Plane Motion of Rigid Bodies: Forces and Accelerations. Contents

Transcription

1 Plane Motion of Rigid odies: Forces and Accelerations Contents Introduction Equations of Motion of a Rigid ody Angular Momentum of a Rigid ody in Plane Motion Plane Motion of a Rigid ody: d Alembert s Principle Axioms of the Mechanics of Rigid odies Problems Involving the Motion of a Rigid ody Sample Problem 6. Sample Problem 6. Sample Problem 6.3 Sample Problem 6.4 Sample Problem 6.5 Constrained Plane Motion Constrained Plane Motion: Noncentroidal Rotation Constrained Plane Motion: Rolling Motion Sample Problem 6.6 Sample Problem 6.8 Sample Problem 6.9 Sample Problem

2 Introduction In this chapter and in Chapters 7 and 8, we will be concerned with the kinetics of rigid bodies, i.e., relations between the forces acting on a rigid body, the shape and mass of the body, and the motion produced. Results of this chapter will be restricted to: - plane motion of rigid bodies, and - rigid bodies consisting of plane slabs or bodies which are symmetrical with respect to the reference plane. Our approach will be to consider rigid bodies as made of large numbers of particles and to use the results of Chapter 4 for the motion of systems of particles. Specifically, r r r &r F ma and M G H G D Alembert s principle is applied to prove that the external forces acting on a rigid body are equivalent a vector ma ṛ attached to the mass center and a couple of moment Iα 6-3 Equations of Motion for a Rigid ody Consider a rigid body acted upon by several external forces. Assume that the body is made of a large number of particles. For the motion of the mass center G of the body with respect to the Newtonian frame Oxyz, r F r ma For the motion of the body with respect to the centroidal frame Gx y z, r M G H &r G System of external forces is equipollent to the r system consisting of ma and H &r. G 6-4

3 Angular Momentum of a Rigid ody in Plane Motion Angular momentum of the slab may be computed by r n r H ( r v Δm ) G i i i n r [ ri ( ω ri ) Δmi ] i r ω ( ri Δmi ) r Iω After differentiation, r H &r G I &r ω Iα i Consider a rigid slab in plane motion. Results are also valid for plane motion of bodies which are symmetrical with respect to the reference plane. Results are not valid for asymmetrical bodies or three-dimensional motion. 6-5 Plane Motion of a Rigid ody: D Alembert s Principle Motion of a rigid body in plane motion is completely defined by the resultant and moment resultant about G of the external forces. F ma F ma M Iα x x y y G The external forces and the collective ective forces of the slab particles are equipollent (reduce to the same resultant and moment resultant) and equivalent (have the same ect on the body). d Alembert s Principle: The external forces acting on a rigid body are equivalent to the ective forces of the various particles forming the body. The most general motion of a rigid body that is symmetrical with respect to the reference plane can be replaced by the sum of a translation and a centroidal rotation

4 Axioms of the Mechanics of Rigid odies r r The forces F and F act at different points on a rigid body but but have the same magnitude, direction, and line of action. The forces produce the same moment about any point and are therefore, equipollent external forces. This proves the principle of transmissibility whereas it was previously stated as an axiom. 6-7 Problems Involving the Motion of a Rigid ody The fundamental relation between the forces acting on a rigid body in plane motion and the acceleration of its mass center and the angular acceleration of the body is illustrated in a free-body-diagram equation. The techniques for solving problems of static equilibrium may be applied to solve problems of plane motion by utilizing - d Alembert s principle, or - principle of dynamic equilibrium These techniques may also be applied to problems involving plane motion of connected rigid bodies by drawing a freebody-diagram equation for each body and solving the corresponding equations of motion simultaneously

5 Sample Problem 6. Calculate the acceleration during the skidding stop by assuming uniform acceleration. At a forward speed of 30 ft/s, the truck brakes were applied, causing the wheels to stop rotating. It was observed that the truck to skidded to a stop in 0 ft. Determine the magnitude of the normal reaction and the friction force at each wheel as the truck skidded to a stop. Draw the free-body-diagram equation expressing the equivalence of the external and ective forces. Apply the three corresponding scalar equations to solve for the unknown normal wheel forces at the front and rear and the coicient of friction between the wheels and road surface. 6-9 Sample Problem 6. ft v 30 x s 0 0ft Calculate the acceleration during the skidding stop by assuming uniform acceleration. v v 0 + a ft 0 30 s ( x x ) + a 0 ( 0ft) ft a.5 s Draw a free-body-diagram equation expressing the equivalence of the external and ective forces. Apply the corresponding scalar equations. F y ( F y ) N A + N W 0 F x ( F x ) F F μ ( N + N ) k A A ma μ W k a μk g ( W g) a

6 Sample Problem 6. Apply the corresponding scalar equations. M A ( M A ) N N ( 5ft) W + ( ft) N ( 4ft) W W a 5W + 4 a g g 0.650W N A W N W ( 0. W ) Nrear N A 350 N rear 0. 75W ( 0. W ) N front N V 650 N front 0. 35W ma ( 0.690)( 0. W ) Frear μk Nrear 75 ( 0.690)( 0. W ) F rear 0. W Ffront μk N front 35 F front W 6 - Sample Problem 6. The thin plate of mass 8 kg is held in place as shown. Neglecting the mass of the links, determine immediately after the wire has been cut (a) the acceleration of the plate, and (b) the force in each link. Note that after the wire is cut, all particles of the plate move along parallel circular paths of radius 50 mm. The plate is in curvilinear translation. Draw the free-body-diagram equation expressing the equivalence of the external and ective forces. Resolve into scalar component equations parallel and perpendicular to the path of the mass center. Solve the component equations and the moment equation for the unknown acceleration and link forces. 6-6

7 Sample Problem 6. Note that after the wire is cut, all particles of the plate move along parallel circular paths of radius 50 mm. The plate is in curvilinear translation. Draw the free-body-diagram equation expressing the equivalence of the external and ective forces. Resolve the diagram equation into components parallel and perpendicular to the path of the mass center. F t ( F t ) W cos30 ma mg cos30 ( 9.8m/s ) cos a 30 a 8.50m s 60 o 6-3 Sample Problem 6. a 8.50m s 60 o Solve the component equations and the moment equation for the unknown acceleration and link forces. M ( ) G M G ( FAE sin30 )( 50mm) ( FAE cos30 )( 00mm) ( F sin 30 )( 50mm) + ( F cos30 )( 00mm) 0 F DF DF 38.4 F F F F AE +.6 F 0.85F AE DF F n ( F n ) AE AE AE + F DF 0.85F W sin 30 0 AE ( 8kg)( 9.8m s ) FDF 0.85( 47.9 N) DF W sin 30 0 F AE F DF 47.9 N 8.70 N T C 6-4 7

8 Sample Problem 6.3 Determine the direction of rotation by evaluating the net moment on the pulley due to the two blocks. Relate the acceleration of the blocks to the angular acceleration of the pulley. A pulley weighing lb and having a radius of gyration of 8 in. is connected to two blocks as shown. Assuming no axle friction, determine the angular acceleration of the pulley and the acceleration of each block. Draw the free-body-diagram equation expressing the equivalence of the external and ective forces on the complete pulley plus blocks system. Solve the corresponding moment equation for the pulley angular acceleration. 6-5 Sample Problem 6.3 Determine the direction of rotation by evaluating the net moment on the pulley due to the two blocks. note: M G W I mk k g lb 3.ft s Relate the acceleration of the blocks to the angular acceleration of the pulley. a A r A α ( 0 ft)α ( 0lb)( 6in) ( 5lb)( 0in) 0in lb rotation is counterclockwise lb ft s 8 ft a r α ( 6 ft)α 6-6 8

9 Sample Problem 6.3 Draw the free-body-diagram equation expressing the equivalence of the external and ective forces on the complete pulley and blocks system. Solve the corresponding moment equation for the pulley angular acceleration. M G ( M G ) ( )( 6 ) ( )( 0 ) ( 6 ) ( 0 0lb ft 5lb ft Iα + m ) a ft m AaA ft ( )( 6 ) ()( 0) ( ) ( 0 )( 6 )( 6 ) ( 5 )( 0)( α + α ) I lb ft s a a A ( 0 α ) ft s ( 6 α ) ft s Then, α a A r A α a r ( 0 ft )(.374rad s ) ( 6 ft )(.374rad s ) α.374rad s a A.978ft s a.87ft s 6-7 Sample Problem 6.4 A cord is wrapped around a homogeneous disk of mass 5 kg. The cord is pulled upwards with a force T 80 N. Determine: (a) the acceleration of the center of the disk, (b) the angular acceleration of the disk, and (c) the acceleration of the cord. Draw the free-body-diagram equation expressing the equivalence of the external and ective forces on the disk. Solve the three corresponding scalar equilibrium equations for the horizontal, vertical, and angular accelerations of the disk. Determine the acceleration of the cord by evaluating the tangential acceleration of the point A on the disk

10 Sample Problem 6.4 Draw the free-body-diagram equation expressing the equivalence of the external and ective forces on the disk. Solve the three scalar equilibrium equations. F x ( F x ) 0 ma x a x 0 F y ( F y ) T W ma a y y T W 80 N - m M G ( M G ) Tr Iα T α mr ( mr ) ( 5kg)( 9.8m s ) 5kg α ( 80 N) ( 5kg)( 0.5m) a y.9m s α 48.0rad s 6-9 Sample Problem 6.4 Determine the acceleration of the cord by evaluating the tangential acceleration of the point A on the disk. a r cord ( aa) t a + ( aa G ) t.9m s ( 0.5m)( 48 rad s ) + s 6.m a cord a x 0 a y.9m s α 48.0rad s 6-0 0

11 Sample Problem 6.5 A uniform sphere of mass m and radius r is projected along a rough horizontal surface with a linear velocity v 0. The coicient of kinetic friction between the sphere and the surface is μ k. Determine: (a) the time t at which the sphere will start rolling without sliding, and (b) the linear and angular velocities of the sphere at time t. Draw the free-body-diagram equation expressing the equivalence of the external and ective forces on the sphere. Solve the three corresponding scalar equilibrium equations for the normal reaction from the surface and the linear and angular accelerations of the sphere. Apply the kinematic relations for uniformly accelerated motion to determine the time at which the tangential velocity of the sphere at the surface is zero, i.e., when the sphere stops sliding. 6 - Sample Problem 6.5 Draw the free-body-diagram equation expressing the equivalence of external and ective forces on the sphere. Solve the three scalar equilibrium equations. F y ( F y ) N W 0 F x ( F x ) F ma μ mg k M G ( M G ) Fr Iα ( μ mg) r ( mr )α k 3 N W mg a μ g k 5 μ α k g r NOTE: As long as the sphere both rotates and slides, its linear and angular motions are uniformly accelerated. 6 -

12 Sample Problem 6.5 a μk g 5 μ α k g r Apply the kinematic relations for uniformly accelerated motion to determine the time at which the tangential velocity of the sphere at the surface is zero, i.e., when the sphere stops sliding. v v + at v + ( μ g)t μk g ω ω0 + αt 0 + t r 5 μk g v0 μk gt r t r 5 μk g t r k At the instant t when the sphere stops sliding, v rω 5 μk g v r 7 μk g 0 ω 5 v r 7 r t v0 7 g μ k v 0 5 ω 7 r 0 5 v rω v v Constrained Plane Motion Most engineering applications involve rigid bodies which are moving under given constraints, e.g., cranks, connecting rods, and non-slipping wheels. Constrained plane motion: motions with definite relations between the components of acceleration of the mass center and the angular acceleration of the body. Solution of a problem involving constrained plane motion begins with a kinematic analysis. e.g., given θ, ω, and α, find P, N A, and N. - kinematic analysis yields a x and a y. - application of d Alembert s principle yields P, N A, and N. 6-4

13 Constrained Motion: Noncentroidal Rotation Noncentroidal rotation: motion of a body is constrained to rotate about a fixed axis that does not pass through its mass center. Kinematic relation between the motion of the mass center G and the motion of the body about G, a rα a rω t n The kinematic relations are used to eliminate a t and a n from equations derived from d Alembert s principle or from the method of dynamic equilibrium. 6-5 Constrained Plane Motion: Rolling Motion For a balanced disk constrained to roll without sliding, x rθ a rα Rolling, no sliding: F μs N a rα Rolling, sliding impending: F μs N a rα Rotating and sliding: F μk N a, rα independent For the geometric center of an unbalanced disk, a O rα The acceleration of the mass center, r r r ag ao + ag O r r r a + a + a O ( G O ) ( G O ) t n 6-6 3

14 Sample Problem 6.6 me 4 kg ke 85 mm mo 3 kg The portion AO of the mechanism is actuated by gear D and at the instant shown has a clockwise angular velocity of 8 rad/s and a counterclockwise angular acceleration of 40 rad/s. Determine: a) tangential force exerted by gear D, and b) components of the reaction at shaft O. Draw the free-body-equation for AO, expressing the equivalence of the external and ective forces. Evaluate the external forces due to the weights of gear E and arm O and the ective forces associated with the angular velocity and acceleration. Solve the three scalar equations derived from the free-body-equation for the tangential force at A and the horizontal and vertical components of reaction at shaft O. 6-7 Sample Problem 6.6 me 4 kg ke 85 mm mo 3 kg α 40rad s ω 8 rad/s Draw the free-body-equation for AO. Evaluate the external forces due to the weights of gear E and arm O and the ective forces. WE WO ( 4kg)( 9.8m s ) 39. N ( 3kg)( 9.8m s ) 9.4 N m k ( 4kg)( 0.085m) ( 40rad s ) IEα E Eα.56 N m m m I O O O ( a ) m ( rα ) ( 3kg)( 0.00m)( 40rad s ) O t O 4.0 N ( a ) m ( rω ) ( 3kg)( 0.00m)( 8rad s) O n O 38.4 N ( m L ) ( 3kg)( 0.400m) ( 40rad s ) α α O.600 N m 6-8 4

15 Sample Problem 6.6 Solve the three scalar equations derived from the freebody-equation for the tangential force at A and the horizontal and vertical components of reaction at O. F ( ) M O M O ( 0.0m) IEα + mo( ao ) t ( 0.00m) + IOα.56 N m + ( 4.0N)( 0.00m) +.600N m F 63.0 N W W E O 39. N 9.4 N I E α.56 N m m O( a O ) t 4.0 N ( ) 38.4 N m O a O I O α.600 N m n ( ) F x F x Rx mo O t y y ( a ) 4.0 N ( ) Fy F y R F W E W O m O ( a ) O R 63.0 N 39. N 9.4 N 38.4 N R x 4.0 N R y 4.0 N 6-9 Sample Problem 6.8 A sphere of weight W is released with no initial velocity and rolls without slipping on the incline. Determine: a) the minimum value of the coicient of friction, b) the velocity of G after the sphere has rolled 0 ft and c) the velocity of G if the sphere were to move 0 ft down a frictionless incline. Draw the free-body-equation for the sphere, expressing the equivalence of the external and ective forces. With the linear and angular accelerations related, solve the three scalar equations derived from the free-body-equation for the angular acceleration and the normal and tangential reactions at C. Calculate the friction coicient required for the indicated tangential reaction at C. Calculate the velocity after 0 ft of uniformly accelerated motion. Assuming no friction, calculate the linear acceleration down the incline and the corresponding velocity after 0 ft

16 Sample Problem 6.8 a rα Draw the free-body-equation for the sphere, expressing the equivalence of the external and ective forces. With the linear and angular accelerations related, solve the three scalar equations derived from the free-bodyequation for the angular acceleration and the normal and tangential reactions at C. M C ( M C ) ( W sinθ ) r ( ma ) r + Iα 5g sin30 a rα 7 ( mrα ) r + ( mr ) α W W rα r + r g 5 g ( s ) 5 3.ft 7 5 sin 30 α a.50ft s 5g sinθ α 7r 6-3 Sample Problem 6.8 5g sinθ α 7r a rα.50ft s Solve the three scalar equations derived from the freebody-equation for the angular acceleration and the normal and tangential reactions at C. F x ( F x ) W sinθ F ma W 5g sinθ g 7 F y ( F y ) N W cosθ 0 F W sin W 7 N W cos W Calculate the friction coicient required for the indicated tangential reaction at C. F μ N s F 0.43W μs N 0.866W μ s

17 Sample Problem 6.8 Calculate the velocity after 0 ft of uniformly accelerated motion. a( x 0 ) ( s )( 0ft) v v0 + x ft v r 5.7ft s 5g sinθ α 7r a rα.50ft s Assuming no friction, calculate the linear acceleration and the corresponding velocity after 0 ft. ( ) MG M G 0 Iα α 0 F x ( F x ) a( x 0 ) ( s )( 0ft) v v0 + x ft W W sinθ ma a g a ( 3.ft s ) sin30 6.ft s v r 7.94ft s 6-33 Sample Problem 6.9 A cord is wrapped around the inner hub of a wheel and pulled horizontally with a force of 00 N. The wheel has a mass of 50 kg and a radius of gyration of 70 mm. Knowing μ s 0.0 and μ k 0.5, determine the acceleration of G and the angular acceleration of the wheel. Draw the free-body-equation for the wheel, expressing the equivalence of the external and ective forces. Assuming rolling without slipping and therefore, related linear and angular accelerations, solve the scalar equations for the acceleration and the normal and tangential reactions at the ground. Compare the required tangential reaction to the maximum possible friction force. If slipping occurs, calculate the kinetic friction force and then solve the scalar equations for the linear and angular accelerations

18 Sample Problem 6.9 I mk 0.45kg m ( 50kg)( 0.70m) Assume rolling without slipping, a rα ( 0.00m)α Draw the free-body-equation for the wheel,. Assuming rolling without slipping, solve the scalar equations for the acceleration and ground reactions. M C ( M C ) 00N 0.040m ma 0.00m + Iα ( )( ) ( ) 8.0 N m α rad s a ( 50kg)( 0.00m) α + ( 0.45kg m ) ( 0.00m)( 0.74rad s ).074m s F x ( F x ) F + 00 N ma F 46.3N ( ) F x F x N W 0 N mg ( 50kg)(.074m s ) ( 50kg)(.074m s ) N α 6-35 Sample Problem 6.9 Compare the required tangential reaction to the maximum possible friction force. ( N) 98.N Fmax μsn 0.0 F > F max, rolling without slipping is impossible. Without slipping, F 46.3N N 490.5N Calculate the friction force with slipping and solve the scalar equations for linear and angular accelerations. ( N) 73.6 N F F μ N 0.5 F x ( F x ) k 00 N 73.6 N k ( 50kg)a a.53m s M G ( M G ) ( 73.6 N)( 0.00m) ( 00 N)( m) α ( 0.45kg m ) 8.94rad s α α 8.94rad s

19 Sample Problem 6.0 ased on the kinematics of the constrained motion, express the accelerations of A,, and G in terms of the angular acceleration. The extremities of a 4-ft rod weighing 50 lb can move freely and with no friction along two straight tracks. The rod is released with no velocity from the position shown. Determine: a) the angular acceleration of the rod, and b) the reactions at A and. Draw the free-body-equation for the rod, expressing the equivalence of the external and ective forces. Solve the three corresponding scalar equations for the angular acceleration and the reactions at A and Sample Problem 6.0 ased on the kinematics of the constrained motion, express the accelerations of A,, and G in terms of the angular acceleration. Express the acceleration of as r r r a a + a A A With a A 4α, the corresponding vector triangle and the law of signs yields a A 5.46α a 4. 90α The acceleration of G is now obtained from r r r r a a a + a where α G A G A a G A Resolving into x and y components, a 5.46α α cos α a x y α sin 60.73α

20 Sample Problem 6.0 Draw the free-body-equation for the rod, expressing the equivalence of the external and ective forces. I.07lb ft s Iα.07α ma ma x y ml 50lb 3.ft s ( 4ft) 50 ( 4.46α ) 6.93α (.73α ).69α 3. Solve the three corresponding scalar equations for the angular acceleration and the reactions at A and. M E ( M E ) ( 50)(.73) ( 6.93α )( 4.46) + (.69α )(.73) α +.30rad s F x ( F x ) R sin 45 ( 6.93)(.30) R.5lb F y ( F y ) + R A α.30rad s R r.5lb (.5) cos45 50 (.69)(.30) R A 7.9lb +.07α 45 o