Ch. 3: Kinetics of Particles

Transcription

1 3.0 Outline Introduction Newton s Second Law Equations of Motion Rectilinear Motion Curvilinear Motion 3.0 Outline

2 3.1 Introduction Ch. 3: Kinetics of Particles Kinetics is the study of the relations between the forces and the motion. Here we will not seriously concern whether the forces cause the motion or the motion generates the forces (causality). In this chapter, the focus is on the particles. That is the body whose physical dimensions are so small compared with the radius of curvature of its path. There are at least 3 approaches toe the solution of kinetic problems: (a) Newton s second law (b) work and energy method (c) impulse and momentum method. 3.1 Introduction

3 3. Newton s Second Law Ch. 3: Kinetics of Particles F= ma m = mass (resistance to rate of change of velocity) of the particle F= resultant force acting on the particle a= resulting acceleration measured in a nonaccelerating frame of reference For most engineering problems on earth, the acceleration measured w.r.t. reference frame fixed to the earth s surface may be treated as absolute. And Newton s nd law of motion holds. Newton s nd law breaks when the velocities of the order of the speed of light are involved theory of relativity 3. Newton s Second Law

4 3.3 Equation of Motion and Solution of Problems F= ma --- equation of motion scalar components decomposition according to a specified coordinate Two problems of dynamics (1) specified kinematic conditions, find forces straightforward application of Newton s law as algebraic equations () specified forces, find motion Difficulty depends on the form of force function (t, s, v, a), as the solutions are found by solving a system of differential equations. For simple functions, we can find closed form solutions of motion as in rectilinear motion (sec..). 3.3 Equation of Motion and Solution

5 Unconstrained motion Motion of the particle is determined by its initial motion and the forces from external sources. It is free of constraints and so has three degrees of freedom to specify the position. Three scalar equations of motion would have to be applied and integrated to obtain the motion. Constrained motion Motion of the particle is partially or totally determined by restraining guides, other than its initial motion and the forces from external sources. Therefore, all forces, both applied and reactive, that act on the particle must be accounted for in Newton s law. The number of d.o.f. and equations are reduced regarding to the type of constraints. 3.3 Equation of Motion and Solution

6 Free body diagram All forces acting on the particle needed to be accounted in the equations of motion. Free body diagram unveils every force that acts on the isolated particle. Only after the FBD has been completed should the equations of motion be written. The appropriate coordinate axes and directions should be indicated and consistently used throughout the problem. Treatment of the body as particle is valid when the forces may be treated as concurrent through the mass center. 3.3 Equation of Motion and Solution

7 3.4 Rectilinear Motion Ch. 3: Kinetics of Particles If the x-axis is the direction of the rectilinear motion, F = ma F = 0 F = 0 x x y z If we are not free to choose a coordinate direction along the motion, the nonzero acceleration component will be shown up in all equations: F = ma F = ma F = ma x x y y z z Other coordinate system such as n-t or r-θ a may be determined via the use of relative motion For pure translating moving reference frame a =a +a /B A B A 3.4 Rectilinear Motion

8 P. 3/17 The coefficient of static friction between the flat bed of the truck and the crate it carries is Determine the minimum stopping distance s that the truck can have from a speed of 70 km/h with constant deceleration if the crate is not to slip forward. 3.4 Rectilinear Motion

9 P. 3/17 mg +x F < 0.3N N If the crate is not to slip, crate and truck must have same acceleration. If the crate is not to slip, friction = static friction at impending status. Minimum stopping distance when the deceleration is the max allowable value. Fx = ma x 0.3mg = ma x, a x = 0.3g constant for minimum distance 10 v = vo + a ( s s o) 0 = 70 + ( 0.3g) s, s = 64. m Rectilinear Motion

10 P. 3/18 If the truck of Prob. 3/17 comes to stop from an initial forward speed of 70 km/h in a distance of 50 m with uniform deceleration, determine whether or not the crate strikes the wall at the forward end of the flat bed. If the crate does strike the wall, calculate its speed relative to the truck as the impact occurs. Use the friction coefficients μ s = 0.3 and μ k = Rectilinear Motion

11 P. 3/18 Ch. 3: Kinetics of Particles stopping distance = 50 m, which is less than minimum value 64. m the crate slips 10 = o + ( o) = + truck truck = v v a s s 0 70 a 50, a m/s v = vo + a ( t t o) 0 = t, tstop = s 36 Friction force: F = 0.3mg =.943m and F = 0.5mg =.45m k crate crate s Assume crate and truck go together a = a ( ) Fx = ma x F = m required friction = 3.781m > Fs the crate slips and F = F F = ma, a k =.45 m/s truck k crate [ ] ( ) crate/truck crate truck crate/truck a = a a a = = m/s the crate slips forward but will it strike the wall? 1 s = so + vo( t to) + a ( t t o) relative motion calculation 1 3 = t, tstrike =.13 s < t stop crate will strike the wall before the truck stops v = vo + a ( t t o) relative motion calculation v = =.86 m/s crate/truck +x F < 0.3N N 3.4 Rectilinear Motion

12 P. 3/3 If the coefficients of static and kinetic friction between the 0-kg block A and the 100-kg cart B are both essentially the same value of 0.50, determine the acceleration of each part for (a) P = 60 N and (b) P = 40 N. 3.4 Rectilinear Motion

13 P. 3/3 A max A B B Ch. 3: Kinetics of Particles (a) N = 0g, F = 0.5N = 98.1 N < 10 N block A moves forward relative to B Fx = ma x = 0a A, a A = m/s 98.1 = 100a, a = m/s 0g N A F N A 100g P F (b) F > 80 N block A does not move relative to B F max x = ma x A & B move together 80 = 10a, a = m/s Find developed friction by isolated FBD at A or B 80 F = 0a, F = N < F assumption is valid F = 100a, F = N max N B 3.4 Rectilinear Motion

14 P. 3/4 A simple pendulum is pivoted at O and is free to swing in the vertical plane of the plate. If the plate is given a constant acceleration a up the incline θ, write an expression for the steady angle β assumed by the pendulum after all initial start-up oscillations have ceased. Neglect the mass of the slender supporting rod. 3.4 Rectilinear Motion

15 P. 3/4 T y β x mg θ β Fy = 0 Tcosβ mgcosθ = 0 Fx = ma x Tsinβ mgsinθ = ma tan a+ gsinθ gcosθ 1 = 3.4 Rectilinear Motion

16 P. 3/8 For the friction coefficients μ s = 0.5 and μ k = 0.0, calculate the acceleration of each body and the tension T in the cable. 3.4 Rectilinear Motion

17 P. 3/8 60g T T F N s + s + c= l a + a = 0 A B A B N = 60gcos30, F = μ N = 17.4 N max s Assume motion impends at block A F = F and equilibrium F = 0 60gsin30 Fmax T = 0, T = N but cylinder B will not be in equilibrium 0g T < 0 move Assum block A slides down and block B moves up max F = ma 60gsin30 F T = 60a = 10a k A B 0g ( up) 0g T = 0a, T = N, a = 0.75 m/s, a = 1.45 m/s B B A 3.4 Rectilinear Motion

18 P. 3/35 A bar of length l and negligible mass connects the cart of mass M and the particle of mass m. If the cart is subjected to a constant acceleration a to the right, what is the resulting steady-state angle θthat the freely pivoting bar makes with the vertical? Determine the net force P (not shown) that must be applied to the cart to cause the specified acceleration. 3.4 Rectilinear Motion

19 P. 3/35 y P Mg θ T T N x mg From the given statements, pendulum and cart have same acceleration At the pendulum, Fy = 0 Tcosθ mg = 0, T = mg/cosθ Fx ma Tsinθ ma, θ tan At the cart, a g 1 = x = = Fx = ma x P Tsinθ = Ma, P = ( m+ ) M gtanθ 3.4 Rectilinear Motion

20 P. 3/36 Determine the accelerations of bodies A and B and the tension in the cable due to the application of the 50 N force. Neglect all friction and the masses of the pulleys. 3.4 Rectilinear Motion

21 P. 3/36 Ch. 3: Kinetics of Particles 70g 35g T 3T 300 N N A s A s B N B s + 3s + c = l a + 3a = 0 A A B A B F = ma T = 70a and 300 3T = 35a x x A B a =.34 m/s, a = 1.56 m/s, T = 81.8 N B 3.4 Rectilinear Motion

22 P. 3/44 The sliders A and B are connected by a light rigid bar and move with negligible friction in the slots, both of which lie in a horizontal plane. For the position shown, the velocity of A is 0.4 m/s to the right. Determine the acceleration of each slider and the force in the bar at this instant. 3.4 Rectilinear Motion

23 P. 3/44 N B s B T T N A s A 40 N Kinematics: triangle OAB s = s and 0.5 = s cos15 + s cos15, s = s = m l A B A B A B = s + s s s cos150 A B A B A ( ) diff: 0 = s v + s v cos150 s v + s v A A B B A B B A given: v = 0.4 m/s v = 0.4 m B diff: 0 = v + s a + v + s a cos150 s a + s a A A A A B B B A B B A B B ( A + vavb) () 0 = a a 1 Kinetics: F = ma 40 Tcos15 = a A and Tcos15 = 3a B into 1 a = 7.95 m/s, a = 8.04 m/s, T = 5.0 N ( ) 3.4 Rectilinear Motion

24 P. 3/46 With the blocks initially at rest, the force P is increased slowly from zero to 60 N. Plot the accelerations of both masses as functions of P. 3.4 Rectilinear Motion

25 P. 3/46 35g F A NA N = 35g, N = N + 4g = 77g A B A F = 0.N = N, F = 0.15N = N A A B B max F = 0.15N = 51.5 N, F = 0.10N = N A A B B k k max 4g N A F A N B F B P Three possible situations: no motion, B & A move together, and B & A move separately. Two impossible situations: B moves alone then F will 0 A will move eventually and A moves alone ( not jump right to FA ) max A P is applied at block B and force P is increased slowly from zero 1) 0 P F : F will be developed to cancel with the applied P, B max B F will stay zero, and so there is no motion a = 0 & a = 0 A A B 3.4 Rectilinear Motion

26 P. 3/46 ) assume both A and B go together in this phase F F and F = F F = ma FA = 35a & P FA FB = 4a k A max A A B B ( ) ( ) max ( elative to each other ), P = 6.6 N and a = 1.96 m/s at P = P = F increased slowly, a = 0.49 m/s and F = N jumping min B A at F = F about to slip r max A P FB k between these extremum values, a = : linear function of P 77 F < P 6.6 : a = a which varies linearly from 0.49 to 1.96 m/s B A B ( ) 3) A slide backward P > 6.6 N makes A slips F = F relative to B increasing P makes B accelerates more and more A A k F = ma FA = 35a k A & P FA F k B = 4a k B P a A = 1.47 m/s constant and a B = : linear function of P < P 60.0 : a = 1.47 m/s constant and.37 < a m/s jumping A max B 3.4 Rectilinear Motion k ( )

27 P. 3/ Rectilinear Motion

28 P. 3/48 The system is released from rest in the position shown. Calculate the tension T in the cord and the acceleration a of the 30 kg block. The small pulley attached to the block has negligible mass and friction. (Suggestion: First establish the kinematic relationship between the accelerations of the two bodies.) 3.4 Rectilinear Motion

29 P. 3/48 Kinematics: b = c + x and b + y = l diff: bb = xx and b + y = 0 () ( ) b + bb = x + xx and b + y = 0 1 at this instant: x/b = 4 / 5, x = 0, b = 0 initially rest assume cylinder moves down, hence block moves to the left Kinetics: for 30 kg block F = ma T 3/5 T 30g + N = 0, N = 30g + T/5 F T 4/5 = 30x assume the block moves F = 0.5N for 15 kg cylinder 15g T = 15y = 15b x 4 b ( T/15 g) 30 recall () 1, = = =, T = N b 5 x 7.5g 0.7T 7.5g 0.7T x = = m/s 30 c T F +b +x +y 30g T N T 15g 3.4 Rectilinear Motion

30 3.5 Curvilinear Motion Ch. 3: Kinetics of Particles Choose appropriate coordinate system (x-y, n-t, or r-θ) for the given problem. Determine the motion along those axes. Then set up the Newton s law along those axes. The positive sense of the force and acceleration must be consistent. x-y system: F = mx F = my n x r y ( ) ρβ ( ρ ) t ( ) ( ) θ n-t system: F = m = m v / F = mv r- θ system: F = m r r θ F = m rθ + rθ 3.5 Curvilinear Motion

31 P. 3/55 The member OA rotates about a horizontal axis through O with a constant counterclockwise velocity ω= 3 rad/s. As it passes the position θ= 0, a small block of mass m is placed on it at a radial distance r = 450 mm. If the block is observed to slip at θ= 50, determine the coefficient of static friction μ s between the block and the member. 3.5 Curvilinear Motion

32 P. 3/55 use n-t coordinate system s s ( ) given: ρ = 0.45 m, β = 50, ρ = 0 no slip until β = 50, β = 3 rad/s, β = 0 ( ρβ ρβ ) ( ρβ ) Ft = ma t N mgcos50 = m +, N = mgcos50 Fn = ma n mgsin50 F= m At 50, F = F = μ N and directs upward because gsin50 > ρβ which means bar OA rotates too slow than required to keep the block stays on the bar. The friction will develop to resist ( ρβ ) the block from sliding down or to match F with. If the bar rotates very very slow, friction force cannot make ρβ ( > ρβ ) F to match F cannot be reduced any more, F. And then the block will slide down, hence ρ decreases, to the position ρ where v / lar ge enough to match F (i.e., to satisfy Newton's law) t n F mg N s ( ) mgsin50 μ mgcos50 = m ρβ, μ = s 3.5 Curvilinear Motion

33 P. 3/67 A kg sphere S is being moved in a vertical plane by a robotic arm. When the arm angle θis 30, its angular velocity about a horizontal axis through O is 50 deg/s CW and its angular acceleration is 00 deg/s CCW. In addition, the hydraulic element is being shortened at the constant rate of 500 mm/s. Determine the necessary minimum gripping force P if the coefficient of static friction between the sphere and the gripping surfaces is 0.5. Compare P to the minimum gripping force P s required to hold the sphere in static equilibrium in the 30 position. 3.5 Curvilinear Motion

34 P. 3/67 θ mg r mg F r F θ F s given: m = 1 kg, r = 1 m, r = 0.5 m/s, r = 0 π π θ = 30, θ = 50 = rad/s, θ = 00 = 3.49 rad/s ( θ ) ( ) Fr = ma r mgsinθ + Ff = m r r, F r f = N r Fθ = ma θ Ff mgcosθ = m rθ + r θ, Ff = N θ θ F = F + F = = μ P P = 7.0 N f f f s r θ static equilibrium: F = μ P = mg P = 19.6 N s s s s 3.5 Curvilinear Motion

35 P. 3/69 A flatbed truck going 100 km/h rounds a horizontal curve of 300 m radius inwardly banked at 10. The coefficient of static friction between the truck bed and the 00 kg crate it carries is Calculate the friction force F acting on the crate. 3.5 Curvilinear Motion

36 P. 3/69 y mg n F assume the crate tends to slide up the truck bed friction directs downslope N the crate has absolute curve motion into the paper on the horizontal plane m 10 Fn = ma n Nsin10 + Fcos10 = Fy = 0 mg + Ncos10 Fsin10 = 0 N = 01.5 N and F = N check if this friction can be provided F = 0.7N= 1415 N> F max the crate tends to slide up due to high speed curved motion but still too far from sliding up (can increase the truck speed yet the crate does not move relative to the truck bed) 3.5 Curvilinear Motion

37 P. 3/69 The flatbed truck starts from rest on a road whose constant radius of curvature is 30 m and whose bank angle is 10. If the constant forward acceleration of the truck is m/s, determine the time t after the start of motion at which the crate on the bed begins to slide. The coefficient of static friction between the crate and truck bed is μ s = 0.3, and the truck motion occurs in a horizontal plane. 3.5 Curvilinear Motion

38 P. 3/69 static motion y 00g 00g n F sn F N F st (inward) N 3.5 Curvilinear Motion

39 P. 3/69 Ch. 3: Kinetics of Particles Static case: N = 00gcos10 = 193. N, F = 0.3N = N s s s F = 00gsin10 = N upward to prevent sliding down the incline, and < F s Slipping when friction = F but in what direction? s s F can be divided in two components: along n- and t-axis F points in positive t (inward the paper) to match the positive a s s t F points down the incline to match the component of a down the incline n When the truck moves, N > N to match the positive component of a up the truck bed s n t n given: ρ = 30 m, ρ = 0, ρ = 0, a = m/s [ t ] y a = v v = a t = t n t F = 0 Ncos10 00g Fs sin10 = 0 1 n 4t Fn = ma n Fs cos10 + Nsin10 = 00 n 30 Ft = ma t Fs = 00 = 400 N t n t ( ) Fs + F n s = F t s Fs + F n s = 0.3N t () Fs = 0.09N and substitute into 1 s s () ( ) N = , N but < N which is impossible N = N, F = N, F = 400 N, t = 5.58 s s t 3.5 Curvilinear Motion

40 P. 3/7 The small object is placed on the inner surface of the conical dish at the radius shown. If the coefficient of static friction between the object and the conical surface is 0.30, for what range of angular velocities ωabout the vertical axis will the block remain on the dish without slipping? Assume that speed changes are made slowly so that any angular acceleration may be neglected. 3.5 Curvilinear Motion

41 P. 3/7 Ch. 3: Kinetics of Particles mg mg ω min ω max F s F s N N given: ω = 0, ρ = 0. m, ρ = 0, ρ = 0 ω ω ω ω min n s n max n s n min max causes small a F upward to reduce F causes large a F downward to increase F : Fy = 0 Ncos Nsin30 mg = 0 Fn = ma n Nsin30 0.3Ncos30 = m 0. ωmin, ωmin = rad/s ( ) : Fy = 0 Ncos30 0.3Nsin30 mg = 0 ( ω ) Fn = ma n Nsin Ncos30 = m 0. max, ωmax = 7.14 rad/s < ω < 7.14 rad/s 3.5 Curvilinear Motion

42 P. 3/74 The kg slider fits loosely in the smooth slot of the disk, which rotates about a vertical axis through point O. The slider is free to move slightly along the slot before one of the wires becomes taut. If the disk starts from rest at time t = 0 and has a constant clockwise angular acceleration of 0.5 rad/s, plot the tensions in wires 1 and and the magnitude N of the force normal to the slot as functions of time t for the interval 0<=t<=5 s. 3.5 Curvilinear Motion

43 P. 3/74 given: θ = 0.5 rad/s constant θ = 0.5t, θ = 0.5t r = 0.1 m & free to move slightly move with the disk r = 0, r = 0 assume N and T to be in the indicated direction and use r- θ coordinate ( ( ) ) ( ) Fr = ma r Ncos45 Tcos45 = t = 0.05t Fθ = ma θ Nsin45 Tsin45 = = t t 0.1 N = T = N is always positive the assumed direction is correct T will be negative for t < s 0, 0 t s t, 0 t < s T1 = 0.05t 0.1 and T =, t > s 0, t s T N θ r 45 θ 3.5 Curvilinear Motion

44 P. 3/ Curvilinear Motion

45 P. 3/86 A small rocket-propelled vehicle of mass m travels down the circular path of effective radius r under the action of its weight and a constant thrust T from its rocket motor. If the vehicle starts from rest at A, determine its speed v when it reaches B and the magnitude N of the force exerted by the guide on the wheels just prior to reaching B. Neglect any friction and any loss of mass of the rocket. 3.5 Curvilinear Motion

46 P. 3/86 mg T n t N Fn = ma n N mgsinθ = mv /r Ft = ma t T + mgcosθ = ma t, a t = θ t 0 [ ] ( θ ) T+ mgcosθ m Tθ vdv = a tds v / = a rd, v = r + gsinθ m N= 3mgsinθ + Tθ πt vθ= π/ = r + g Nθ= π/= 3mg + Tπ m 3.5 Curvilinear Motion

47 P. 3/97 A hollow tube rotates about the horizontal axis through point O with constant angular velocity ω o. A particle of mass m is introduced with zero relative velocity at r = 0 when θ= 0 and slides outward through the smooth tube. Determine r as a function of θ. 3.5 Curvilinear Motion

48 P. 3/97 given: θ = ω, θ = 0 p () () o p h o o Ch. 3: Kinetics of Particles () at t = 0, r = 0, r= 0, θ = 0 θ t = ω t ( θ ) Fr = ma r mgsinθ = m r r r ω r = gsinω t differential equation of r t r t = r + r o () particular solution r is a solution of r ω r = gsinω t p o o r t = forced response of gsinω t = Cgsinω t h () ( ) o sub. into diff. eq. Cω gsinω t Cω gsinω t = gsinω t C = o o o o o o ωo homogeneous solution r is a solution of r ω r = 0 r t = free natural response = Ae sub. into diff. eq. As e () () t ωot = + r h Ae Be ω t = + = Ae + Be h st o st st o o o o ωot ωot r t r r gsinω t, which must satisfy i.c. p h o ωo o o o Aω e = 0, s = ω, ω g r( 0) = 0 = A+ B and r ( 0) = 0= Aωo Bωo ω g g g A=, B= r = sinh sin 4ω 4ω ω 1 ( θ θ) o 1 mg N r θ 3.5 Curvilinear Motion

49 P. 3/98 The small pendulum of mass m is suspended from a trolley that runs on a horizontal rail. The trolley and pendulum are initially at rest with θ= 0. If the trolley is given a constant acceleration a = g, determine the maximum angle θ max through which the pendulum swings. Also find the tension T in the cord in terms of θ. 3.5 Curvilinear Motion

50 P. 3/98 lθe t lθ e n t T n a P gi mg use n-t coordinate to avoid unknown T in t-direction [ a = a + a ] P C P/C a = gi a = l θ e + l θe C P/C n t translating axes attached to the cart to observe pendulum ( l ) Ft = ma t mgsinθ = m gcosθ + θ g θ = ( cos θ sin θ) a s function of θ l θ g g d = d / = cos sin d, = sin + cos 1 ( θ θ ) ( ) ( ) θ θ θ θ θ θ θ θ θ θ θ l l 0 θ or θ when θ = 0 sinθ + cosθ = 1 θ = π / max min max ( l ) Fn = ma n T mgcosθ = m gsinθ + θ T = mg 3sin + 3cos 3.5 Curvilinear Motion

51 P. 3/99 A small object is released from rest at A and slides with friction down the circular path. If the coefficient of friction is 0., determine the velocity of the object as it passes B. (Hint: Write the equations of motion in the n- and t- directions, eliminate N, and substitute vdv = a t rdθ. The resulting equation is a linear nonhomogeneous differential equation of the form, the solution of which is well known.) ( ) ( ) dy/dx + f x y= g x 3.5 Curvilinear Motion

52 P. 3/99 Fn = ma n N mgsin = m 3 Ft = ma t mgcosθ 0.N = m 3 eliminate N: gcosθ 0. gsinθ + 3θ = 3θ ( θ ) Ch. 3: Kinetics of Particles θ ( θ ) ( θ) ( ) { ( )} ( ) 1 1 θd θ = θd θ θd θ = gcosθ 0. gsinθ + 3 θ dθ = d θ 3 d + 0.4( θ ) = g ( cosθ 0.sin θ), θ as a function of θ dθ 3 let θ = u θ and to solve the differential equation for u θ u = u + u p ( ) ( ) h up ( θ) = forced response of g( cosθ 0.sinθ) = Acosθ + Bsinθ 3 sub. into diff. eq. Asinθ + Bcosθ + 0.4( Acosθ + Bsinθ) = g ( cosθ 0.sinθ) match the coeff. of sin θ and cos θ: A= g B= g u h Ce ( θ ) = solution of the homogeneous equation = Ce + = = sθ sθ s 0.4 Ce 0, s θ u ( θ) = gcosθ + gsin θ + Ce with u ( 0) = g C 0 u θ = θ = gcosθ + gsinθ ge at θ = π /, θ = 3.38 v = r θ = 5.5 m/s = ( ) B Real world where friction exists makes the phenomena difficult sθ θ mg n F=0.N N t 3.5 Curvilinear Motion

53 P. 3/100 A small collar of mass m is given an initial velocity of magnitude v o on the horizontal circular track fabricated from a slender rod. If the coefficient of kinetic friciton is μ k, determine the distance traveled before the collar comes to rest. (Hint: Recognize that the friction force depends on the net normal force.) 3.5 Curvilinear Motion

54 P. 3/100 n mg y N h Normal force has component N and N Fy = 0 Nv = mg Fn = ma n Nh = m r [ ] k 4 vdv a tds vdv r m g m v ds 0 s ( ) v o μ 0 k k r g v v Ft = ma t F= μk Nv + Nh = mat μ = = + mr rdv r v + v + r g = ds, s = ln + μ rg v h F t 4 o o N v 3.5 Curvilinear Motion

55 P. 3/101 The slotted arm OB rotates in a horizontal plane about point O of the fixed circular cam with constant angular velocity θ = 15 rad/s. The spring has a stiffness of 5 kn/m and is uncompressed when θ= 0. The smooth roller A has a mass of 0.5 kg. Determine the normal force N that the cam exerts on A and also the force R exerted on A by the sides of the slot when θ= 45. All surfaces are smooth. Neglect the small diameter of the roller. 3.5 Curvilinear Motion

56 P. 3/ β θ r θ F r Kinematics: 0. = r + 0.rcosθ diff: 0 = rr + 0.rcos θ 0.r θsinθ 0 = r + rr + 0.rcosθ 0.rθsinθ 0.r sin 0.r sin 0.r cos θ θ θ θ θ θ given: θ = π / 4, θ = 15 rad/s, θ = 0 N β R r = m, r = 0.66 m/s, r = m/s =, β = 0.7 sin135 sin β ( θ ) ( θ θ) ( ) Kinetics: spring force at θ = π / 4 : F = 5000 r 0.1 compressed Fr = ma r F + Ncos0.7 = m r r Fθ = ma θ R Nsin0.7 = m r + r N = 81.7 N R = 38.7 N 3.5 Curvilinear Motion

57 P. 3/10 The small cart is nudged with negligible velocity from its horizontal position at A onto the parabolic path that lies in a vertical plane. Neglect friction and show that the cart maintains contact with the path for all values of k. 3.5 Curvilinear Motion

58 P. 3/10 If the cart maintains contact, N > 0 use n-t coordinate since N aligns with the n-axis 3/ 1 ( y ') + dy d y ρ = y kx k ( x) tan θ k y '' = = = = dx dx 3/ 1+ 4k x ρ = k v Fn = ma n N + mgcosθ = m ρ 1+ tan [ ] θ = sec θ cosθ = F = ma mgsinθ = ma t t t t N mkgx 0 3/ 3/ k x vdv = a ds vdv = gsinθds = gdy, v = gy = kgx mg k mg = = > 1+ 4k x 1+ 4k x 1+ 4k x n N mg dx ds θ dy t 3.5 Curvilinear Motion