1 Kinematics 1. 2 Particle Dynamics Planar Dynamics 42

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1 Dynamics: Example Problems Contents 1 Kinematics 1 Particle Dynamics 13 3 Planar Dynamics 4 1 Kinematics Problem 1: 0 pts. The two blocks shown to the right are constrained the move in orthogonal slots and are connected by a rod of length l = 10 cm. If the velocity of A is constant, so that v A = v ĵ, find the velocity and acceleration of block B, as a function of v, when x = 6 cm. A v A ĵ ê 1 ê î x θ l z B a We define the angle θ of the link and the displacement of block as z, as shown in the figure. Thus ê 1 = cos θ î sinθ ĵ, ê = sin θ ĵ + sinθ ĵ. In addition, at this instant θ = and z = 8 cm. Given the velocity of A, the velocity of B can be written as v B = v A + ω β/ r BA. which reduces to v B = v ĵ + = l θ sin θ θ ˆk lê 1, î + v l θ cos θ ĵ. The velocity of block B only lies in the î direction so that the ĵ component must vanish, which implies that θ = v l cos θ. 1

2 Returning this to the velocity v B = lsin θ v î = v tan θ î. l cos θ inally, at this instant tanθ = 3/4, so that v B = 3v/4î. b The acceleration of the lower block can be written as a B = a A + α β/ r BA + ω β/ ω β/ r BA. The acceleration of A vanishes, so that a A = 0, and with the angular acceleration of the bar identified as α β/ = θ ˆk, this reduces to a B = l θ sin θ l θ cos θ î + l θ cos θ + l θ sin θ ĵ. As for the velocity, the component of the acceleration in the ĵ direction must vanish, so that θ = θ tan θ. inally, returning to the acceleration of B a B = l θ cos θ î = v l cos 3 θî. inally, at this instant a B = 5v 64 î. Problem : 0 pts. In the mechanism shown, the inner arm length l = 5 cm spins at a constant angular speed ω = θ = 10 rad/s, while the disk, attached to the arm at A, spins at constant rate ω 3 = θ 3 = 5 rad/s with respect to the ground. inally, the particle P with mass m = 1.5 kg slides in a smooth slot on the disk and is attached to a spring of stiffness k = 50 N/m. Assume that the spring is unstretched when the particle is at the center of the disk. At this instant, θ = 60, θ 3 = 45, and x = 8 cm: x A P ĵ 3 θ 3 î 3 a Determine the acceleration of A, the center of the disk, with respect to the ground; b Determine the acceleration of P with respect to the disk at this instant, that is, relative to a frame of reference fixed in the disk in the direction of the slot. Hint: you must use linear momentum balance on the particle. ĵ î θ ĵ î

3 The above directions can be related as î = cos θ î + sinθ ĵ, î 3 = cos θ 3 î + sin θ 3 ĵ, ĵ = sin θ î + cos θ ĵ, ĵ 3 = sin θ 3 î + cos θ 3 ĵ. a In terms of the coordinates provided and the directions defined above, the acceleration of A can be expressed as a A = ω β/ ω β/ r AO = θ ˆk θ ˆk lî = l θ î. Therefore, at this instant a A = 5 m/sî = 1.5 m/sî 1.6 m/sĵ. b The acceleration of P with respect to the disk is 3 a P = ẍî 3, which can be related to the response of the system through linear momentum balance, = m a P, with a P = a A + 3 a P + α D/ r P A + ω D/ ω D/ r P A + ω β/ 3 v P In this the angular acceleration of the disk vanishes and the velocity of P with respect to the disk is 3 v P = ẋ î 3, so that the acceleration of P becomes a P = l θ î + ẍ î 3 + θ3 ˆk θ3 ˆk xî 3 + θ3 ˆk ẋî 3, = l θ î + ẍ î 3 x θ 3 î 3 ẋ θ 3 ĵ3. An appropriate free body diagram is shown to the right, where the only forces that act on the particle are due to the spring and the normal force arising from the constrained motion within the slot. inally, noting that î = cosθ θ 3 î 3 + sinθ θ 3 ĵ 3, linear momentum balance can be written in the î 3,ĵ 3 directions as k xî 3 N ĵ 3 = m a P, k xî 3 + N ĵ 3 = m ẍ x θ 3 l θ cosθ θ 3 î 3 +m ẋ θ 3 l θ sinθ θ 3 ĵ 3. The component in the î 3 direction may be solved for ẍ as ẍ = θ 3 k x + l m θ cosθ θ 3. or the given values we find that ẍ =.95 m/s. 3

4 Problem 3: A particle moves along a circular path of radius r = 10 m. If the speed v increases uniformly from rest with v = 0.50 m/s a ind the speed and the acceleration of the particle after t = 4 s; b How far does the particle travel during this time? c If the mass of the particle is m = kg, how long in time does the particle travel before the magnitude of the force acting on the particle is = 16 N? r a With v = 0.50 m/s the speed of the particle when released from rest can be expressed as vt = 0.50 m/s t. In addition, using normal and tangential coordinates, the acceleration of the particle may be written as a P = v ê t + v r ên = 0.50 m/s ê t m/s 4 t ê n. Therefore, at t = 4 s, the speed and acceleration of the particle are v4 s =.00 m/s, a P 4 s = 0.50 m/s ê t m/s ê n. Note that the magnitude of the acceleration at t = 4 s is a P 4 s = 0.64 m/s. b The displacement st of the particle can be found by integrating the speed again, so that st = 0.5 m/s t, and at t = 4 s, the distance travelled by the particle is s4 s = 4 m. c rom momentum balance, the magnitude of the external force equals the mass times the magnitude of the acceleration, that is = m a P. 4

5 Using the acceleration of the particle determined previously, the magnitude of the force becomes = m v v + = m v v t +, r r = kg 0.50 m/s m/s 4 t. Therefore, solving this expression for t with = 16 N, we find that m v r t 4 = m v = s 4, t = s. Problem 4: A disk rotates with constant angular velocity θ = ω = 0.5 rad/s. The particle P moves along a radial direction relative to the disk at constant rate ṙ = v = 1 m/s. ê θ ê r a ind the acceleration of the particle when r = 0.75 m in terms of radial and transverse directions ê r and ê θ. rt b What force must act on the particle at this instant to achieve this motion? ω a In terms of polar coordinates, the acceleration of the particle can be written as a P = r r θ ê r + r θ + ṙ θ ê θ. or this system, we find that so that the acceleration of P becomes ṙ = v, r = 0, θ = ω, θ = 0, a P = r ω ê r + v ω ê θ = m/s ê r + m/s ê θ. b Linear momentum balance implies that = m a P. Since the acceleration is known from above, the force required to act on the particle must be = m a P = m [ 364 ] 1 m/s ê r + ê m/s θ, where m is the mass of the particle. 5

6 Problem 5: The particle P moves along a spiral path, defined by: rθ = 1 π θ. If the angular position of the particle is defined as θt = ω t, find the position and acceleration of P with respect to the ground after t = s if ω = 4 rad/s. ê θ ĵ ê r î We measure the position of the particle P with respect to the origin O in terms of polar coordinates: r P O = rt ê r. With this, the velocity and acceleration of P become: v P = ṙt ê r + rt θt ê θ, a P = rt rt θ t ê r + rt θt + ṙt θt ê θ. or the given path: rt = ω t π, θt = ω t, ṙt = ω π, θt = ω, rt = 0, θt = 0, so that the position, velocity not required, and acceleration at t = s become: r P O = 8 π êr, v P = 4 π êr + 3 π êθ, inally, at this instant ê r and ê θ may be related to î and ĵ as: a P = 18 π êr + 3 π êθ. ê r = cos θt ê r + sinθt ê θ, ê θ = sin θt ê r + cos θt ê θ, = ê r ê θ, = ê r ê θ. Problem 6: The telescoping shaft is extending at a constant rate ṙ = 0. m/s and is rotating with an angular speed given by θ = ωt = e t rad/s. ind the acceleration at time t = 10 s. You may leave your answer in the ê r and ê θ directions. 6

7 a If we define ê r to be along the shaft and ê θ to be in the direction of increasing θ as is standard, we find: a P = r r θ ê r + r θ + ṙ θ ê θ. With ṙ = 0. m/s and θ = e t, we see that, if r0 = 0: rt = 0. m/st, ṙ = 0. m/s, θ = e t s rad/s, r = 0, θ = 4te t s rad/s 3. So at t = 10 s, the acceleration becomes: a P = m/s ê r m/s ê θ. Awfully big, huh? Problem 7: A bug is walking outward on a record player at a constant velocity, with respect to the vinyl, of v B = vî + ĵ. At this instant, the bug is at a position r BO = rî. ind a the acceleration of the bug with respect to the vinyl; b the acceleration of the bug with respect to the ground. a Note that the velocity of the bug with respect to the vinyl is constant. Therefore a B = 0. b We simply use the acceleration formula: a B = a O + a B + α / r BO + ω / ω / r BO + ω / v B. and identify each term as follows: a O = 0, ω / = ωˆk, a B = 0, α / = αˆk, so that: a B = αˆk rî + ωˆk ωˆk rî + ωˆk vî + ĵ, = rω + ωv î + rα + ωv ĵ. Note that we have expressed the acceleration of the bug with respect to the ground in terms of unit directions fixed in the disk. However, the resulting expression is nonetheless a B. 7

8 Problem 8: or the two-link system shown to the right, the base link CO of length l = 4 m spins with constant angular velocity θ ω = π/ rad/s. Meanwhile the attached link PQ of length d = m oscillates, with the angle φ given as φt = φ 0 sinµt, or φ 0 = π/6 rad and µ = π rad/s, find C d P φt ĵ 3 ĵ î î 3 a the velocity of P, and b the acceleration of P Q l ĵ with respect to the ground at time t = s. θ î The basis directions in and 3 can be related to ground as î = cos θ î + sin θ ĵ, î 3 = cos φî + sinφĵ, ĵ = sin θ î + cos θ ĵ, ĵ 3 = sin φ î + cos φ ĵ. O In general θt = θ t, φt = µφ0 cosµt, and φt = µ φ 0 sinµt, so that at t = s, θ s = π rad, φ s = 0 rad, φ s = π /6 s 1, φ s = 0 s. a The velocity of P can be written as v P = v C + ω β3 / r P C, where ω β3 / is the angular velocity of link PQ and the velocity of C can be written as v C = v O + ω β / r CO. The various components of these expressions can be expressed as v O = 0, ω β / = θ ˆk, ω β3 / = r CO = lî, r P C = d î 3. φ ˆk, As a result, the velocity of P becomes v P = θ ˆk lî + θ ˆk d î 3, = l θ ĵ + d φĵ 3. Therefore at this instant ĵ = ĵ and ĵ 3 = ĵ, so that v P s = π + π m/s ĵ = 4.64 m/sĵ. 6 8

9 b The acceleration of P can be expressed as a P = a C + α β3 / r P C + ω β3 / ω β3 / r P C, where α β3 / is the angular acceleration of link PQ and the acceleration of C is a C = a O + α β / r CO + ω β / ω β / r CO. In these equations a O = 0, α β / = θ ˆk = 0, and α β3 / = φ ˆk, so that the acceleration of P becomes a P = θ θ ˆk lî + φ ˆk d î3 + φ φ ˆk d î3, = l θ î d φ î 3 + d φ ĵ 3. At this instant î = î, and î 3 = î, so that a P s = π + π4 36 m/s î = 7.16 m/s î. Problem 9: With respect to a disk, the particle P moves along a spiral with path r P C = xtî + ytĵ, with ĵ î P xt = ρt cosω t, yt = ρt sinω t, where ρ = 0.10 m/s and ω = 1 rad/s. In addition the disk is translating with constant velocity v C = v î = 0.50 m/s. ind the acceleration of the particle at t = 5 s. ĵ vî î C We can identify an intermediate frame of reference fixed at the center of the disk. Thus, the acceleration of P can be written as a P = a C + α / r P C + ω / ω / r P C + ω / v P + a P. However, the disk is not rotating and moreover the acceleration of C vanishes because its velocity is constant. Therefore a C = 0, α / = 0, ω / = 0, so that a P = a P. In essence is is an inertial reference frame. We thus find that a P = ẍtî + ÿtĵ, 9

10 where ẍt = ρt cosω t ρω sinω t, ÿt = ρt sinω t + ρω cosω t. At t = 5 s, this reduces to a P = m/s î m/s ĵ. Problem 10: In the mechanism shown, the inner arm length l = 5 cm spins at a constant angular speed ω = 5 rad/s, while the disk radius r = 10 cm, attached to the arm at A, spins at constant rate ω 3 = 10 rad/s. At this instant, θ = 90 and θ 3 = 45 : a Deterine the velocity of A, the center of the disk; ĵ 3 P θ 3 ω 3 A î 3 b ind the velocity and acceleration of P, at the edge of the disk, with respect to the ground. ω î θ ĵ ĵ î a The velocity of A can be expressed as: v A = ω / r AO, = ω ˆk l î, = lω ĵ. At this instant in time, ĵ = î, so that with the values given above: v A = 1.5 m/s î. b Likewise, the velocity of P can be expressed as: v P = v A + ω 3/ r P A, = lω ĵ + ω 3 ˆk r î3, = lω ĵ + rω 3 ĵ 3. At this instant in time, ĵ 3 = 1/ î + 1/ ĵ, so that with the values given above: v P = 1.5 m/s î + 1 m/s 1 î + 1 ĵ, = 1.96 m/s î m/s ĵ 10

11 inally, the acceleration of P is: a P = a A + α 3/ r P A + ω 3/ ω 3/ r P A, with α 3/ = 0, and: a A = α / r AO + ω / ω / r AO. Therefore, with α / = 0: a P = ω / ω / r AO + ω 3/ ω 3/ r P A, = ω ˆk ω ˆk l î + ω 3 ˆk ω 3 ˆk r î3, = lω î rω3 î3, = 6.5 m/s î + 10 m/s î 3. At this instant: î = ĵ, î 3 = 1 î + 1 ĵ, so that: a P = 7.07 m/s î m/s ĵ. Problem 11: The body shown to the right rotates with constant angular speed θ = 5 rad/s and r AO = 0.50 m î. In addition, the position of the point P relative to A is described by r P A = ut î + vt ĵ, with: π ut = 0.10 m sin t, vt = t 8 m/s. ĵ,vt P A î,ut If, at this instant, θ = 60 and t = s: a Determine v P, the velocity of P relative to the rotating body; b ind the velocity and acceleration of P with respect to the ground. θ O a In terms of a, fixed in the rotating body, the velocity of P is found to be: v P = d dt r P A = d ut î + vt ĵ, dt = ut î + vt ĵ. Therefore, with: π ut = 0.05 π m/s cos t, vt = 0.15 m/s, 11

12 at t = s, v P becomes: v P = 0.05 π m/s î m/s ĵ, = m/s î m/s ĵ. Similarly, we may also find a P to be: a P = d dt r P A = d dt ut î + vt ĵ, = üt î + vt ĵ, and at this instant, a P = 0. b Since the point O is fixed in the body as well as the ground, the velocity and acceleration of P with respect to the ground may, in general, be written as: v P = v O + v P + ω / r P O, a P = a O + a P + α / r P O + ω / ω / r P O + ω / v P. or this system, we find that: v O = 0, a O = 0, ω / = ω ˆk = 5 rad/s ˆk, α / = 0. In addition, at this instant: a P = 0, r P O = 0.50 m î m ĵ. inally, the velocity and acceleration of P become: v P = v O + v P + ω / r P O, = m/s î m/s ĵ + 5 rad/s ˆk 0.50 m î m ĵ, = m/s î +.65 m/s ĵ, a P = a O + a P + α / r P O + ω / ω / r P O + ω / v P, = rad/s ˆk 5 rad/s ˆk 0.50 m î m ĵ + 5 rad/s ˆk m/s î m/s ĵ = m/s î m/s ĵ. Note that the velocity and acceleration of P could be expressed through the intermediate point A, instead of O. In either case the final result is the same. Problem 1: The rod AB is confined to move along the inclined planes as shown. If the acceleration of point A has a magnitude of 5.0 m/s and a speed.0 m/s, both directed upward when the bar is horizontal, find the angular acceleration of the bar at this instant. 1

13 Problem 13: At time t = 0 the disk of mass m and radius r has an angular velocity of ω ˆk. If a block of mass m is attached to the cord, find: a the angular acceleration of the disk at t = 0 + ; b the distance the block falls after t seconds. Particle Dynamics Problem 14: A particle of mass m moves along a frictionless circular hoop of radius r, and is subject to gravity and a transverse force ê θ. a ind the angular acceleration θ of the particle as a function of θ; ĵ î ê θ ê r ê θ b If the particle is released from rest at θ = 0, determine its speed when θ = 90 if = mg. θ g a Using the coordinate θ as provided, the acceleration of the particle can be written as a G = r θ ê r + r θ ê θ. In addition, a free body diagram for the particle can be constructed as shown to the right. The force N ê r represents the normal force acting between the hoop and the particle while the gravitational force is in the ĵ direction. inally, the î and ĵ directions can be written in terms of ê r and ê θ as î = cos θ ê r sinθ ê θ, ĵ = sin θ ê r + cos θ ê θ. N ê r ê θ mg ĵ Applying linear momentum balance to this particle, the equations of motion can be written 13

14 as = N mg sin θ ê r + mg cos θ ê θ = m a G mr θ ê r + mr θ ê θ. Therefore, the component of this equation in the ê θ direction may be solved for θ to yield θ = mg cos θ. mr b In the above result, the second derivative of θ with respect to time is a function of θ alone. Thus we can relate θ and θ as t { } mg cos θ θ = θ dt, 0 mr t t θ θ mg cos θ dt = θ dt, 0 0 mr θ t θ 0 = θt θ0 g sinθt sinθ0. mr r When the particle is released from rest at θ = 0, θ0 = 0 and θ0 = 0, so that this reduces to θ t = mr θt g r sin θt. With = mg, when θt f = π/, the value of θt f becomes θt f = π 1 g r. inally, the speed of the particle is ṡ = r θ, and at this instant ṡt f = r θt f = π 1 g r. Problem 15: The upper block rests on a rough surface with coefficient of friction µ and is subject to a force = 0 î with 0 > 0. x a ind the acceleration of the lower block assuming the upper block slides to the right. b If the system is released from rest, for what values of the coefficient of friction µ will the system remain stationary? 0 î m G 1 µ g ĵ y î m G 14

15 a We begin by defining two coordinates, x and y, as shown in the figure, which measure the displacement of the two masses. As the upper block moves to the right by some distance x, the pulley moves down by half the distance, so that y = x/. inally, the acceleration of each block is a G1 = ẍî, a G = ÿ ĵ. An appropriate free body diagram for each mass is shown to the right, as well as one for the massless pulley. Notice that the forces acting on the pulley must equilibrate because this element is massless. Also, the unknown tension in the cable is T, while the friction force acting on the upper block is f r î. When the block is sliding the magnitude of this force is related to the velocity of the upper block as f r = µn sgnẋ, 0 î N ĵ mg î f r î T î T ĵ T ĵ T ĵ T ĵ so that if the block is sliding to the right, ẋ > 0, f r < 0, and the friction force is orientated in the î direction. mg ĵ Applying linear momentum balance to each block yields the following equations of motion = fr + T 0 î + N mg ĵ = = m ẍ î = m a G1, = T mg ĵ = m ÿ ĵ = m a G. Therefore N = mg and, substituting in for the friction force with ẋ > 0, the remaining equations reduce to µmg + T 0 = m ẍ, T mg = m ÿ. Eliminating the unknown tension T from these equations leads to mg µmg 0 = m ẍ + ÿ. inally, using the kinematic relation between x and y, we may solve for ÿ as ÿ = g 1 µ 0 m. b If the block is stationary, the acceleration of each block vanishes, so that ẍ = ÿ = 0, and the equations of motion become = fr + T 0 î + N mg ĵ = 0, = T mg ĵ = 0. 15

16 Eliminating the unknown tension, these equations of motion reduce to f r + mg 0 = 0, N mg = 0, and solving for the friction force and the normal load f r = 0 mg, N = mg. Therefore the stationary state is valid provided f r µn, or µ 0 mg 1 Problem 16: A pair of blocks, each of mass m = kg is being pulled up a smooth surface inclined at an angle φ = 30 by a force = 0 ê 1, with = 10 N. ê a ind the acceleration of the blocks; ê 1 0 ê 1 b What is the tension in the cable connecting the two masses? x m φ m G 1 ĵ G g î a We begin by defining the directions ê 1 and ê normal and tangential to the surface, while the coordinate x measures the displacement of the blocks up the plane, in the ê 1 direction. In addition, the directions î and ĵ can be expressed in the ê 1 and ê directions as î = cos φ ê 1 sin φ ê, ĵ = sin φ ê 1 + cos φ ê. m g ĵ 0 ê 1 A free body diagram is shown to the right for each mass. Note that the acceleration of each block is identical, so that a G1 = a G = ẍê 1. Applying linear momentum balance to each mass leads to m g ĵ T ê 1 T ê 1 N 1 ê 0 Tê 1 + N 1 ê mg ĵ = mẍê 1, T ê 1 + N ê mg ĵ = mẍê 1. N ê If these equations are expressed only in the ê 1 and ê directions 0 T mg sin φ ê 1 + N 1 mg cos φ ê = mẍê 1, T mg sin φ ê 1 + N mg cos φ ê = mẍê 1, 16

17 Eliminating T from these equations of motion in the ê 1 direction yields mẍ = 0 mg sin φ, so that solving for the acceleration ẍ = 0 m g sin φ ẍ = 1.5 m/s. b Likewise, from the equations of motion the tension T can be determined to be T = 0 T = 5 N. Problem 17: A projectile is launched with initial speed v at an angle of inclination of φ = 30. ind the range of v P initial speeds so that the particle striks the wall φ of height h = 3 m that is located a distance d = ψ O 5 m down the surface that is inclined ψ = 15. h ĵ ê î d If the position of the projectile is measured as r P O = xtî + ytĵ, then its velocity is v P = ẋî + ẏ ĵ, so that ẋ0 = v 0 cos φ, ẏ0 = v 0 sin φ, ê 1 where v 0 is the initial speed of the particle. urther, the acceleration is a P = ẍî + ÿ ĵ and since the only force that acts on the particle is due to gravity, the velocity and position of the particle may be determined to be ẍt = 0, ÿt = g, ẋt = v 0 cos φ, ẏt = g t + v 0 sin φ, xt = v 0 cos φ t, yt = g t + v 0 sin φ t. Therefore, the displacement in the ĵ direction may be expressed in terms of x as yx = g x + sin φ v 0 cos φ cos φ x. Solving for v 0, the initial speed necessary for the particle to pass through a position r P0 = xî+y ĵ yields g v 0 = x cos φ x sin φ y cos φ. The final position of the particle as it hits the wall is r P O t f = d ê 1 + z ĵ = d cos ψ î + z d sin ψ ĵ, 17

18 where at the base of the wall z = 0 and at the top of the wall z = h. Therefore the initial speed required to hit the wall at height z can be written as g v 0 = d cos ψ cos φ d sinφ + ψ z cos φ. inally, substituting numerical values into this expression, the minimum velocity required to hit the wall occurs for z = 0, and is v 0,min = m/s. Likewise, the maximum velocity required to hit the wall occurs for z = h = 3 m, and is v 0,max = m/s. Thus the range of initial speeds required to strike the wall is m/s = v 0,min v 0 v 0,max = m/s. Problem 18: As illustrated in the figure, the projectile, subject to only the influence of gravity, is launched with initial speed v 0 at an angle θ 0. or what initial inclinations will the projectile pass through the opening of width d at a minimum height h and a distance l from the initial location? Assume that g = 10 m/s and l = 10 m, h = 3 m, d = 0.5 m, v 0 = 3.81 m/s. v 0,θ 0 d h l Problem 19: A block of mass m A is initially at rest on a rough surface with coefficient of friction µ. A particle of mass m B is moving with velocity v B î, strikes the block and is embedded, so that the coefficient of restitution is e = 0. ind the distance the combined block and particle slide after the impact before coming to rest. v A,0 B 18

19 Problem 0: A pendulum with of mass m and length l is released horizontally from rest. As this mass swings through an angle φ it impacts a massless spring with stiffness k. a What is the maximum compression of the spring? m l φ g b If the mass remains attached to the spring, what is the amplitude of the resulting oscillations? k φ φ φ δ a As we are interested in the state of the system at some final time given the initial state, we formulate the problem in terms of work and energy. Moreover, since all external forces are conservative we use conservation of energy. We measure the compression of the spring as δ as shown in the figure above. Therefore the potential energy of the system can be written as V = mg h + 1 k δ, where h is the height of the mass above the ground. In this problem the system is released from rest and at the system is again at rest at the maximum compression of the spring. Therefore both the initial and final kinetic energy is zero. 19

20 The initial height and spring displacement are h i = 0, δ i = 0, while the final height and displacement are h f = l sinφ δ max cos φ, δ f = δ max. l sin φ Therefore conservation of energy yields T i + V i = T f + V f, 0 = mg l sin φ + δ max cos φ + 1 k δ max. δ max cos φ inally, solving for δ max yields two solutions mg mg cos φ cos φ δ max = ± k k mg l sin φ. k Therefore the maximum compression of the spring is the largest of these two, or mg mg cos φ cos φ mg l sin φ δ max = +. k k k b The second solution to the above quadratic equation physically corresponds to the extension of the spring during the rebound, assuming the mass remains attached to the spring. Thus the amplitude A of the resulting oscillations is half the distance between these two extrema, or mg cos φ A = k mg l sinφ. k Problem 1: The point mass m shown to the right is supported on the rotating bar by a spring with unstretched length l and stiffness k. If the bar is rotating at constant angular speed θ = ω: ĵ î d ê θ ê r a find the acceleration of the mass relative to the bar in the radial direction as a function of rt, the radial displacement; rt l b when the mass is in equilibrium relative to the bar in the radial direction, what is the steady-state compression of the spring? θ δ a In addition to the coordinates rt and θt as shown in the figure, we also identify the stretch in the spring as δt, related to zt as δt = d l rt. 0

21 The directions ê r and ê θ are radial and tangent to the rotating bar. Neglecting gravity, a free-body diagram for this system is shown to the right. The force N ê θ represents the force from the bar applied to the point mass while the force from the spring is k δtê r. inally, in terms of the coordinates rt and θt, the acceleration of the mass can be written as k δtê r a G = rt rtω ê r + ṙtω ê θ, where we have made use of the constant angular velocity of the bar, so that θt = 0. Ntê θ Applying momentum balance to this system = k δt êr + Nt ê θ = m rt rtω ê r + ṙtω ê θ = m a G. Thus, taking the component in the ê r direction k δt = m rt rtω. inally, solving for the acceleration along the bar, we obtain rt = k m d l rt + rtω, = kd l m k mω rt. m b The equilibrium position of the mass relative to the bar implies that rt 0, so that the above equation can be solved to yield k d l r eq = k mω. Problem : The two blocks shown to the right, each of mass m = kg, rest on a rough surface with coefficient of friction µ = 0.5. The initial velocity of the left block is v î, with v = 1.50 m/s while the second block, located a distance d = 0.5 m away is initially at rest. If the coefficient of restitution between the blocks is e = 0.80, find the distance the right block slides after the impact before coming to rest. vî m A d m B µ The motion of this system can be divided into three phases. In the first, the block A slides to the right. The second phase describes the impact between blocks A and B, while the final phase describes the motion of block B as it comes to rest. The first and third phase can be described using the work-energy formulation while the collision is best described using an impulse-momentum formulation. 1

22 Because the blocks are identical, a free-body diagram of either block is shown to the right. Since the motion is only in the î direction, only the force in the î direction contributes to the work. The work-energy formulation for either sliding block yields xt 0 µmg î dxî = m ẋ t ẋ 0, µmg î m N î so that the velocity of block A at the impact, ẋ A,1 is determined from µmg d = m ẋ A,1 v. Solving for the velocity of block A just before the impact, ẋ A,1 ẋ A,1 = v µg d. Across the impact conservation of momentum holds m A ẋ A,1 + m B ẋ B,1 = m A ẋ A, + m B ẋ B,, ẋ A,1 = ẋ A, + ẋ B,. Together with the definition of the coefficient of restitution e = ẋb, ẋ A, ẋ A,1 ẋ B,1, eẋ A,1 = ẋ B, ẋ A,. Therefore, solving for the velocity of block B after the collision, ẋ B, ẋ B, = 1 + e ẋ A,1. Over the final interval of the motion we again use a work-energy formulation for block B. If the distance traveled by block B before coming to rest is δ, then ẋ B,3 = 0 and using the relation obtained above µmg δ = m ẋ B,, and solving for δ δ = = = 1 µg ẋ B,, 1 + e ẋ 8µg A,1, 1 + e v µg d = 8µg 1 + e 4 v µg d. inally, with the supplied values, we find that block B slides a distance δ = 0.17 m before coming to rest.

23 Problem 3: In the system shown to the right, the two masses m = 1 kg are connected by a cable wrapped around a massless pulley. If the system is released from rest, with the spring k = 50 N/m compressed by a distance d = 0.15 m, from its unstretched position: a find the velocity of the mass on the right when the spring is unstretched; g b what is the maximum velocity of the mass on the left? x m k 4m x ĵ î We choose a work-energy formulation to describe the motion of this system. Specifically, all forces are conservative so that the total energy of the system is conserved T i + V i = T f + V f, where T and V are the kinetic and potential energies of the system. The displacement of each mass relative to its position when the spring is unstretched is described by x, so that the kinetic and potential energies can be written as T = 1 mẋ + 1 4mẋ = 1 5mẋ, V = 1 k x + mg x 4mg x = 1 k x 3mg x. Notice that the potential energy has been taken relative to the configuration of the system when the spring is unstretched. or an initial compression in the spring of d, so that x i = d, conservation of energy yields T i + V i = T f + V f, k d + 3mg d = 5m ẋ + k x 3mg x. a When the spring is unstretched, x f = 0 and solving the above for ẋ f yields k d + 6mg d ẋ f =. 5m Using the provided parameter values, ẋ f = 1.70 m/s. 3

24 b The maximum kinetic energy of the system occurs when the potential energy is minimized. Thus finding the location x = x m at which this occurs, V min dv dx = k x m 3mg = 0, x m = 3mg k. Therefore, at this position the velocity of the masses can be determined to be k d + 6mgk d + 9mg ẋ m =, 5k m = k d + 3mg 5k m Using the numerical values for this problem, ẋ m = 1.89 m/s. Problem 4: A particle of mass m moves along a frictionless circular bar of radius r, and is subject to gravity and a horizontal force î. a What is the angular acceleration θ of the particle? ĵ î î ê θ ê r b ind the magnitude of the force that the bar applies to the particle. θ Note that your answers may or may not depend on θ and θ. a With the coordinate θ as defined above, the acceleration of the mass P can be written as a P = r θ ê r + r θ ê θ. N ê r A free-body diagram for the mass is shown to the right. In addition to the gravitational force and the applied force, a normal force in the ê r direction represents the interaction between the frictionless hoop and the mass. Therefore the equations of motion can be written as î = m a P, î mg ĵ + N ê r = m r θ ê r + r θ ê θ. mg ĵ Relating the î,ĵ directions to the ê r,ê θ directions yields î = cos θ ê r sin θ ê θ, ĵ = sinθ ê r + cos θ ê θ, 4

25 Therefore, to determine the angular acceleration of the particle, we take the components of the this equation in the ê θ direction, which eliminates the unknown normal force N. This becomes sinθ mg cos θ = mr θ. inally, solving for θ θ = sinθ + mg cos θ. mr b The force between the bar and the particle, described as N unive r can be determined from the equations of motion in the ê r direction as cos θ mg sin θ + N = mr θ, so that solving for N yields N = mg sin θ cos θ mr θ. Problem 5: The two blocks of mass m are connected by a cable. The left surface is rough with a coefficient of friction µ = 0.5 while the right surface is smooth µ right = 0. The left surface is inclined at an angle φ and the angle between the surfaces is 90. x ĵ î a for what range of the angle φ do the blocks slip? b assuming the blocks do slip, what are their accelerations? µ G l φ x G r ĵ µ right = 0 î As shown in the figure above, we measure the displacment of each block by the coordiante xt and identify the directions î,ĵ aligned with the surface. Therefore the acceleration of the block resting on the rough surface is a Gl = ẍî, while the acceleration of the block on the smooth surface is a Gr = ẍĵ. 5

26 ree-body diagrams for each block are shown to the right. Notice that a force due to friction acts only the block resting on the rough surface, although both blocks are subject to forces normal to their respective surfaces. On the left block the equations of motion can therefore be written as l = m a Gl, mg ĵ T î f l î T î + f l î + N l ĵ mg ĵ = mẍî, while on the right block r = m a Gr, N l ĵ T ĵ T ĵ + N r î mg ĵ = m ẍĵ. The direction of the gravitational force ĵ can be written in terms of î,ĵ as ĵ = sinφî + cos φ ĵ. N r î mg ĵ inally, taking components of these equations in terms of î,ĵ, we obtain the four scalar equations T + f l mg sin φ = m ẍ, N l mg cos φ = 0, N r mg sinφ = 0, T mg cos φ = m ẍ. We can immediately solve for the unknown normal forces acting on the two blocks as N l = mg cos θ, N r = mg sin θ. Eliminating the unknown tension yields mẍ = mg cos φ sinφ + f l, where f l is the as yet unknown friction force acting on the right block. If the blocks slip ẋ 0 and f l,slip = µn l sgnẋ = µmg cos φ sgnẋ, while if the blocks are stationary then ẋ = 0 and f l µn l = µmg cos φ. a The blocks slip provided the friction force f l is unable to maintain static equilibrium, which implies that for equilibrium f l,eq = mg cos φ sinφ µmg cos φ, or µmg cos φ mg cos φ sin φ µmg cos φ. 6

27 This can be solved for φ so that static equilibrium requires 1 µ tan φ 1 + µ. Therefore, the system slides for φ outside this interval, which for these parameter values yields φ < 36.9, and 51.3 < φ < 90. b If the blocks slip then the friction force is f l = µmg cos φ sgnẋ, so that the ẍ may be determined to be ẍ = g 1 µ sgnẋ cos φ sin φ. Problem 6: As illustrated in the figure, the projectile, subject to only the influence of gravity, is launched with initial speed v 0, an angle θ 0 = 60, and a distance l = 10 m from the edge. or what range of initial speeds will the mass will reach the lower platform? v 0 θ 0 r P O = xtî + ytĵ ĵ l l î l a As shown in the figure, the position of the projectile P with respect to the firing point O is r P O = xtî+ytĵ, so that xt measures the horizontal distance travelled while yt measures the vertical displacement of P. The initial position of the projectile is r P O 0 = x0î+y0ĵ = 0, while the initial velocity is v P 0 = ẋ0î + ẏ0ĵ = v 0 cos θ î + v 0 sinθ ĵ. The acceleration of the projectile is then determined to be a P = ẍî + ÿ ĵ. A suitable free-body diagram for the projectile is shown to the right, where only the force due to gravity acts on P. Therefore the equations of motion can be written as = m a P, mg ĵ = mẍî + ÿ ĵ. or taking components in the î and ĵ directions mg ĵ ẍ = 0, ÿ = g. 7

28 Using the initial conditions x0 = 0, ẋ0 = v 0 cos θ, y0 = 0, ẏ0 = v 0 sin θ, these equations may be solved to yield xt = v 0 cos θ t, yt = g t + v 0 sin θ t. Notice that we can eliminate t from these two equations to directly relate x and y as g y = v 0 cos θ x + sin θ v cos θ x = 0 sinθ cos θ g x x v0. cos θ At the minimum initial speed v 0 = v min the projectile just clears upper edge at x,y = l,0, so that v 0 = min sin θ cos θ g l l vmin. cos θ Solving for v min vmin = g l sin θ cos θ = g l sinθ. Likewise, at the maximum initial speed the projectile reaches the lower edge, which is located at x, y = 3 l/, l/, so that l v = max sin θ cos θ 3 g l 3 l vmax cos, θ and solving for v max 9g l v max = 43 cos θ sin θ + cos θ = 9g l 1 + cos θ + 3sin θ. inally, the mass reaches the lower platform for v min < v 0 < v max, or, with the values given above m/s < v 0 < m/s. Problem 7: As illustrated in the figure, the projectile, subject to only the influence of gravity, is launched with initial speed v 0 at an angle θ 0. The surface is inclined at an angle φ. or what initial velocity v 0 will the range up the surface of the projectile be d? v 0,θ 0 φ 8

29 Problem 8: Blocks A and B identical of mass m = kg rest on a rough surface, with coefficient of restitution µ = 0.40, inclined at an angle of φ = 0. Block A has an initial speed of v A,0 = 4 m/s up the surface, while block B is initially at rest, a distance d = 0.50 m up the surface. The coefficient of restitution between blocks A and B is e = ind the distance up the surface traveled by block B before it comes to rest hint: use work-energy to describe the motion of the blocks and conservation of momentum to describe the collision. v A,0 A d B Problem 9: In the system shown to the right, the plunger mass m 1 = kg is pulled back to compress the spring stiffness k by a distance δ = 5 cm and released from rest. As the plunger returns to its unstretched position, it impacts a ball of mass m = 6 kg with a coefficient of restitution e = /3. inally, the plane of motion is inclined at an angle φ = 30 and is assumed to be frictionless. a If the ball travels a distance l = 1 m, find the stiffness of the spring k; k m 1 δ φ m b Using the stiffness that you determined above, find the required initial compression of the spring so that the ball travels a distance d = 0.5 m up the plane. The response of this system can be broken down into three phases. The first describes the return of the plunger from its initial compressed position at time t = t 0, back to the point an instant before it contacts the ball at time t = t 1. The second phase describes the instantaneous collision between the plunger and ball, and spans from t = t 1 to t = t+ 1. In the final phase of the motion the ball moves up the inclined plane, reaching its maximum travel at time t = t. Over the first and third phase the system is conservative and we apply conservation of energy to relate the response at the beginning and end of each phase. The middle phase involves a collision, during which we apply conservation of momentum and the given coefficient of restitution. During the initial phase of the motion the total energy of the system is conserved, so that T 0 + V 0 = T 1 + V 1, with: T 0 = 0, T 1 = 1 m 1 v 1,1, V 0 = 1 k δ m 1 g δ sinφ, V 1 = 0. In this, v 1,1 represents the velocity of the plunger just before impact. Solving for this velocity in terms of δ we obtain: k v 1,1 = δ δ g sinφ. m 1 9

30 During the second phase of motion the linear momentum of the system is conserved, so that: m 1 v 1,1 + m v,1 = m 1 v 1,1 + + m v,1 +, where v,1 represents the velocity of the ball just before impact, equal to zero, while v 1,1 + and v,1 + described the velocities of the plunger and ball respectively just after the impact. This, together with the coefficient of restitution, defined as: e = v,1 + v 1,1 +, v 1,1 v,1 can be used to solve for the post-collision velocity of the ball, yielding: v,1 + = m 1 m 1 + m 1 + e v 1,1, where v 1,1 is known from the previous phase of the motion. In the final phase the total energy of the ball is again conserved, so that T V 1 + = T + V. Note that in this phase we only consider the motion of the ball. Although the total energy of the plunger is also constant, the motion of the plunger no longer affects the dynamics of the ball. At the final state t = t, the velocity is constant and the ball moves up the plane a distance z, so that: T 1 + = 1 m v,1 +, T = 0, V 1 + = 0 V = m g z sin φ. Therefore, solving for z in terms of the post-collision velocity we find that: z = v,1 + g sin φ. inally, we can relate the initial compression of the spring to the distance the ball travels, which yields: z = 1 + e m 1δ kδ m 1 + m m 1 g sin φ 1. 1 a Using Eq. 1, with z = l, we can solve for k as: k = m 1g sinφ δ m1 + m l 1 + e m 1 δ + 1. or the numerical values given in the problem statement, we find that k = N/m b If the ball moves up the plane a distance d, then the initial compression x c can be determined from Eq. 1. Substituting in the value of k determined above, this reduces to: d = l x c d δ δ e m 1δ x c x c δ m 1 + m, δ which can be written as a quadratic equation in x c : [ ] 0 = lm 1 + m e m 1δ x c [ 1 + e m 1δ ] x c [ m 1 + m dδ ]. 30

31 With the parameter values given, this can be solved for x c as: x c = m. Problem 30: A block of mass m slides down a rough inclined plane, with inclination φ and unknown coefficient of friction µ. The block is released with initial speed v 0 down the plane and after it has traveled a distance d the speed is measured to be v 1. a ind the coefficient of friction µ. b How far does the block slide before it comes to rest? v 0 µ ĵ ê φ d v 1 î ê 1 We can use the work-energy formulation to relate the motion of the block between times t = t 0 and t = t 1. or this systemm a free-body diagram is shown to the right. In addition, relating ĵ to the directions normal and tangential to the surface: ĵ = sin φ ê 1 + cos φ ê. Therefore, linear momentum balance provides: = m a G, f r + mg sinφ ê 1 + N mg cos φ ê = v ê 1. N ê mg ĵ f r ê 1 Becuase there is no acceleration in the ê direction, the forces must balance in this direction, so that: N = mg cos φ. Therefore, assuming the block slides down the surface as given in the problem statement, the friction force is: f r ê 1 = µmg cos φ ê 1. Applying the equations of work-energy to this system as it slides a distance x obtaining a velocity v: x 0 { } f r + mg sinφ ê 1 + N mg cos φ ê ds ê 1 = m mg µ cos φ + sinφ x = m v v0, v v0. 31

32 a If, after sliding a distance d the speed of the block is v 1, we can solve the above equation for µ as: µ = v 0 v1 + gd sin φ. gd cos φ b The total distance the block slides, say x = x f is reached as the final velocity vanishes, so that v = 0. Again the above equation can be solved to give: x f = v 0. g µ cos φ sinφ Using the value of µ found in the part a, this can be written as: d x f =. v 1 1 v 0 Problem 31: As illustrated in the figure, the projectile, subject to only the influence of gravity, is launched with initial speed v 0 at an angle θ 0. or what range of distances l will the projectile clear the barrier of height d? v 0 ĵ î θ 0 l We identify the origin as O and the projectile as P, with the position of the projectile measured as: r P O = xt î + zt ĵ, and the equations of motion for this system reduce to: ẍ = 0, z = g. Upon integrating with respect to time, these reduce to: Using the initial conditions: ẍt = 0, ẋt = ẋ0, xt = ẋ0 t + x0, zt = g, żt = g t + ż0, zt = g t + ż0 t + z0. x0 = 0, z0 = 0, ẋ0 = v 0 cos θ, ż0 = v 0 sinθ, The position of the projectile reduces to: r P O = v 0 cos θ t î + 3 g t + v 0 sinθ t ĵ.

33 rom this, we may solve for the time t where zt = d as: t = v 0 sinθ ± v 0 sin θ g d. g Therefore there exist two times for which the projectile is at height d, at which time the horizontal displacement is found to be: xt = v 0 cos θ t = v 0 cos θ g v 0 sinθ ± v 0 sin θ g d. So, for: v 0 cos θ g v 0 sin θ v 0 sin θ g d l v 0 cos θ g v 0 sin θ + v 0 sinθ g d, the projectile clears the barrier. Problem 3: A girl throws a ball with an initial speed v 0 at an angle of θ = 36.9 = sin from a height of d = 1.5 m. ind the initial speed so that the ball travels l = 10 m before striking the ground. we choose to describe the position of the ball using the coordinates x,z which are measure the displacement of the ball in the î and ˆk directions respectively. The initial conditions on the motion of the ball can be written as r BO 0 = x0î + z0 ˆk = 0î + dˆk, v B = ẋ0î + ż0 ˆk = v 0 cos θ î + sin θ ˆk, while the final conditions, which occur when the ball strikes the ground at, say, some unknown time t f, are r BO t f = xt f î + zt f ˆk = l î + 0 ˆk. Although we may now use our expressions for the motion of particles in free-fall, it is more instructive to develop these equations from Newton s laws. The only force acting on the particle is due to gravity, mgˆk, and the acceleration is written as a B = ẍî + zˆk. Thus we write the force-acceleration relationship as = m a P, mgˆk = mẍî + z ˆk, or, taking components in the î and ˆk directions ẍt = 0, zt = g, ẋt = ẋ0, żt = gt + ż0, xt = ẋ0t + x0, zt = g t + ż0t + z0. 33

34 inally, substituting in our initial and final conditions ẋt = v 0 cos θ, żt = gt + v 0 sin θ, xt = v 0 cos θt, zt = g t + v 0 sin θt + d, xt f = v 0 cos θt f = l, zt f = g t f + v 0 sin θt f + d = 0. rom the two equations governing the position at time t f, we find that t f = l v 0 cos θ, or substituting into the equation for zt f 0 = g l l + v 0 sin θ v 0 cos θ v 0 cos θ + d, which can be solved for v 0 to yield gl v 0 = = 9.3 m/s cos θl sinθ + d cos θ Problem 33: A crate of mass m is being carried on the back of the flat-bed truck. The coefficient of friction between the crate and the bed is µ while the gravitational acceleration is g. In addition, the truck is traveling on a hill inclined at an angle of φ. a ind the maximum deceleration so that the crate does not hit the cab of the truck; b What is the maximum angle φ = φ max of the hill so that the crate does not slide forward when the truck is traveling at constant velocity? a Let us assume a basis with î directed along the path of the truck, and ˆk normal to the surface, so that x measures the position of the crate with respect to the origin. In addition, we let y represent the position of the truck with respect to the origin. Note that î is at an inclination φ with respect to horizontal. Clearly, the acceleration of the crate is simply a C = ẍ î. The acceleration of the truck, which is assumed known, can be expressed as a T = ÿ î = aî. Three forces act upon the crate: the gravitational force, the normal force, and the frictional force. As a side note, these three types of forces also act on the truck. Although when considering the truck which we are not in this problem, only the crate you might be tempted to include the force that is stopping the truck from the brakes. However, in reality this is precisely the role of the frictional force. After all, brakes stop your wheels, but your wheels stop your car. Application of Newton s second law to the crate yields: = m a C, mg sinφ fî + N mg cos φˆk = mẍî. 34

35 We immediately see that: N = mg cos φ, f = mg sin φ ẍ b The crate will not hit the cab of the truck if and only if it remains stationary with respect to the truck, so that xt = yt and therefore ẍ = ÿ = a. Therefore, the frictional force is sufficiently large enough to overcome the decceleration and the component of gravity along the bed. However, f µn. Therefore: f = mg sinφ ẍ µn, mg sin φ a µmg cos φ, mgsinφ µcos φ a. Thus the decceleration which is a must be less than mgµcos φ sinφ. c Again, we require that the relative velocity between the crate and the truck be zero, so that, in this situation ẍ = 0. As above: f = mg sin φ ẍ µn, mg sin φ µmg cos φ, tan φ µ, or finally φ arctan µ. Problem 34: A mass m, under the influence of gravity, is swinging around a vertical axis at the end of a massless string of length l. The angular speed of the mass is given as ω = constant. If the mass is supported by a frictionless platform that is located at a height h below the support for the string: a find the tension in the string in terms of ω, m, and l; b what is the maximum value of ω for fixed m and l so that the mass remains in contact with the table? Hint: use ˆk as well as the ê r and ê θ directions. We choose cylindrical coordinates ê r,ê θ, ˆk, where position of the mass with respect to the center of the platform, say C, is r P C = rê r. ˆk is directed upwards, so that, if O is the point of support for the string r P O = l h ê r hˆk. As a result, with constant angular speed ω, the acceleration of the mass is a P = ω l h ê r. 35

36 The total force acting on the mass is composed of the normal force, the force due to gravity, and the tension in the string = T cos θêr + N + T sin θ mg ˆk, where, assuming that the mass is on the table, sin θ = h l. So, Newton s second law can be written as = m a P, l h T ê r + N + T hl l mg ˆk = m ω l h ê r, or, taking components in the ê r and ˆk directions T l h l = mω l h, N + T h l mg = 0. a The tension in the string is found by solveing the above equations for T, to yield T = mlω. b inally, the normal force required to keep the ball on the table is N = mg T h l = m g hω, So, if N < 0, this implies that the mass will lift off the table and swing freely. The critical angular speed is obtained by setting N = 0, to yield g ω cr = h. Interestingly, the tension is independent of h and the critical speed is independent of l. Problem 35: A cannonball is fired with an initial speed of 50 m/s. ind the maximum distance the ball can travel if it s height never exceeds 5 m. Problem 36: The upper block with mass m sits on a rough surface with coefficient of friction µ. If the block suspended by the cable is assumed to have mass αm, with α > 1, find: a the acceleration of the upper block; b the minimum value of α, defined as α 0, so that the suspended block falls; c what happens if α < α 0? 36

37 Problem 37: In the pulley system shown, the mass of each block is identical, say m. If a constant force pushes up on block B, find the speed of each block after B has moved down a distance l assume that < mg so that motion occurs in the downward direction if the system starts from rest. a Let x measure the position of A positive to the left and y denote the position of B positive down. Because we assume that all forces are conservative i.e. no friction, the total energy of the system is conserved throughout the motion, so that T 1 + V 1 = T + V. In addition, we assume that the potential energy at state 1 is zero the potential energy datum is located at the initial state. Consequently: T 1 = 0, V 1 = 0, T = m ẋ + ẏ, V = mgy, inally, we see that the string enforces the following constraint: x + y = constant, ẋ + ẏ = 0. So E = 0 provides: 0 = m ẏ + ẏ + mgy. So solving for ẋ and ẏ, after block B has fallen a distance l, the speed of each block is: ẋ = g l, 5 m 8 ẏ = g l. 5 m Problem 38: Two blocks, each of mass m, are connected by a spring with spring constant k. If the blocks are placed on a smooth surface, the spring is initially stretched by a distance d, and released from rest, find: a the speed of each block when the spring is unstretched; b the maximum compression in the spring. 37

38 We combine both conservation of energy and conservation of momentum to approach this problem. Clearly, the system is conservative so that E = 0, and because no external forces act in the horzontal direction, we have L î = 0, where î is directed along the ground. Let x 1 and x denote the position of masses 1 and, respectively, as measured from some origin fixerd in the ground. Thus: T = m ẋ 1 + ẋ, V = k x x 1 l 0, where l 0 is the unstretched length of the spring, so that x x 1 l 0 represents the stretch in the spring. Considering the linear momentum of the system, which is conserved: L = mẋ 1 î + mẋ î, so that, because the system starts from rest, L = 0. As a result, ẋ 1 = ẋ, and conservation of energy provides: k d = m ẋ 1 + ẋ k + x x 1 l 0, k d = mẋ 1 + k x x 1 l 0 a If the spring is uncompressed, x x 1 l 0 = 0, so that: k d = mẋ 1, or: ẋ 1 = k m d. b When the spring is at maximum compression, the relative velocity between the two masses is zero, so that: k d = k x x 1 l 0, or the maximum compression is simply d, identically equal to the initial stretch. Problem 39: A massless rod of length l is connected to a particle with mass m, and is released from rest in a horizontal position. ind the velocity of the particle with mass m, which is initially at rest on the ground, after the collision. Assume that the coefficient of restitution is ε = 0.8 and gravity is the only external force. Consider two regimes of the motion: falling of the pendulum, in which the energy is conserved, and impact, which conserves linear momentum. Thus, in phase one, E = 0, so that T 1 + V 1 = 38