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# Download Deformation Rotational Motion Rotation of Rigid Objects

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Deformation Rotational Motion Rotation of Rigid Objects Rotational Kinematics Torque Lana Sheridan De Anza College Feb 26, 2015 Overview • deforming systems • rotation • rotational kinematics • relating rotational and translational quantities • torque Example 9.14 - Exploding Rocket A rocket is fired vertically upward. At the instant it reaches an altitude of 1000 m and a speed of vi = 300 m/s, it explodes into three fragments having equal mass. One fragment moves upward with a speed of v1 = 450 m/s following the explosion. The second fragment has a speed of v2 = 240 m/s and is moving east right after the explosion. What is the velocity of the third fragment immediately after the explosion? Example 9.14 - Exploding Rocket pi = pf ⇒ M vi = M (v1 + v2 + v3 ) 3 Let j point in the the upward vertical direction, and i point east. v3 = 3vi − v1 − v2 = 3 × 300j − 450j − 240i = (−240i + 450j) m/s Or, 510 m/s at an angle of 62◦ above the horizontal to the west. Deformable Systems Some systems will change the distribution of their mass during their motion. Deformable Systems Some systems will change the distribution of their mass during their motion. 1 Image from http://northdallasgazette.com Deformable Systems In a system that is deformed, the positions of the masses of particles may change relative to the center of mass, but we can still study the system by considering what happens to the center of mass. leave the floor? (d) Does it make sense to say that this momentum came from the floor? Explain. (e) With what kinetic energy does the person leave the floor? (f) Does it make sense to say that this energy came from the floor? Explain. Deformable Systems Problem Page 288, #59 59. Figure P9.59a shows an overhead view of the initial S configuration of two pucks of mass m on frictionless ice. The pucks are tied together with a string of length , and negligible mass. At time t 5 0, a constant force of magnitude F begins to pull to the right on the center point of the string. At time t, the moving pucks strike each other and stick together. At this time, the force has moved through a distance d, and the pucks have attained a speed v (Fig. P9.59b). (a) What is v in terms of F, d, ,, and m? (b) How much of the energy transferred into the system by work done by the force has been transformed to internal energy? (a) It has an exhaust speed dizer is requi design could amount of fue same task? (c (b) is 2.50 tim why the requ factor of 2.50. 64. A rocket has 330 kg of fu it starts from engine at tim ative speed v 2.50 kg/s. Th Mf /k 5 330 k ing the burn time is given b m d CM , F v F d m t!0 a (b) Make a gr tion of time fo that the accel t!t b (d) Graph th the person? (c) With what momentum does the person leave the floor? (d) Does it make sense to say that this Deformable Systems momentum came Problem from the floor? Explain. (e) With what kinetic energy does the person leave the floor? Does it make sense to say that this energy came Page (f) 288, #59 from the floor? Explain. 59. Figure P9.59a shows an overhead view of the initial S configuration of two pucks of mass m on frictionless ice. The pucks are tied together with a string of length , and negligible mass. At time t 5 0, a constant force of magnitude F begins to pull to the right on the center point of the string. At time t, the moving pucks strike each other and stick together. At this time, the force has moved through a distance d, and the pucks have attained a speed v (Fig. P9.59b). (a) What is v in terms of F, d, ,, and m? (b) How much of the energy transferred into the system by work done by the force has been transformed to internal energy? m d CM v en (a ex di de am sa (b wh fa 64. A 33 it en at 2. M in tim each other and stick together. At this time, the force Deformable Systems Problem has moved through a distance d, and the pucks have attained a speed v (Fig. P9.59b). (a) What is v in terms of F, d, ,, and m? (b) How much of the energy transferred into the system by work done by the force has Pagebeen 288,transformed #59 to internal energy? 3 i e a 2 M i t m d CM , F v F d m t!0 ( t t t!t a b Figure P9.59 Section 9.9 Rocket Propulsion 60. A model rocket engine has an average thrust of 5.26 N. ( ( Deformable Systems (a) Speed of pucks, v , after collision? Deformable Systems (a) Speed of pucks, v , after collision? F = M aCM Deformable Systems (a) Speed of pucks, v , after collision? F = M aCM F is constant ⇒ aCM is constant. Deformable Systems (a) Speed of pucks, v , after collision? F = M aCM F is constant ⇒ aCM is constant. aCM = F 2m vf2 = vi2 + 2aCM x F ` 2 vf = 0 + 2 d− 2m 2 r 2Fd − F ` vf = 2m Deformable Systems (b) Increase in internal energy? Deformable Systems (b) Increase in internal energy? ∆Eint = W − Kf Deformable Systems (b) Increase in internal energy? ∆Eint = W − Kf 1 = Fd − (2m)v 2 2 1 2Fd − F ` = Fd − (2m) 2 2m ` = Fd − Fd + F 2 1 = F` 2 Rotation of Rigid Objects Now we understand that while we can treat a collection of particles as a single point particle at the center of mass, we do not have to do that. This will allow us to describe another important kind of motion: rotation. Rotation of Rigid Objects Now we understand that while we can treat a collection of particles as a single point particle at the center of mass, we do not have to do that. This will allow us to describe another important kind of motion: rotation. Begin with rotational kinematics. here fordisc rotational motion. As the rotates, a particle at 10.1 illustrates P Figure moves through an arc lengthan ove s onrotates a circularabout path ofaradius disc fixedr.axis p The angular position of P isthe u. disc through the center of ticle at P is at a fixed distance r radius r. (InPfact, every element o s convenientr to represent the pos u r the distance from the origin to P O P Reference Reference ence lineOfixed in space as show line line changes in time while r remai cle from the reference line, wh b a length s as in Figure 10.1b. The s s = r θ ; θrelationship = Figure r 10.1 A compact disc As the disc rotates, a particle at for the disc, a fixed reference Rotation of Rigid Objects line is chosen. A particle at P is located at a distance r from To begin, consider a rotating disc. the rotation axis through O. rotating about a fixed axis Pr moves through arcfor length is constant in an time a rigid object. through O perpendicular to the s on a circular path of radius r. plane of the figure. The angular position of P is u. P Pitfall Prevention 10.1 Remember the Radian In rota- here fordisc rotational motion. As the rotates, a particle at 10.1 illustrates P Figure moves through an arc lengthan ove s onrotates a circularabout path ofaradius disc fixedr.axis p The angular position of P isthe u. disc through the center of ticle at P is at a fixed distance r radius r. (InPfact, every element o s convenientr to represent the pos u r the distance from the origin to P O P Reference Reference ence lineOfixed in space as show line line changes in time while r remai cle from the reference line, wh b a length s as in Figure 10.1b. The s s = r θ ; θrelationship = Figure r 10.1 A compact disc As the disc rotates, a particle at for the disc, a fixed reference Rotation of Rigid Objects line is chosen. A particle at P is located at a distance r from To begin, consider a rotating disc. the rotation axis through O. rotating about a fixed axis Pr moves through arcfor length is constant in an time a rigid object. through O perpendicular to the s on a circular path of radius r. plane of the figure. The angular position of P is u. Units for θ: radians. Often written as “rad”. But notice, that a dimensional analysis gives [m] [m] = 1, unitless! The radian is an artificial unit. In fact, angles given inPitfall radians are dimensionless. Prevention 10.1 P Remember the Radian In rota- Rotation of Don’t fall into the trap of using angles measured in degrees in Rigid Objects rotational equations. position of a rigi the object, such angular position the object and th Such identificati lational motion which is the orig motion that the As the particl tion " in a time sweeps out an an placement of the How does the angle advance in time? y ", t f r !,ti uf ui O Figure 10.2 x A particle on a rotating rigid object moves from ! to " along the circle. ∆θarc =ofθfa − θi In the time interval Dt 5 tf 2 ti , the radial The rate at whic spins rapidly, th slowly, this displ rates can be qua Angular Speed Rate at which the angle advances is a speed: the angular speed, ω. Average angular speed: ωavg = θf − θi ∆θ = tf − ti ∆t Instantaneous angular speed: ω= dθ dt Angular Acceleration Rate at which the angular speed changes: the angular acceleration, α. Average angular acceleration: αavg = ωf − ωi ∆ω = tf − ti ∆t Instantaneous angular acceleration: α= dω dt Rotation of Rigid Objects and Vector Quantities We can also define these quantities as vectors! This might seem a bit strange because the direction of the motion any point (other than the axis) on a rotating disc is always changing. Rotation of Rigid Objects and Vector Quantities We can also define these quantities as vectors! This might seem a bit strange because the direction of the motion any point (other than the axis) on a rotating disc is always changing. However, the angle can be positive or negative depending on whether it is clockwise or counterclockwise from the reference point. This is the same way that a 1-dimensional displacement x can be positive or negative based on whether it is on the left or right of the origin. Rotation of Rigid Objects and Vector Quantities We can also define these quantities as vectors! This might seem a bit strange because the direction of the motion any point (other than the axis) on a rotating disc is always changing. However, the angle can be positive or negative depending on whether it is clockwise or counterclockwise from the reference point. This is the same way that a 1-dimensional displacement x can be positive or negative based on whether it is on the left or right of the origin. In that case we could write x, and the direction of the vector was indicated by the sign of the quantity x. Rotation of Rigid Objects and Vector Quantities By convention, we define the counterclockwise direction to be Chapter Rotation of athe Rigid Object About a Fixed Axis The vector itself10is drawn along axis of rotation. 296 positive. S direction of v for the particle is out o is counterclockwise and into the pla wise. To illustrate this convention, it strated in Figure 10.3. When the fou direction of rotation, the extended r S v direction of S a follows from its definit S if the angular speed is increasing in v Figure 10.3 The in right-hand speed is decreasing time. rule S v for determining the direction of the angular velocity vector. Then we can write: ω dω 10.2 α Analysis Model: R = dt Constant Angular where n̂ is a unit vector perpendicular to the plane of rotation. θ= S v s n̂ ; r ω= θ dθ ; dt In our study of translational motion, Comparison of Linear and Rotational quantities Linear Quantities Rotational Quantities θ x v= dx dt ω= dθ dt a= dv dt α= dω dt Rotational Kinematics If α is constant, we have basically the same kinematics equations as before, but the relations are between the new quantities. ωf = ωi + αt 1 θf = θi + ωi t + αt 2 2 ω2f = ω2i + 2α · ∆θ 1 ωavg = (ωi + ωf ) 2 1 θf = θi + (ωi + ωf )t 2 Kinematics Comparison Linear Quantities Rotational Quantities vf = vi + at ωf = ωi + αt xf = xi + vi t + 21 at 2 θf = θi + ωi t + 12 αt 2 vf2 = vi2 + 2a · ∆x ω2f = ω2i + 2α · ∆θ vavg = 12 (vi + vf ) ωavg = 21 (ωi + ωf ) xf = xi + 21 (vi + vf )t θf = θi + 21 (ωi + ωf )t and for the velocity, vf 5 vi 1 at 5 2.00 Relating Rotational Quantities to Translation of m/s 1 (3 Points There is no translational analog to part (B) because transl Consider a point on the rotating object. How does its speed relate to the angular speed? y 10.3 Angular a S In this section, we derive acceleration of a rotating P of a point in the object. s r rotates about a fixed axi u circle whose center is on x O Because point P in Figu is always tangent to the ci nitude of the tangential v v 5 ds/dt, where s is the d Figure 10.4 As a rigid object that s 5 r We know s = r θ, so rotates since about the object’s speed speedRecalling along the the fixed axis (the is its path. v path s, z axis) through O, the point P S v that is has a tangential dsvelocitydθ v = to the = rcircular path always tangent dt of radius r. dt Relating Rotational Quantities to Translation of Points Since ω = dθ dt , that gives us and expression for (tangential) speed v = rω And differentiating both sides with respect to t again: at = r α Relating Rotational Quantities to Translation of Points Since ω = dθ dt , that gives us and expression for (tangential) speed v = rω And differentiating both sides with respect to t again: at = r α Notice that the above equation gives the rate of change of speed, which is the tangential acceleration. Centripetal Acceleration Remember: at = dv dt where v is the speed, not velocity. So, at = r α But of course, in order for a mass at that point, radius r , to continue moving in a circle, there must be a centripetal component of acceleration also. v2 ac = = ω2 r r Centripetal Acceleration Remember: at = dv dt where v is the speed, not velocity. So, at = r α But of course, in order for a mass at that point, radius r , to continue moving in a circle, there must be a centripetal component of acceleration also. v2 ac = = ω2 r r For a rigid object, the force that supplies this acceleration will be some internal forces between the mass at the rotating point and the other masses in the object. Those are the forces that hold the object together. Linear and Rotational Kinematics To get an idea of another way to confirm the relation between the linear and rotational kinematics equations: vf vf r ωf = vi + at vi a = + t r r = ωi + αt Linear and Rotational Kinematics To get an idea of another way to confirm the relation between the linear and rotational kinematics equations: vf vf r ωf = vi + at vi a = + t r r = ωi + αt The scalar versions of the equations can all be recovered this way. bout 3.00 s. pe Example of 1 4. A bar on a hinge starts from rest and rotates with an angular acceleration a 5 10 1 6t, where a is in rad/s2 nguand t is in seconds. Determine the angle in radians rage through which the bar turns in the first 4.00 s. ond. ring Page Section 10.2#5 Analysis Model: Rigid Object 325, eed? Under Constant Angular Acceleration 5. A wheel starts from rest and rotates with constant ition 0t 1 W angular acceleration to reach an angular speed of 12.0 rad/s in 3.00 s. Find (a) the magnitude of the angueterlar acceleration of the wheel and (b) the angle in radians through which it rotates in this time interval. 6. A centrifuge in a medical laboratory rotates at an angular speed of 3 600 rev/min. When switched off, it rotates through 50.0 revolutions before coming to rest. Find the constant angular acceleration of the centrifuge. 7. An electric motor rotating a workshop grinding wheel 2 M at 1.00 3 10 rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magni- w ef a fr e o Secti 15. A Example 1 Known: ωi , ωf , t (a) Angular accel., α? Example 1 Known: ωi , ωf , t (a) Angular accel., α? ωf = ωi + αt Example 1 Known: ωi , ωf , t (a) Angular accel., α? ωf = ωi + αt α = ωf − ωi t α = 12.0 rad/s − 0 3.00 s α = 4.00 rad s−2 Example 1 Known: ωi , ωf , t, α (b) Angle in radians, ∆θ? Example 1 Known: ωi , ωf , t, α (b) Angle in radians, ∆θ? Either use 1 1 ∆θ = ωi t + αt 2 = 0 + (4)(3)2 = 18 radians 2 2 or use ∆θ = 1 (ωi + ωf )t 2 Example 1 Known: ωi , ωf , t, α (b) Angle in radians, ∆θ? Either use 1 1 ∆θ = ωi t + αt 2 = 0 + (4)(3)2 = 18 radians 2 2 or use ∆θ = 1 (ωi + ωf )t 2 ∆θ = 1 (0 + 12.0)(3.00) 2 = 18.0 radians the constant angular acceleration of the centrifuge. 7. An 2electric motor rotating a workshop grinding wheel Example 2 M at 1.00 3 10 rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magnitude 2.00 rad/s2. (a) How long does it take the grinding stop? (b) Through how many radians has the Pagewheel 325,to#8 wheel turned during the time interval found in part (a)? 8. A machine part rotates at an angular speed of Q/C 0.060 rad/s; its speed is then increased to 2.2 rad/s at an angular acceleration of 0.70 rad/s2. (a) Find the angle through which the part rotates before reaching this final speed. (b) If both the initial and final angular speeds are doubled and the angular acceleration remains the same, by what factor is the angular displacement changed? Why? 9. A dentist’s drill starts from rest. After 3.20 s of constant angular acceleration, it turns at a rate of 2.51 3 104 rev/min. (a) Find the drill’s angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period. Secti 15. A A 4 tu 16. M o tu e 17. A W d th o la o to fo 18. F h Example 2 Known: ωi , ωf , α (a) ∆θ? Example 2 Known: ωi , ωf , α (a) ∆θ? Example 2 Known: ωi , ωf , α (a) ∆θ? ω2f = ω2i + 2α∆θ Example 2 Known: ωi , ωf , α (a) ∆θ? ω2f = ω2i + 2α∆θ ∆θ = = ω2f − ω2i 2α (2.2)2 − (0.060)2 2 × 0.70 = 3.5 rad Example 2 (b) If both ωi and ωf are doubled, α kept constant, what happens to ∆θ? ∆θ 0 = (2 × ωf )2 − (2 × ωi )2 2α = 4 ω2f − ω2i 2α = 4 ∆θ Torque Torque is a measure of force-causing-rotation. It is not a force, but it is related. It depends on a force vector and its point of application relative to an axis of rotation. Torque is given by: τ=r×F That is: the cross product between • a vector r, the displacement of the point of application of the force from the axis of rotation, and • an the force vector F Torque Torque is a measure of force-causing-rotation. It is not a force, but it is related. It depends on a force vector and its point of application relative to an axis of rotation. Torque is given by: τ=r×F That is: the cross product between • a vector r, the displacement of the point of application of the force from the axis of rotation, and • an the force vector F Units: N m Newton-meters. These are not Joules! 10.4 Torqu The component F sin f tends to rotate the wrench about an axis through O. Torque F sin f S F r f O S d Figure 10.7 r F cos f f Line of action S The force F has a greater rotating tendency about an axis through O as F increases τ=r× F= rF sin φ and n̂ as the moment arm d increases. In our study of tra we studied the cau What is the cause Imagine trying ular to the door s hinges. You will ac force near the doo When a force is to rotate about th axis is measured b but we will consid Chapter 11. Consider the wr perpendicular to t where φ is the angle between r and F, and n̂ is the unit vector perpendicular to r and F, as determined by the right-hand rule. Torque Diagram also The component F sin f tends to rotate the wrench illustrates two points of O. view about an axis through 10.4 Torque about torque: F sin f S F r f O S d r F cos f f Line of action S Figure 10.7 The force F has a or greater rotating tendency τ = r (F sin φ)about n̂ an axis through O as F increases and as the moment arm d increases. In our study of translation we studied the cause of c What is the cause of chang Imagine trying to rotat ular to the door surface n hinges. You will achieve a force near the doorknob t When a force is exerted to rotate about that axis. axis is measured by a quan but we will consider only Chapter 11. Consider the wrench in perpendicular to the page τ = F (r sin φ) n̂ In the diagram, the distance d = r sin φ and is called the “moment arm” or “lever arm” of the torque. Torque Torque: when a force is applied above the center of mass. CM The system rotates clockwise when a force is applied a above the center of mass. CM The system rotates counteraCM clockwise when a force is applied below the center of mass. Torque: a system rotates counterThe clockwise when a force is applied below CM the center of mass. The system rotates counterclockwise when a force is applied below the center of mass. b CM No torque: The system moves in the bCM direction of the force without rotating when a force is applied at the center of mass. b The system moves in the direction of the force without rotating when a force is applied The system moves in the at the center of mass. CM direction of the force without rotating when a force is applied at the center of mass. c CM the rotating x axis and liesFig. somewhe out (see 9.13c) this procedure. is applied at the center of ma center(see of mass of the outThe rotating Fig. 9.13c) the xprocedure. axis and lies somewhe this For example, if x 1 5 0, x 2 The center mass of the center of massoflies closer to the axis and lies somewher thexcenter of mass lies midw can extend ForWe example, if x 1 this 5 0,conc x2 5 dimensions. The coordin center of mass liesxcloser to the example, center of mass lies0,midw For if x 1 5 x2 5 m 2xconce m 1xlies We can extend 1 1this 2 1 center of ; mass closer to x CM dimensions. The x coordina m 1 lies 1 mmidw the center of mass 2 1 We can extend this conce m 2x 2 1 m dimensions.mThe x coordina 1x 1 1 x CMx ; where x coordinate i is the m 1 1 m2 1 m the sum runs all n par m 1x over 1 1 m 2x 2 1 m x CM ;defined by the equ similarly m1 1 m2 1 m where xi is the x coordinate the sum runs over all nyCM part ; similarly by the equo where xi isdefined the x coordinate of mass the The sumcenter runs over all ncan partb yCMequ ; The components ofthe this vec similarly defined by 9.30. Therefore, The center of massycan b CM ; S ^i 1 r CM 5ofx CM yC The components this vect 9.30. TheTherefore, center of mass can be Question B A A torque is supplied by applying a force at point A. To produce the same torque, the force applied at point B must be: (A) greater (B) less (C) the same 1 Image from Harbor Freight Tools, www.harborfreight.com the rotation axis to the line of action of nary line extending out both ends of the S line extending from the tail of F in Fig. om the right triangle in Figure 10.7 that that d 5 r sin f. The quantity d is called rotation axis. Net Torque S Moment Object that can rotate about anarmaxis at O: f F that tends to cause rotation of the n f, the component perpendicular to a oint of application of the force. The horie of action passes through O, has no tenpassing through O. From the definition s as F increases and as d increases, which we push at the doorknob rather than at a apply our push as closely perpendicular o 908. Pushing sideways on the doorknob S F1 d1 O d2 ject as in Figure 10.8, each tends to proS In this example, F2 tends to rotate the counterclockwise. We use the convention S F2 a force is positive if the turning tendency S tive if the turning tendency is clockwise. Figure 10.8 The force F 1 tends S sulting from F1, which has a moment arm to rotate the object counterclockS are but the twoO, and torques ue from F2There is negative andtwo equalforces to 2F 2d 2.acting, wise about an axis through S F 2 tends to rotate it clockwise. gh O is τ2 point in opposite directions. 5 F 1d1 2 F 2d 2 rce. Forces can cause a change in translaτ1 would produce a counterclockwise rotation cond law. Forces can also cause a change ess of the τforces in causing this change a clockwise rotation 2 would produce orces and the moment arms of the forces, has units of force times length—newton e reported in these units. Do not confuse produced, τ1 and om the right triangle in Figure 10.7 that that d 5 r sin f. The quantity d is called Net Torque Moment arm S f F that tends to cause rotation of the n f, the component perpendicular to a oint of application of the force. The horie of action passes through O, has no tenpassing through O. From the definition s as F increases and as d increases, which we push at the doorknob rather than at a apply our push as closely perpendicular o 908. Pushing sideways on the doorknob ject as in Figure 10.8, each tends to proS In this example, F2 tends to rotate the counterclockwise. We use the convention a force is positive if the turning tendency tive if the turning tendency is clockwise. S sulting from F1, which has a moment arm SThe net torque is2Fthe ue from F2 is negative and equal to 2d 2. gh O is 5 F 1d1 2 F 2d 2 S F1 d1 O d2 S F2 Figure 10.8 S The force F 1 tends to rotate the object counterclocksum of the torques acting on the object: wise about an axis through O, and S F 2 tends to rotate it clockwise. X rce. Forces can cause a change in translaτnet = τi cond law. Forces can also cause a change ess of the forces in causing this change i orces and the moment arms of the forces, has units of force times length—newton this units. case,Dowith n̂ pointing out of the slide: e reportedIn in these not confuse its but are very different concepts. τnet oosen a stubborn screw from a piece of d you find a screwdriver for which the = τ1 + τ2 = (F1 d1 − F2 d2 )n̂ Summary • applying kinematics • torque • moment of inertia 3rd Collected Homework! due Monday, Feb 29. Next test Friday, Mar 4. (Uncollected) Homework Serway & Jewett, • Read ahead in Chapter 10. • Ch 10, onward from page 288. Probs: 3, 7, 11, 15, 17, 19, 21, 25