treat the Earth as a particle orbiting the Sun). Given the Earth’s mass = 6.0 x 1024 kg, radius = 6.4 x 106 m and is 1.5 x 108 km from the Sun. ANS. : 7.1 x 1033 kg m2 s1; 2.7 x 1040 kg m2 s1 3. Calculate the magnitude of the angular momentum of the second hand on a clock about an axis through the centre of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a thin rod rotating with angular velocity about one end. (Given the moment of inertia of thin rod about the axis through the CM is 1 ML2 ) 12 ANS. : 4.71 x 106 kg m2 s1 113 PHYSICS CHAPTER 8 Summary: Linear Motion ds v dt dv a dt Relationship Rotational Motion a r n m F ma p mv I mi ri 2 i 1 rF sin L rp sin W Fs P Fv 114