os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e D. Gross • W. Hauger J. Schröder • W.A. Wall S. Govindjee Engineering Mechanics 3 Dynamics Solutions to Supplementary Problems Gr The numbers of the problems and the figures correspond to the numbers in the textbook Gross et al., Engineering Mechanics 3, Dynamics, 2nd Edition, Springer 2013 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e Gr Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e Gr Chapter 1 1 E1.17 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 A point P moves on a given path from A to B (Fig. 1.42). Its velocity v decreases linearly with arc-length s from a value v0 at A to zero at B. How long does it take P to reach point B? Example 1.17 s=l B v v v0 P s A l s Fig. 1.42 Solution First we write down the velocity v as a linear function of the arc-length s: s v(s) = v0 1 − . l Separation of variables ds =v dt → v0 ds = dt 1 − sl and indeterminate integration lead to Z Z v0 t ds = v dt → s = l 1 − C exp − . 0 1 − sl l The integration constant C is determined from the initial condition: v0 t s(0) = 0 : 0 = 1 − C → s = l 1 − exp − . l Point B is reached when s = l. This yields the corresponding time tB : Gr s(tB ) = l → tB → ∞ . 3 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.18 A radar screen tracks a rocket which rises vertically with a constant acceleration a (Fig. 1.43). The rocket is launched at t = 0. Determine the angular velocity ϕ̇ and the angular acceleration ϕ̈ of the radar screen. Calculate the maximum angular velocity ϕ̇ and the corresponding angle ϕ. ϕ 111111111111 000000000000 l Fig. 1.43 Solution Since the acceleration a is constant, the velocity v and the position x of the rocket are x v = at + v(0) , x = at2 /2 + v(0)t + x(0) . Applying the initial conditions yields v(0) = 0 → x(0) = 0 → v = at, at2 x= . 2 The angle ϕ of the radar screen follows as 2 x at tan ϕ = → ϕ(t) = arctan . l 2l Differentiation leads to the angular velocity " 2 2 # at at ϕ̇(t) = / 1 + l 2l Gr and the angular acceleration " 2 2 #2 a 3a3 t4 at ϕ̈(t) = − / 1+ . l 4l3 2l ϕ E1.18 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 The time t∗ of the maximum angular velocity ϕ̇max is obtained from 2 1/4 4l ϕ̈(t∗ ) = 0 → t∗ = . 3a2 Thus, ϕ̇max = ϕ̇(t∗ ) → ϕ̇max s √ 3 3a = 8l and Gr √ ϕ(t∗ ) = arctan(1/ 3) → ϕ(t∗ ) = 30◦ . 5 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.19 Two point masses P1 and P2 start at point A with zero initial velocities and travel on a circular path. P1 moves with a uniform tangential acceleration at1 and P2 moves with a given uniform angular velocity ω2 . a) What value must at1 have in order for the two masses P1 r to meet at point B? b) What is the angular velociB A ty of P1 at B? c) What are the normal acceP2 lerations of the two masses Fig. 1.44 at B? Solution The velocity and position of P1 follow from at1 = const = v̇ with the initial conditions s01 = 0 and v01 = 0 as v1 = at1 t , s1 = 1 at1 t2 . 2 Similarly, for point P2 we obtain from at2 = 0 with s02 = 0 and v02 = rω2 : v2 = rω2 , s2 = rω2 t . a) Both points meet at time tB at point B. Thus, πr = 1 at1 t2B , 2 πr = rω2 tB . This leads to Gr tB = π ω2 → at1 = 2πr 2rω22 . = 2 tB π E1.19 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 6 b) The tangential acceleration at1 and the time tB are now known. Hence, we can calculate the angular velocity of P1 at B: v1 (tB ) = at1 tB = rω1 (tB ) → ω1 (tB ) = 2rω22 π at1 tB = = 2ω2 . r πrω2 c) The normal accelerations at B follow from an = rω 2 : Gr an1 = rω12 (tB ) = 4rω22 , an2 = rω22 . 7 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.20 A child of mass m jumps up and down on a trampo- line in a periodic manner. The child’s jumping velocity (upwards upon leaving the trampoline) is v0 and during the contact time ∆t the contact force F (t) has a triangular form. Find the necessary contact force amplitude F0 and the jumping period T0 . F F0 z t T0 ∆t Fig. 1.45 Solution The child is subjected to its constant weight W = mg and, during the contact with the trampoline, to the contact force F (t). While the child is in the air, the equation of motion is given by F F0 mg F 0 t1 t2 t T0 ↑: mz̈ = −mg. Integration between t = t0 = 0 (end of a contact) and t = t1 (beginning of a new contact) yields mv1 − mv0 = −mgt1 , where v1 = v(t1 ) and v0 = v(0). Recall that the velocity at the beginning of a vertical motion and the velocity at the end of a free fall of a body are equal in magnitude: v1 = −v0 (note the different signs). Therefore, we obtain Gr 2v0 = gt1 → t1 = 2 v0 . g E1.20 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 8 The jumping period T0 follows with ∆t = t2 − t1 as T0 = t1 + ∆t → T0 = 2v0 + ∆t . g We now apply the Impulse Law between time t0 and time t2 (see the figure): 1 ↑ : m v2 − m v0 = −mgT0 + F0 (t2 − t1 ) . 2 Gr In order to have a periodic process, the velocities v2 and v0 have to coincide: v2 = v0 . With this condition the Impulse Law yields 2v 2T0 1 0 mg = 2 + 1 mg . −mgT0 + F0 ∆t = 0 → F0 = 2 ∆t g∆t 9 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.21 A car is travelling in a circular arc with radius R s and velocity v0 when it starts to brake. v0 If the tangential deceleration is at (v) = −(a0 + κv), where a0 and κ are given constants, find the time to brake tB , the stopping distance sB , and the normal acceleration an during the braking. R Fig. 1.46 Solution The acceleration at is given as a function of the velocity: at (v) = v̇ = −(a0 + κv) . We separate the variables and integrate (initial condition: v(0) = v0 ): Z v 1 a0 + κv0 dv̄ t(v) = − = ln . a + κv̄ κ a0 + κv v0 0 The time tB when the car comes to a stop follows from the condition v = 0: 1 κv0 tB = t(v = 0) = ln 1 + . κ a0 Now we determine the inverse function of t(v): i a0 + κv0 a0 h κv0 −κt eκt = → v(t) = 1+ e −1 . a0 + κv κ a0 Integration with the initial condition s(0) = 0 leads to the position Z t i κv0 a0 h s(t) = v(t̄)dt̄ = 2 1 + 1 − e−κt − κt . κ a0 0 Gr The stopping distance sB is obtained for t = tB : " ! # a0 κv0 1 κv0 sB = s(tB ) = 2 1 + 1− − ln 1 + 0 κ a0 a0 1 + κv a0 h i a0 κv0 κv0 = 2 − ln 1 + . κ a0 a0 E1.21 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 10 The normal acceleration an (t) during the breaking is found as 2 a20 v2 κv0 −κt = 1+ e −1 . an = R Rκ2 a0 Gr As a check on the correctness of the result we calculate an for t = tB and obtain an = 0. 11 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.22 A point P moves on 2 a parabola y = b(x/a) from A to B. Its position as a function of the time is given by the angle ϕ(t) = arctan ω0 t (see Fig. 1.47). Determine the magnitude of velocity v(t) of point P . How much time elapses until P reaches point B? Calculate its velocity at B. E1.22 y b B P r ϕ A a x Fig. 1.47 Solution The parabola y = b(x/a)2 and the angle ϕ(t) = arctan ω0 t are given. The angle ϕ can also be expressed as ϕ = arctan y/x. Thus, y/x = ω0 t. Solving for x and y yields the position of point P : x(t) = a2 ω0 t , b y(t) = a2 (ω0 t)2 . b To obtain the velocity of P , we differentiate: ẋ(t) = v= p a2 ω0 , b ẋ2 + ẏ 2 a2 2 ω t, b 0 a2 p v(t) = ω0 1 + 4ω02 t2 . b ẏ(t) = 2 → Point B is reached at time tB when x = a: x(tB ) = a → tB = b . aω0 Gr Thus, the velocity at B is obtained as s a2 b2 vB = v(tB ) → vB = ω0 1 + 4 2 . b a E1.23 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 12 Example 1.23 A rod with length l rota- tes about support A with angular position given by ϕ(t) = κ t2 . A body G slides along the rod with position r(t) = l(1 − κ t2 ). a) Find the magnitude of velocity and acceleration of G when ϕ = 45◦ . b) At what angle ϕ does G hit the support? Given: l = 2 m, κ = 0.2 s−2 . B l r G ϕ(t) A Fig. 1.48 Solution a) First we calculate the time t = t1 that it takes the rod to reach the position ϕ = ϕ1 = 45◦ : q ϕ1 = π/4 = κt21 → t1 = π/(4κ) = 1.98 s . Then we determine the derivatives of the given functions r(t) and ϕ(t): r = l(1 − κt2 ) , ṙ = −2κlt , r̈ = −2κl , ϕ = κt2 , ϕ̈ = 2κ . ϕ̇ = 2κt , vϕ With t = t1 we obtain the velocity vr = ṙ = −2κlt1 = −1.58 m/s , κt21 )2κt1 Gr vϕ = rϕ̇ = l(1 − = 0.34 m/s , q v = vr2 + vϕ2 = 1.62 m/s v vr π 4 13 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass and the acceleration ar = r̈ − rϕ̇2 = −2κl − l(1 − κt21 )4κ2 t21 = −1.07 m/s2 , aϕ = rϕ̈ + 2ṙ ϕ̇ = l(1 − κt21 )2κ − 2 · 2κlt1 2κt1 2 = −2.34 m/s , q a = a2r + a2ϕ = 2.57 m/s2 . ar π 4 b) Body G reaches the support (r = 0) at time tE : q r = 0 = l(1 − κt2E ) → tE = 1/κ = 2.24 s . This yields the corresponding angle ϕE : Gr ϕE = κt2E = 1 (= b 57.3◦) . aϕ a E1.24 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 14 Example 1.24 A mouse sits in a tower (with radius R) at point A and a cat sits at the center 0. If the mouse runs at a constant velocity vM along the tower wall and ϕ r the cat chases it in an Archimedian 0 spiral r(ϕ) = R ϕ/π, what must the H A R cat’s constant velocity vC be in order to catch the mouse just as the mouse reaches its escape hole H? At what Fig. 1.49 time does it catch the mouse? Solution We determine the components of the velocity of the cat from the given equation r(ϕ) = Rϕ/π of the Archimedian spiral: vr = ṙ = dr dϕ R = ϕ̇ dϕ dt π and vϕ = rϕ̇ = R ϕϕ̇ . π Thus, the constant velocity vC can be written as q R p R dϕ p 1 + ϕ2 . vC = vr2 + vϕ2 = ϕ̇ 1 + ϕ2 = π π dt Separation of variables and integration lead to vC Zt 0 R dt̄ = vC t = π Zϕ p 1 + ϕ̄2 dϕ̄ . 0 With the integral Z p i 1h p 1 + x2 dx = x 1 + x2 + arsinh x 2 Gr this results in R p vC t = ϕ 1 + ϕ2 + arcsinh ϕ . 2π 15 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass The time T when both the cat and the mouse reach the hole follows from the constant velocity vM of the mouse along the wall: πR = vM T → T = πR . vM Gr Introduction of T and ϕ(T ) = π into the expression for vC t yields the velocity of the cat: vM p vC = 2 π 1 + π 2 + arcsinh π = 0.62 vM . 2π E1.25 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 16 A soccer player kicks the ball (mass m) so that it leaves the ground at an angle α0 with an initial velocity v0 (Fig. 1.50). The air exerts a drag force Fd = k v on the ball; it acts in the direction opposite to the velocity. Determine the velocity v(t) of the ball. Calculate the horizontal component vH of v when the ball reaches the teammate at a distance l. Example 1.25 x z v0 α0 l Fig. 1.50 Solution The equations of motion are mẍ = −Fd cos α , mz̈ = mg + Fd sin α . We introduce the kinematic relations ẍ = dẋ , dt z̈ = x v dż dt α Fd z mg and the drag force Fd = kv. Noting that the components of the velocity are given by ẋ = v cos α , ż = −v sin α we obtain m dẋ = −k ẋ , dt m dż = mg − k ż . dt Separation of variables gives Gr dẋ k = − dt , ẋ m dż k mg = − m dt ż − k 17 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass and integration yields ẋ k ln =− t, C1 m ln mg k =−kt. C2 m ż − The constants of integration C1 and C2 can be determined from the initial conditions. For t = 0 the left-hand sides of the equations above have to be zero. Thus, the arguments of the ln-functions have to be equal to one (numerator and denominator have to be equal): C1 = ẋ(0) = v0 cos α0 , C2 = ż(0) − mg mg = −v0 sin α0 − . k k This leads to the components of the velocity: mg mg ẋ(t) = v0 cos α0 e−kt/m , ż(t) = − + v0 sin α0 e−kt/m . k k We now integrate the x-component to obtain the horizontal position of the ball: m x(t) = − v0 cos α0 e−kt/m + C . k Exploiting the initial condition m x(0) = 0 → C = v0 cos α0 k yields x(t) = m v0 cos α0 1 − e−kt/m . k The time t∗ when the ball reaches the teammate at the distance l is found to be m mv0 cos α0 x(t∗ ) = l → t∗ = ln . k mv0 cos α0 − kl This leads to the corresponding horizontal component of the velocity: Gr ẋ(t∗ ) = v0 cos α0 − kl . m E1.26 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 18 Example 1.26 A body with mass m starts at a height h at time t = 0 with an initial horizontal velocity v0 . If the wind resistance can be approximated by a horizontal force H = c0 m ẋ2 , at what time tB and location xB does it hit the ground? z m v0 h B x Fig. 1.51 Solution The equations of motion are given by → : mẍ = −mc0 ẋ2 , H z mg ↑ : mz̈ = −mg . x We integrate and apply the initial conditions x(0) = 0, z(0) = h, ẋ(0) = v0 , ż(0) = 0 to obtain Z ẋ Z t dx̄˙ 1 , = −c dt̄ → ẋ = 0 2 1 ˙ v0 x̄ 0 v0 + c0 t Z x Z t 1 1 dx̄ = dt̄ → x = ln(1 + c0 v0 t) , 1 c0 0 0 v0 + c0 t̄ g ż = −gt , z = − t2 + h . 2 The time tB when the body hits the ground follows from zB = 0: s 2h tB = . g Thus, the location of point B is obtained as s ! 2h 1 xB = x(t = tB ) = ln 1 + c0 v0 . c0 g Gr In the case of a vanishing wind resistance (c0 = 0) this result reduces to s s ! 2h 2h 1 xB = lim ln 1 + c0 v0 = v0 . c0 →0 c0 g g 19 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.27 A mass m slides on a E1.27 rotating frictionless and massless rod S such that it is pressed against a rough circular wall (coefficient of friction µ). If the mass starts in contact with the wall at a velocity v0 , how many rotations will it take for its velocity to drop to v0 /10? µ r S m g Fig. 1.52 Solution If we express the accelera- tion vector in terms of the SerretFrenet frame, then the equations of motion are written as տ : mat = −R , ւ : man = N . at an N R Note that the weight of the mass does not influence the motion since the motion takes place in a horizontal plane. If we use the kinematic relations at = v̇, an = v 2 /r, and the friction law R = µN we can write the first equation of motion in the form mv̇ = −µm v2 . r Separation of variables and integration yield Z v Z t dv̄ µ v0 . = − dt̄ → v(t) = µv 2 1 + r0 t v0 v̄ 0 r Gr We integrate again to obtain the distance traveled: Z t Z s dt̄ r µv0 ds̄ = v0 v0 t̄ → s(t) = µ ln (1 + r t) . 1 + µ 0 0 r 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 20 Now we calculate the time t1 that it takes for the velocity to drop to v0 /10: v0 v0 = µv 10 1 + r 0 t1 → t1 = 9r . µv0 The corresponding value s(t1 ) is found as s1 = s(t1 ) = r µv0 r ln (1 + t1 ) = ln 10 . µ r µ This yields the corresponding number of rotations: Gr n= s1 ln 10 = . 2πr 2πµ 21 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.28 A car (mass m) is traveling with constant velocity v along a banked circular curve (radius r, angle of slope α), see Fig. 1.53. The coefficient of static friction µ0 between the tires of the car and the surface of the road is given. r 0000000 1111111 0000000 1111111 0000000 1111111 1111111 0000000 0000000 1111111 m µ0 α Fig. 1.53 Determine the region of the allowable velocities so that sliding (down or up the slope) does not take place. Solution We model the car as a point mass and express the ac- celeration vector in terms of the Serret-Frenet frame. Then the equation of motion in the direction of the normal vector is (see the free-body diagram) man = N sin α + H cos α . 2.0 µ0 = 0.3 g en et r 1.5 v v 2 /gr M 1.0 0.5 mg r M en H 0 N π/6 π/3 α α Since the car does not move in the vertical direction we can apply the equilibrium condition 0 = N cos α − H sin α − mg . Gr ↑: E1.28 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 22 We now solve these two equations for the normal force N and the force of static friction H: H = man cos α − mg sin α , N = man sin α + mg cos α . The car does not slide if the condition of static friction |H| ≤ µ0 N is satisfied. With an = v 2 /r this leads to the allowable region of the velocity: tan α + µ0 v2 tan α − µ0 ≤ ≤ . 1 + µ0 tan α gr 1 − µ0 tan α Gr This is displayed for µ = 0.3 as a function of the angle α in the figure. 23 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.29 A car (mass m) has a velocity v0 at the beginning of a curve (Fig. 1.54). Then it slows down with constant tangential acceleration at = −a0 . The coefficient of static friction between the road and the tires is µ0 . s Calculate the velocity v v0 of the car as a function of the arc-length s. What is the m necessary radius of curvature ρ(s) ρ(s) of the road so that the M car does not slide? Fig. 1.54 Solution We model the car as a point mass. Its velocity v can be determined from the given constant tangential acceleration at through integration (initial condition v(0) = v0 ): v̇ = at = −a0 → v(t) = v0 − a0 t . The distance traveled follows from ṡ = v → s(t) = s0 + v0 t − a0 t2 /2 . With the initial condition s(0) = s0 = 0 we obtain s(t) = v0 t − a0 t2 /2 . Gr Elimination of the time t yields the velocity as a function of the arc-length: q v(s) = v02 − 2a0 s . E1.29 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 24 In order to determine the necessary radius of curvature ρ(s) we write down the equations of motion: mat = Ht → man = Hn → 0 = N − mg → Ht = −ma0 , v2 Hn = m , ρ N = mg . The static friction force H = (Ht2 +Hn2 )1/2 and the normal force N have to satisfy the condition of static friction to avoid sliding of the car: |H| ≤ µ0 N → a20 + v4 ≤ µ20 g 2 . ρ2 Solving for ρ yields v 2 − 2a0 s ρ(s) ≥ p0 2 . µ0 g 2 − a20 This condition can not be satisfied if a0 > µ0 g. mg et Ht Hn ̺ Gr en M N 25 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.30 A bowling ball (mass m) moves with constant ve- locity v0 on the frictionless return of a bowling alley. It is lifted on a circular path (radius r) to the height 2r at the end of the return. The upper part of the circular path has a frictionless guide of length r ϕG (Fig. 1.55). Given the angle ϕG , determine the velocity v0 such that the bowling ball reaches the upper level. g ϕG r v0 m Fig. 1.55 Solution The velocity v0 has to be large enough so that the bow- ling ball reaches the upper level (height 2r) with a velocity v ≥ 0. It is convenient to use the Conservation of Energy Law to determine the relation between v0 and v. With T0 = mv02 /2, V0 = 0, T1 = mv 2 /2 and V1 = 2mgr we obtain T0 + V0 = T1 + V1 → mv02 /2 + 0 = mv 2 /2 + 2mgr . This yields a first requirement on v0 : v02 = v 2 + 4gr → v02 ≥ 4gr . In addition, the velocity v0 has to be large enough so that the normal force between the bowling ball and the lower part of the circular path does not become zero before the ball reaches the guide of length rϕG . The necessary velocity v(ϕG ) follows from the equation of motion in the normal direction (note that ϕ is measured from the upper level): Gr ց: m v 2 (ϕ) = N + mg cos ϕ . r E1.30 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 26 The condition N ≥ 0 for ϕ = ϕG leads to v 2 (ϕG ) ≥ gr cos ϕG . In order to calculate the corresponding velocity v0 we use the Conservation of Energy Law between the lower level and the beginning of the guide: T0 + V0 = T2 + V2 mv02 /2 + 0 = mv 2 (ϕG )/2 + mgr(1 + cos ϕG ) . → Thus, we obtain the second requirement v02 ≥ (2 + 3 cos ϕG )gr . Both requirements are plotted in the figure, which shows that the minimum velocity v0 is obtained as ( (2 + 3 cos ϕG )gr for ϕG < ϕ∗G = arccos 2/3 , 2 v0 = 4gr for ϕG > ϕ∗G . 8 ϕ 6 N mg 4 v02 /gr 2+3 cos ϕG 2 ϕ∗G 0 0.2π 0.4π 0.6π 0.8π Gr ϕG π 27 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.31 A point mass m is E1.31 subjected to a central force F = mk 2 r, where k is a constant and r is the distance of the mass from the origin 0. At time t = 0 the mass is located at P0 and has velocity components vx = v0 and vy = 0. Find the trajectory of the mass. y v0 P0 m F r0 r α 0 x Fig. 1.56 Solution The equations of motion in the directions x and y are →: mẍ = −F cos α , ↑ : mÿ = −F sin α . With F = mk 2 r, x = r cos α and y = r sin α we obtain ẍ + k 2 x = 0 , ÿ + k 2 y = 0 . Both equations are equivalent to the differential equation that describes undamped free vibrations of a point mass (see Chapter 5). The solutions are x = A cos kt + B sin kt , y = C cos kt + D sin kt . The constants of integration are calculated from the initial conditions: → A=0, ẋ(0) = v0 → B= Gr x(0) = 0 y(0) = r0 v0 , ẏ(0) = 0 k → C = r0 , → D=0. 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 28 We now have a parametric representation of the curve describing the motion: v0 sin kt , y = r0 cos kt . x= k To eliminate the time, these equations are squared and then added. Thus, we finally obtain 2 2 x y + =1. v0 /k r0 Gr The point mass moves along an ellipse which reduces to a circle for k = v0 /r0 . 29 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.32 A centrifuge with radius r rotates with constant an- gular velocity ω0 . A mass m is to be placed at rest in the centrifuge and accelerated within a time t1 to the angular velocity ω0 . What will be the needed (constant) moment M acting on the mass? What is the power P of this moment? ω0 m r Fig. 1.57 Solution The centrifuge rotates with a constant angular velocity. Thus, the driving torque M needed to accelerate the point mass is equal to the moment of the friction force R which acts between the mass and the cylinder: at R N N R M = rR . The equation of motion in the tangential direction for the mass is ↑ : mat = R , where at = rω̇. Thus, mr ω̇ = M r → ω̇ = M . mr2 Integration with the initial condition ω(0) = 0 yields M t. r2 m Gr ω= E1.32 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 30 The condition ω(t1 ) = ω0 leads to the required moment: M= r2 mω0 . t1 The corresponding power is P = M · ω 0 = M ω0 = r2 mω02 . t1 Gr Note that the moment M has to be reduced to zero after time t1 . Otherwise the mass would be further accelerated. 31 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.33 A skier (mass m) has the velocity vA = v0 at point A of the cross country course (Fig. 1.58). Although he tries hard not to lose velocity skiing uphill, he reaches point B with only a velocity vB = 2 v0 /5. Skiing downhill between point B and the finish C he again gains speed and reaches C with vC = 4 v0 . Between B and C assume that a constant friction force acts due to the soft snow in this region; the drag force from the air on the skier can be neglected. Calculate the work done by the skier on the path from A to B (here the friction force is negligible). Determine the coefficient of kinetic friction between B and C. 0000000000000000000000000 1111111111111111111111111 00000000000 11111111111 0000000000000000000000000 1111111111111111111111111 00000000000 11111111111 0000000000000000000000000 1111111111111111111111111 00000000000 11111111111 0000000000000000000000000 1111111111111111111111111 00000000000 11111111111 0000000000000000000000000 1111111111111111111111111 00000000000 11111111111 0000000000000000000000000 1111111111111111111111111 00000000000 11111111111 B h f inish 3h A C 10h Fig. 1.58 Solution In order to calculate the work done by the skier on the path from A to B we use the work-energy theorem: TB + VB = TA + VA + U → Gr → 2 2 mvB /2 + mgh = mvA /2 + 0 + U U = m(gh − 21v02 /50) . E1.33 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 32 Now we apply the work-energy theorem between points B and C, where the work of the friction force is given by −RlBC : TC + VC = TB + VB + U → 2 2 mvC /2 + 0 = mvB /2 + 3mgh − RlBC . With the normal force N = mg cos α → N = mg 10h lBC mg and the law of kinetic friction R = µN → R = µmg R 10h lBC N α we obtain µ= 2 3 v 2 − vB − C 10 20gh → µ≈ Gr The result is valid only for µ ≥ 0. 4v 2 3 − 0 . 10 5gh 33 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.34 A circular disk (radius R) rotates with the constant angular velocity Ω. A point P moves along a straight guide; its distance from the center of the disk is given by ξ = R sin ω t where ω = const (Fig. 1.59). Determine the velocity and the acceleration of P . E1.34 R ξ P Ω Fig. 1.59 Solution We use polar coordinates r, ϕ to solve the problem. From Ω = const we find ϕ̇ = Ω = const → ϕ = Ωt , ϕ̈ = 0 . eϕ P er r ϕ The position vector is written as r = ξ er → r = R sin ωt er . Differentiation yields the velocity vector v = ξ˙ er + ξ ϕ̇ eϕ → v = Rω cos ωt er + RΩ sin ωt eϕ and the acceleration vector a = (ξ¨ − ξ ϕ̇2 ) er + (ξ ϕ̈ + 2ξ̇ ϕ̇) eϕ Gr → a = −R(ω 2 + Ω2 ) sin ωt er + 2RωΩ cos ωt eϕ . 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 34 Gr The paths of point P are displayed for several values of Ω/ω in the following figures. Ω/ω = 0.25 Ω/ω = 0.5 Ω/ω = 1.0 Ω/ω = 2.0 35 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Example 1.35 A chain with length l and mass m hangs over the edge of a frictionless table by an amount e. If the chain starts with zero initial velocity, find the position of the end of the chain as a function of time. 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 E1.35 e Fig. 1.60 Solution All the links of the chain have the same displacement, velocity and acceleration. The corner only produces a change of direction. We therefore consider the chain to be a single mass with an applied force that depends on the length x of the overhanging part. Thus, with a = ẍ, the equation of motion is g x ma = m g → ẍ − x = 0 . x l l x 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 This differential equation of second order 0000000 1111111 with constant coefficients has the solution x(t) = A cosh r g t + B sinh l r m g l g t. l We calculate the integration constants from the initial conditions: ) r ẋ(0) = 0 → B = 0 g → x(t) = e cosh t. l x(0) = e → A = e This solution is valid only for x ≤ l. x−l We may also solve the problem by m l making an imaginary section cut at the corner. Then the equations of motion for each part of the chain are d m m d l−x m ẋ = S , xẋ = xg − S . dt l dt l l If we eliminate S we again obtain the differential equation Gr g ẍ − x = 0 . l S S m x l x E1.36 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 36 x Example 1.36 A rotating source of 1111111111 0000000000 light throws a beam of light onto a screen (Fig. 1.61). Point P on the screen should move with constant velocity v0 . Determine the required angular acceleration ϕ̈(ϕ) of the source of light. Sketch ϕ̇(ϕ) and ϕ̈(ϕ). P v0 r0 ϕ Fig. 1.61 Solution The position of point P is given by x = r0 tan ϕ . We write down the inverse function x ϕ = arctan . r0 Differentiation with ẋ = v = v0 = const yields 1 v0 r0 v0 v0 ẋ = 2 = cos2 ϕ , = 2 2 2 r0 r0 + x r0 r0 (1 + tan ϕ) x 1 + r0 v 2 v0 0 ϕ̈ = 2 cos ϕ(− sin ϕ)ϕ̇ = −2 sin ϕ cos3 ϕ . r0 r0 ϕ̇ = ϕ̇ v0 r − π 2 π 2 ϕ ϕ̈ Gr − π 2 π 2 ϕ 37 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass Gr Note: the maximum value of ϕ̈, located at ϕ = ±30◦ , is obtained as 3 √ v0 2 |ϕ̈max | = 3 . 8 r0 E1.37 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 38 Example 1.37 A car travels with velocity v0 = 100 km/h. At time t = 0, the driver fully applies the brakes. At that moment the car starts sliding on a rough road (coefficient of kinetic friction µ). Use the simplest model for the car (point mass) and calculate the time t∗ and the distance x∗ until the car comes to a stop a) on a dry road (µ = 0.8), b) on a wet road (µ = 0.35). Solution The equation of motion in the horizontal direction ← : ma = mẍ = −R , x the equilibrium condition in the vertical direction mg R ↑ : 0 = N − mg N and the law of friction R = µN lead to a = −µg . Integration with v(t = 0) = v0 and x(t = 0) = 0 yields v(t) = v0 − µgt , x(t) = v0 t − 1 µgt2 . 2 The time t∗ follows from the condition v = 0: v0 t∗ = . µg This leads to Gr x∗ = x(t∗ ) = v02 v2 v2 − 0 = 0 . µg 2µg 2µg 39 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass a) On a dry road (µ = 0.8) we obtain 100 · 1000 = 3.55 s , 3600 · 0.8 · 9.81 1 100 2 = 49 m . x∗ = 3.6 2 · 0.8 · 9.81 t∗ = b) On a wet road (µ = 0.35) we are led to 100 · 1000 = 8.1 s , 3600 · 0.35 · 9.81 100 2 1 x∗ = = 112 m . 3.6 2 · 0.35 · 9.81 Gr t∗ = E1.38 1 Motion of a Point Mass os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 40 Example 1.38 In order to determine the coefficient of restitution e experimentally, a ball is dropped from height h0 onto a horizontal rigid surface (Fig. 1.62). After the ball has hit the surface 7 times, it reaches only 20% of the original height h0 . Calculate the coefficient of restitution. Solution With the positive direction of the velocity taken as upwards, the velocity of the ball immediately before the first impact is given by p v = − 2gh0 . The definition h0 h7 = 0.2h0 Fig. 1.62 h0 h1 h2 1111111 0000000 0000000 1111111 h7 v v of the coefficient of restitution in terms of the velocities yields the velocity v̄ immediately after the impact: p v = −ev = e 2gh0 . e=− From the Conservation of Energy Law we obtain the height which the body will reach: 1 mv 2 = mgh1 2 Similarly, we obtain Gr hi = e2 hi−1 → h1 = v2 = e 2 h0 . 2g 41 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 1 Motion of a Point Mass for the subsequent impacts. Thus, h7 = e2 h6 = e4 h5 = . . . = e14 h0 and Gr e= h7 h0 1/14 = 0.21/14 = 0.891 . os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e Gr Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e Gr Chapter 2 2 E2.10 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 44 Example 2.10 Two vehicles (masses m1 and m2 , velocities v1 and v2 ) crash head-on, see Fig. 2.20. After a plastic impact the vehicles are entangled and slide with locked wheels a distance s to the right. v1 v2 The coefficient of kinetic friction between the wheels m2 m1 and the road is µ. Calculate v1 if v2 and s Fig. 2.20 are known. Solution The total momentum of the system remains unchanged during the collision. We assume that positive velocities are directed to the right (note the direction of v2 ). We then obtain the velocity v̄ of the vehicles after the impact: 1111111111111 0000000000000 m1 v1 − m2 v2 = (m1 + m2 )v̄ → v̄ = m1 v1 − m2 v2 . m1 + m2 In order to relate the distance s to the velocity v̄ we apply the work-energy theorem T1 − T0 = U . We insert the kinetic energies T0 = (m1 + m2 )v̄ 2 /2 , T1 = 0 , the work of the friction force U = −Rs and Coulomb’s friction law R = µN → R = µ(m1 + m2 )g . Gr We then obtain 1 − (m1 + m2 )v̄ 2 = −µ(m1 + m2 )gs 2 m2 m2 √ → v1 = v2 + 1 + 2µgs . m1 m1 45 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses Example 2.11 A block (mass m2 ) rests on a horizontal platform (mass m1 ) which is also initially at rest (Fig. 2.21). A constant force F accelerates the l platform (wheels rolling µ m2 without friction) which F m1 causes the block to slide on the rough surface of the platform (coefficient of Fig. 2.21 kinetic friction µ). 11111111111 00000000000 Determine the time t∗ that it takes the block to fall off the platform. Solution We separate the block and the platform and introduce the coordinates x1 and x2 as shown in the free-body diagram. Then the equations of motion for the two bodies are 11 00 11 00 11 00 11 00 R x1 N 1 ① → : m1 ẍ1 = F − R , m2 g 2 x2 R F ② → : m2 ẍ2 = R . m1 g With the friction law R = µN = µm2 g, we obtain F − µm2 g , ẍ2 = µg . ẍ1 = m1 A B Now, we integrate twice. Using the initial conditions x1 (0) = x2 (0) = 0 and ẋ1 (0) = ẋ2 (0) = 0 we obtain F − µm2 g t, m1 x1 = F − µm2 g t2 , m1 2 Gr ẋ1 = ẋ2 = µgt , x2 = µg t2 . 2 E2.11 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 46 The block falls off the platform if x1 − x2 = l . This yields Gr F − µm2 g (t∗ )2 (t∗ )2 − µg =l m1 2 2 → ∗ t = s 2lm1 . F − µg(m1 + m2 ) 47 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses A railroad wagon (mass m1 ) has a velocity v1 (Fig. 2.22). It collides with a wagon (mass m2 ) which is initially at rest. Both wagons roll without friction after the collision. The second wagon is connected via a spring (spring constant k) with a block (mass m3 ) that lies on a rough surface (coefficient of static friction µ0 ). Assume the impact to be plastic and determine the maximum value of v1 so that the block stays at rest. Example 2.12 v1 m1 Fig. 2.22 k m2 m3 µ0 111111111111111111 000000000000000000 Solution The total momentum of the system remains unchanged during the impact. This yields the velocity v̄ of the wagons after the collision: m1 m1 v1 = (m1 + m2 )v̄ → v̄ = v1 . m1 + m2 x m1 m3 g m2 11111111111111 00000000000000 00000000000000 11111111111111 m3 F H N We assume that the block stays at rest. Then we can write down the equilibrium conditions Gr →: H=F , ↑: N = m3 g . E2.12 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 48 The force of static friction H and the normal force N have to satisfy the condition of static friction: H ≤ µ0 N → F ≤ µ0 m 3 g . With F = kx → x≤ µ0 m 3 g k we obtain the maximum compression of the spring: µ0 m 3 g . xmax = k Gr Now we apply the Conservation of Energy Law to determine v1max : r 1 1 2 µ0 m 3 g m 1 + m 2 2 (m1 + m2 )v̄ = kxmax → v1max = . 2 2 m1 k 49 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses 111 000 000 111 Example 2.13 A point mass m1 strikes a point mass m2 which is suspended from a string (length l, negligible mass) as shown in Fig. 2.23. The maximum force S ∗ that the string can sustain is given. Assume an elastic impact and determine the velocity v0 that causes the string to break. l v0 m1 m2 Fig. 2.23 Solution First we formulate the con- servation of linear momentum m1 v0 = m1 v̄1 + m2 v̄2 111 000 and of energy (elastic impact!): ϕ m1 v02 /2 = m1 v̄12 /2 + m2 v̄22 /2 . From these equations we can calculate the velocity v̄2 of mass m2 immediately after the impact: 2m1 v̄2 = v0 . m1 + m2 Now, we write down the equation of motion տ: m2 an = S − m2 g cos ϕ . With an = v22 /l we obtain the force in the string: Gr S = m2 v22 /l + m2 g cos ϕ . S m2 g E2.13 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 50 The maximum force Smax in the string is found for ϕ = 0: Smax = m2 (v̄22 /l + g) . The string breaks if the maximum force Smax is larger than the allowable force S ∗ : Smax > S ∗ . This yields Gr v0 > m1 + m2 q ∗ l(S /m2 − g) . 2m1 51 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses A ball ① (mass m1 ) hits a second ball ② (mass m2 , velocity v2 = 0) with a velocity v1 as shown in Fig. 2.24. Assume that the impact m2 is partially elastic (coefficient of restitution e) and 2 r2 all surfaces are smooth. m 1 1 Given: r2 = 3 r1 , m2 = v1 r1 4 m1 . Determine the velocities of the balls after the Fig. 2.24 collision. Example 2.14 1111111111111 0000000000000 Solution We introduce the auxiliary angle α. Since the surfaces are smooth, the linear impulse Fb acts in the direction of the line of impact and the Impulse Laws are given by ① ② → : m1 (v 1 − v1 ) = −Fb cos α , m2 v 2 = Fb . ր: Note that the weights of the balls can be neglected during impact and that ball ① moves only horizontally. With the hypothesis Fb 1 v1 , v1 e=− v 1x − v 2x v1x − v2x and v1x = v1 cos α , v2x = 0 , x r −r 2 1 α r2 r1 11111111 00000000 2 v2 α b N Fb v 1x = v 1 cos α , α v 2x = v 2 we obtain Gr v 1 = v1 m2 cos2 α m1 , m2 1+ cos2 α m1 1−e v 2 = v1 (1 + e) cos α . m2 1+ cos2 α m1 E2.14 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 52 Introduction of m2 = 4m1 and r2 − r1 1 sin α = = r1 + r2 2 yields Gr 1 − 3e v1 , v1 = 4 → cos α = p √ 3 1 − sin α = 2 √ 3 (1 + e) v1 . v2 = 8 2 53 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses A hunter (mass m1 ) sits in a boat (mass m2 = 2 m1 ) which can move in the water without resistance. The boat is initially at rest. a) Determine the velocity vB1 of the boat after the hunter fires a bullet (mass m3 = m1 /1000) with a velocity v0 = 500 m/s. b) Find the direction of the velocity of the boat after a second shot is fired at an angle of 45◦ with respect to the first one. Example 2.15 Solution a) We introduce a coordinate system and assume that the bullet is fired in the direction of the negative x-axis. Since the linear momentum before the firing of the bullet is zero, the total linear momentum of the system after the firing (velocities positive to the right) also has to be zero: →: (m1 + m2 − m3 )vB1 − m3 v0 = 0 . This yields the velocity vB1 of the boat: m3 vB1 = v0 = m1 + m2 − m3 ≈ 0.167 m . s 1 1000 1 1+2− 1000 500 = 500 2999 The algebraic sign shows that the boat moves in the direction of the positive x-axis. b) Now, we have to formulate the Impulse Laws in the x- and y-direction: m1 + m2 wB2 vB2 45◦ m3 v0 α y x → : (m1 + m2 − m3 )vB1 = (m1 + m2 − 2m3 )vB2 − m3 v0 cos 45◦ , Gr ↑: 0 = (m1 + m2 − 2m3 )wB2 − m3 v0 sin 45◦ . E2.15 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 54 They lead to m1 + m2 − m3 m3 vB1 + v0 cos 45◦ m1 + m2 − 2m3 m1 + m2 − 2m3 m , ≈ 0.285 s vB2 = wB2 = m3 m , v0 sin 45◦ ≈ 0.118 m1 + m2 − 2m3 s Gr → tan α = wB2 = 0.414 vB2 → α = 22.5◦ . 55 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses Example 2.16 A car ② goes into a skid on a wet road and co- mes to a stop sideways across the road as shown in Fig. 2.25. In spite of having fully applied the brakes a distance s1 from car ②, a second car ① (sliding with the coefficient of kinetic s2 2 friction µ) collides with car ②. This caum2 ses car ② to slide an additional distance s2 . Assume a partially elastic central ims1 pact. Given: m1 = 2 m2 , µ = 1/3, e = 0.2, s1 = 50 m, s2 = 10 m. 1 Determine the velocity v0 of car ① bev0 m1 fore the brakes were applied. Fig. 2.25 Solution The velocity v1 of car ① immediately before impact fol- lows from the work-energy theorem: 1 1 m1 v12 − m1 v02 = −(µ m1 g)s1 2 2 → v12 = v02 − 2µ gs1 . The velocity v 2 of car ② immediately after impact can be determined from the conservation of linear momentum (note v2 = 0) m1 v1 = m1 v 1 + m2 v 2 and the hypothesis e= v2 − v1 . v1 We obtain v2 = 1+e m2 v1 . 1+ m1 Now we apply the work-energy theorem to the sliding of car ②: Gr 1 − m2 v 22 = −(µ m2 g)s2 2 → v 22 = 2µ gs2 . E2.16 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 56 If we eliminate v1 and v 2 we obtain m2 ! 2 m1 2µ gs2 + 2µ gs1 1+e 1+ v02 = = or 1.5 2 2 2 · · 9.81 · 10 + · 9.81 · 50 = 429.19 m2/s2 1.2 3 3 v0 = 20.7 m/s → v0 = 74.6 km/h . Gr Note: the velocity of car ① at impact is v1 = 36.4 km/h. 57 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses Example 2.17 Two cars (point masses m1 and m2 ) collide at an intersection with velocities v1 and v2 at an angle α (Fig. 2.26). Assume a perfectly plastic collision. Determine the magnitude and the direction of the velocity immediately after the impact. Calculate the loss of energy during the collision. β v2 m2 Fig. 2.26 Solution Both cars have the same velocity v after the collision (plastic impact). We write down the Impulse Laws in the x- and ydirections: v1 α m1 y m1 +m2 m1 v 1 α v β x v2 m2 → : m1 v1 + m2 v2 cos α = (m1 + m2 )v cos β , ↑: m2 v2 sin α = (m1 + m2 )v sin β . If we square and then add these equations we obtain q 1 v= (m1 v1 )2 + 2m1 m2 v1 v2 cos α + (m2 v2 )2 , m1 + m2 whereas a division leads to Gr tan β = m2 v2 sin α . m1 v1 + m2 v2 cos α E2.17 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 58 The loss of mechanical energy during impact is given by the difference ∆T of the kinetic energies before and after impact: m1 v12 m2 v22 (m1 + m2 )v 2 + − 2 2 2 m1 m2 2 2 (v + v2 − 2v1 v2 cos α). = 2(m1 + m2 ) 1 ∆T = Gr Note that the loss of energy is a maximum if the cars collide headon (α = π). The energy ∆T causes the deformation of the cars. 59 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses Example 2.18 A bullet (mass m) has a velocity v0 (Fig. 2.27). An explosion causes the bullet to break into two parts (point masses m1 and m2 ). The directions α1 and α2 of the two parts and the velocity v1 immediately after the explosion are given. Calculate m1 and v2 . Determine the trajectory of the center of mass of the two parts. E2.18 v1 m1 y m v0 α1 = α2 m2 π 2 x v2 Fig. 2.27 Solution Since no external forces are acting on the system, its linear momentum does not change (linear momentum before the explosion = linear momentum after the explosion). Componentwise, the principle of conservation of linear momentum gives → : mv0 = m2 v2 cos α2 , ↑: 0 = m1 v1 − m2 v2 sin α2 . Thus, we have two equations for the two unknowns m1 and v2 . Solving with m = m1 + m2 yields v0 v0 . m1 = m tan α2 , v2 = v1 cos α2 − vv01 sin α2 We measure the time t from the moment of the explosion. Then, with |yi | = vi t sin αi , the location of the center of mass is given by yc = = 1 (m1 y1 + m2 y2 ) m (−v sin α ) o 1 n v0 v0 0 2 m tan α2 v1 t + m − m tan α2 t v0 m v1 v1 cos α2 − v1 sin α2 = 0. Gr The center of mass of the system continues to move on the original straight line y = 0 (law of motion for the center of mass). E2.19 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 60 Example 2.19 A ball (point mass m1 ) is attached to a cable. It is 1111111 0000000 0000000 1111111 released from rest at a height h1 (Fig. 2.28). After falling to the vertical position at A it collides with a second ball (point mass m2 = 2 m1 ) which is also initially at rest. The com1 efficient of restitution is e = 0.8. Determine the height h2 which the A h1 first ball can reach after the collision m2 and the velocity of the second ball immediately after impact. Fig. 2.28 1111 0000 Solution The velocities of the masses m1 and m2 before the impact are v1 = p 2gh1 , v2 = 0 . 1 The Impulse Laws ① ② → : m1 (v 1 − v1 ) = −Fb , →: and the hypothesis e=− v2 v1 , v1 m1 m2 v 2 = +Fb , Fb Fb m2 v1 − v2 v1 yield the velocities immediately after the impact: p m1 − e m2 1 − 1.6 v 1 = v1 = v1 = −0.2 2gh1 , m1 + m2 1+2 p m1 1 v 2 = v1 (1 + e) = v1 1.8 = 0.6 2gh1 . m1 + m2 1+2 1111 0000 The height h2 follows from the conservation of energy 1 m1 v 21 = m1 gh2 2 as Gr v2 h2 = 1 = 0.04 h1 . 2g h2 v1 m1 m2 v2 2 11111111 00000000 61 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses Example 2.20 The rigid rod (negli- E2.20 gible mass) in Fig. 2.29 carries two point masses. It is struck by an impulsive force Fb at a distance a from the support A. Determine the angular velocity of the rod immediately after the impact and the impulsive reaction at A. Calculate a so that the reaction force at A is zero. A l a m Fb l m Fig. 2.29 Solution We introduce a coordinate system. The y-component of the velocity of the center of mass is zero after the impact. Since the impulsive force Fb also has no y-component, the impulsive reaction at A has only an x-component, i.e. Aby = 0. We apply the principle of linear impulse and momentum →: bx − Fb 2m(v̄c − vc ) = A by A bx A y x 3l/2 C Fb and the principle of angular impulse and momentum (compare Chapter 3) 2 y l bx 3 l − Fb 3 l − a . m(ω̄ − ω) = A C: 2 2 2 2 Inserting the kinematic relation v̄c = −3lω̄/2 and the velocities Gr v̄c = 0 , ω=0 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 62 before the impact we obtain Fb a ω̄ = 2 , 5l m Abx = 3a b 1− F . 5l The reaction force at A is zero (Abx = 0) if the distance a is chosen as Gr a = 5l/3 . 63 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses Example 2.21 A wagon (weight W1 = m1 g) hits a buffer (spring constant k) with velocity v. The collision causes a block (weight W2 = m2 g) to slide on the rough surface (coefficient of static friction µ0 ) of the wagon (Fig. 2.30). Determine a lower bound for the wagon’s speed before the collision. 000000000000 111111111111 000000000000 111111111111 000000000000 111111111111 000000000000 111111111111 000000000000 111111111111 000000000000 111111111111 000000000000 111111111111 µ0 W2 k v W1 Fig. 2.30 Solution We separate the wagon and the block. The orientation of the friction force H is assumed arbitrarily in the free-body diagram. Both bodies have the same acceleration (ẍ1 = ẍ2 = ẍ) as long as the block does not slim2 g de. Therefore, the equations of 2 motion are x H ① ← : m1 ẍ = −H − F , ② ← : m2 ẍ = H . With F = k x we obtain m2 H=− kx . m1 + m2 N H F 1 m1 g 1111111 0000000 0000000 1111111 is obtained at maximal com- Gr The maximum friction force Hmax pression xmax of the spring which follows from the Conservation of Energy Law: r 1 1 m1 + m2 2 2 (m1 + m2 )v = k xmax → xmax = v. 2 2 k E2.21 2 Dynamics of Systems of Point Masses os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 64 The block is on the verge of slipping if the condition of “limiting friction” |Hmax | = µ0 N = µ0 m2 g → k xmax = µ0 g m1 + m2 Gr is satisfied. This yields the velocity that is necessary to cause slippage of the block: r m1 + m2 v = µ0 g . k 65 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 2 Dynamics of Systems of Point Masses Example 2.22 The system shown in Fig. 2.31 consists of two blocks (masses m1 and m2 ), a spring (spring constant k), a massless rope and two massless pulleys. Block 1 lies on a rough surface (coefficient of kinetic friction µ). g x1 At the beginning of the motion (t = 0), the spring k m1 is unstretched and the position of block 1 is given by x1 = 0. µ Determine the velocity x2 m2 ẋ1 as a function of the position x1 . 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 Fig. 2.31 Solution Since we want to find the velocity as a function of the position, we use the work-energy theorem T1 − T0 = U . In moving from the initial position to an arbitrary position, the gravitational force, the force in the spring and the friction force perform work: 1 Uk = − kx21 , 2 UR = −Rx1 = −µN x1 = −µm1 gx1 . UW = m2 gx2 , The kinetic energies are given by T1 = 1 1 m1 ẋ21 + m2 ẋ22 , 2 2 T0 = 0 . With the kinematic relation ẋ2 = 2ẋ1 the work-energy theorem yields Gr 1 1 1 m1 ẋ21 + m2 4ẋ21 = m2 gx2 − kx21 − µm1 gx1 2 2 2 r 2 1 k 2 → ẋ1 = (2 − µ)gx1 − x . 5 5m 1 E2.22 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e Gr Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e Gr Chapter 3 3 E3.24 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 68 er Example 3.24 Point A of the rod in Fig. 3.47 moves with constant velocity vA to the left. Determine the velocity and the acceleration of point B of the rod (point of contact with the step) as a function of the angle ϕ. Find the path y(x) of the instantaneous center of rotation. y h B 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 ϕ e v 0000000 1111111 A 0000000 1111111 0000000 x 1111111 ϕ A Fig. 3.47 Solution We use the coordinate system with the basis vectors er and eϕ as shown in the figure. Then, the velocity of point A is er given by v A = vA cos ϕer − vA sin ϕeϕ . The velocity v B of point B points in the direction of the rod (the rod does not lift off the step). Thus, v B = vB er . arAB vAB 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 vB aϕAB B ϕ vA eϕ A With the kinematic relation v B = v A + v AB where v AB = aϕ̇eϕ and with a= h , sin ϕ we obtain hϕ̇ eϕ sin ϕ vA vB = vA cos ϕ and ϕ̇ = sin2 ϕ . h vB er = vA cos ϕer − vA sin ϕeϕ + Gr → a 69 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies In order to determine the acceleration of point B we use the relation aB = aA + arAB + aϕ AB , where aA = 0 , arAB = −aϕ̇2 er , aϕ AB = aϕ̈eϕ . The angular acceleration of the rod is found by differentiating its angular velocity ϕ̇: ϕ̈ = vA (sin2 ϕ)˙ h → ϕ̈ = 2 2 vA sin3 ϕ cos ϕ . h2 Hence, we obtain aB = 2 vA sin2 ϕ(− sin ϕer + 2 cos ϕeϕ ) . h The instantaneous center of rotation Π is given by the point of intersection of two straight lines which are perpendicular to two velocities. In the present example, the directions of the velocities vA and vB are known. The2.0 refore, the instantaneous cenpath of Π ter of rotation can easily be 1.5 constructed (see the figure). Π Point Π has the coordinates y/h 1.0 x = h cot ϕ , y = h + x cot ϕ . Elimination of the angle ϕ yields the path of Π: x2 +h. h Gr y= 0.5 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 0 -1.0 0.5 0 x/h ϕ 0.5 1.0 E3.25 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 70 The wheel of a crank drive rotates with constant angular velocity ω (Fig. 3.48). Determine the velocity and the acceleration of the piston P . Example 3.25 Solution If we introduce the auxiliary angle ψ, then the position xP of the piston is given by xP = r cos ϕ + l cos ψ . Now we differentiate and obtain (note ϕ̇ = ω = const) l 000 111 r 000 111 000 111 ϕ ψ 00000000 11111111 P ω 111 000 000 111 Fig. 3.48 0 1 0 1 0 1 0 1 0 1 0 1 111111 000000 y r ϕ = ωt x ψ l ẋP = −rω sin ϕ − lψ̇ sin ψ , ẍP = −rω 2 cos ϕ − lψ̈ sin ψ − lψ̇ 2 cos ψ . The quantities sin ψ, cos ψ, ψ̇ and ψ̈ are as yet unknown. They follow from the condition that the piston can move only in the horizontal direction. Therefore, its vertical displacement is zero: yP = 0 = r sin ϕ − l sin ψ . Differentiation yields ẏP = 0 = rω cos ϕ − lψ̇ cos ψ , Gr ÿP = 0 = −rω 2 sin ϕ − lψ̈ cos ψ + lψ̇ 2 sin ψ . P 71 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies Thus, r sin ψ = sin ϕ , l ψ̇ = ω r cos ϕ , l cos ψ cos ψ = r ψ̈ = −ω 2 1− r l 2 sin ϕ , r sin ϕ sin ψ + ψ̇ 2 . l cos ψ cos ψ Hence, we finally obtain r sin ϕ cos ϕ ẋP = −rω sin ϕ + , l cos ψ Gr ẍP = −rω 2 ( " #) r sin2 ϕ cos2 ϕ cos ϕ − − . l cos ψ cos3 ψ E3.26 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 72 Example 3.26 Link MA of the mechanism in Fig. 3.49 rotates with angular velocity ϕ̇(t). Determine the velocities of points B and C, the angular velocity ω and the angular acceleration ω̇ of the angled member ABC at the instant shown. l r ϕ M C 00 11 00 11 00 11 00 11 00 11 l B y z A x α Fig. 3.49 Solution With the given x, y, z-coordinate system, the velocities of points A and B at the instant shown can be written in the form − sin ϕ − cos α v A = rϕ̇ cos ϕ , v B = vB sin α , 0 0 where vB is as yet unknown (note that the direction of v B is known). The angular velocity of the angled member ABC is represented by 0 ω = 0 . ω We now use the kinematic relation Gr v B = v A + ω × rAB 73 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies which, with rAB = (0, −l, 0)T , in coordinates reads − cos α − sin ϕ ωl = rϕ̇ cos ϕ + 0 . vB sin α 0 0 0 Solving yields vB = rϕ̇ cos ϕ , sin α ω= rϕ̇ cos ϕ sin ϕ − . l tan α In an analogous way we obtain the velocity of point C: − sin ϕ cos ϕ v C = v A + ω × r AC → v C = rϕ̇ sin ϕ + cos ϕ − tan α . 0 In order to determine the angular acceleration ω̇ = (0, 0, ω̇)T of the angled member, we first write down the relation aB = aA + ω̇ × r AB + ω × (ω × rAB ) , which in coordinates reads − cos α − sin ϕ cos ϕ ω̇l 0 2 aB sin α = rϕ̈ cos ϕ − rϕ̇ sin ϕ + 0 + ω 2 l . 0 0 0 0 0 Solving for aB and ω̇ yields 1 (rϕ̈ cos ϕ − rϕ̇2 sin ϕ + lω 2 ) , sin α 1 1 ω̇ = rϕ̈ sin ϕ + rϕ̇2 cos ϕ − (rϕ̈ cos ϕ − rϕ̇2 sin ϕ + lω 2 ) . l tan α Gr aB = E3.27 3 Dynamics of Rigid Bodies 00 11 111111111111 000000000000 00Ω 11 111111111111 000000000000 R 111111111111 000000000000 P r 111111111111 000000000000 111111111111 000000000000 111111111111 000000000000 ϕ 111111111111 000000000000 111111111111 000000000000 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 74 Example 3.27 Wheel ① rolls in a gear mechanism without slip along circle ②. The mechanism is driven with a constant angular velocity Ω (Fig. 3.50). Determine the magnitudes of the velocity and the acceleration of point P on the wheel. 1 2 Fig. 3.50 Solution The motion of P is composed of the rotation of B about A with vB = RΩ (vB ⊥ AB) , aB = RΩ2 (aB k AB) and the rotation of P about B. In order to determine the angular velocity ω of wheel ① we consider two positions of the wheel. The figure shows that the relation 00 11 111111111111 000000000000 00 11 111111111111 000000000000 → Rα = rβ 111111111111 000000000000 for the arclengths holds. Diffe- 000000000000 111111111111 000000000000 rentiating and using α̇ = Ω and 111111111111 111111111111 000000000000 β̇ = ω yields 111111111111 000000000000 R 000000000000 RΩ = rω → ω = Ω . 111111111111 r (R + r)α = r(β + α) A α R C′ r B C β B′ α Gr Note that Ω is positive counterclockwise, whereas ω is positive clockwise. 75 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies We can now construct the velocity diagram and the acceleration diagram. Since the velocity diagram is represented by an isosceles triangle, we obtain vP = 2ΩR sin ϕ . 2 vP vB = RΩ Gr Finally, the law of cosines yields 2 2 2 Ω R a2P = (Ω2 R)2 + r 2 2 aB = RΩ2 Ω R cos(π − ϕ) , −2Ω2 R r s 2 R R + 2 cos ϕ . a P = Ω2 R 1 + r r ϕ ϕ aP vP B = rω = Rω anP B = rω 2 Ω2 R2 = r E3.28 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 76 Example 3.28 A disk (mass m, radius r1 ) rests in a frictionless support (ω0 = 0). A second disk r2 2 (mass m, radius r2 ) rotates ω2 m with the angular velocity ω2 . It m 2 is placed on disk 1 as shown in 1 Fig. 3.51. Due to friction, both disks eventually rotate with the r1 same angular velocity ω̄. Determine ω̄. Calculate the Fig. 3.51 change ∆T of the kinetic energy. 1 111111 000000 000000 111111 111111111 000000000 0000 1111 0000 1111 Solution Since there is no moment of external forces acting on the system, the angular momentum is conserved during the process: Θ2 ω2 = (Θ1 + Θ2 )ω̄ . With the mass moments of inertia Θ1 = mr12 /2 , Θ2 = mr22 /2 we obtain the resulting angular velocity ω̄: ω̄ = r22 ω2 . r12 + r22 The change ∆T of the kinetic energy follows as Gr ∆T = 1 1 (Θ1 + Θ2 )ω̄ 2 − Θ2 ω22 2 2 → r2 r2 1 ∆T = − mω22 2 1 2 2 . 4 r1 + r2 77 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies Example 3.29 The door (mass m, moment E3.29 of inertia ΘA ) of a car is open (Fig. 3.52). Its center of mass C has a distance b from the frictionless hinges. The car starts to move with the constant acceleration a0 . Determine the angular velocity of the door when it slams shut. a0 C A m b Fig. 3.52 Solution First, we write down the acceleration of the center of mass C: aC = aA + arAC + aϕ AC An A ϕ C At , where the magnitudes of the individual terms are given by aA = a0 , arAC = bϕ̇2 , aϕ AC = bϕ̈ . We then formulate the principle of linear momentum (in the tdirection, see the free-body diagram) ւ: m(bϕ̈ − a0 cos ϕ) = At and the principle of angular momentum (about the center of mass C): y C: ΘC ϕ̈ = −At b . The mass moment of inertia ΘC follows from the parallel-axis theorem: Gr ΘC = ΘA − mb2 . 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 78 Eliminating the force At yields ϕ̈ = ma0 b cos ϕ . ΘA The angular velocity of the door can be obtained through integration: 1 2 ma0 b ϕ̇ (ϕ) = sin ϕ . 2 ΘA Gr Thus, for ϕ = π/2 (closed door) we find s 2ma0 b ϕ̇(π/2) = . ΘA 79 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies Example 3.30 A child (mass m) runs along the rim of a circular platform (mass M , radius r) starting from point A (Fig. 3.53). The platform is initially at rest; its support is frictionless. Determine the angle of rotation of the platM 0 m form when the child arr rives again at point A. A Fig. 3.53 Solution We apply the principle of angular momentum: the time rate of change of the angular momentum is equal to the moment of the applied forces. Since there are no external applied forces acting in the present example, the angular momentum is constant: dL(0) =0 dt → L(0) = const with L(0) = Θ0 ω + mr2 (ω + ωrel ) . Here, Θ0 = M r2 /2 is the mass moment of inertia of the platform, ω is the angular velocity of the platform and ωrel is the angular velocity of the child relative to the platform. Integration of the principle of angular momentum with the initial condition L(0) (0) = 0 yields Z t L(0) dt̄ = 0 Gr 0 L(0) ≡ 0 → → Θ0 ϕ + mr2 (ϕ + ϕrel ) = 0 . E3.30 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 80 The child arrives again at point A when the relative angle attains the value ϕrel = 2π. Solving for ϕ yields ϕ=− 2π . M +1 2m 2π ϕrel = 2π ϕ |ϕ| ϕchild = ϕ+ϕrel π 0 Gr The result is displayed in a diagram. 2 4 M/m 6 8 10 81 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies A homogeneous triangular plate of weight W = mg is suspended from three strings with negligible mass. Determine the acceleration of the plate and the forces in the strings just after string 3 is cut. Example 3.31 111111111 000000000 1 E3.31 1010 1010 2 α α 3 W l 2l Fig. 3.54 Solution The motion of the plate is a translation after string 3 is cut. Therefore, all the points of the plate have the same acceleration (and the same velocity). We choose point A which rotates about a fixed point to represent the motion of the plate. Its acceleration can be written in the Serret-Frenet frame as a = aA = v̇et + v2 en . ρA Its velocity is zero at the moment of the cut: v=0 → a = v̇et . The principles of linear and angular momentum are then given by (see the free-body diagram) en S1 et 2 l 3 S2 A C α 1 l 3 mg ւ : mv̇ = mg sin α , 0 = S1 + S2 − mg cos α , y C : 0= Gr տ: 2 l l 4 lS1 cos α − S1 sin α − S2 sin α − S2 cos α . 3 3 3 3 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 82 Solving these equations yields v̇ = g sin α, S1 = mg (sin α + 4 cos α), 6 S2 = mg (− sin α + 2 cos α) . 6 Gr Note that S2 = 0 for sin α = 2 cos α (i.e., for tan α = 2 → α = 63.4◦). In this case the line of action of S1 passes through the center of mass C. 83 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies Example 3.32 A sphere (mass E3.32 m1 , radius r) and a cylindrical wheel (mass m2 , radius r) are connected by two bars (mass of each bar m3 /2, length l). They roll down a rough inclined plane (with angle α) without slipping (Fig. 3.55). Find the acceleration of the bars. ϕ x r m1 m3 m2 r l α Fig. 3.55 Solution The only external forces that act on the system are the weights of the individual parts. Therefore, conservation of energy T + V = const leads to the solution. The kinetic energy is given by T = (m1 + m2 + m3 )ẋ2 /2 + (Θ1 + Θ2 )ϕ̇2 /2 and the potential energy is V = −(m1 + m2 + m3 )gx sin α . With the kinematic relation (rolling without slipping) ẋ = rϕ̇ and the mass moments of inertia Gr Θ1 = 2m1 r2 /5 , Θ2 = m2 r2 /2 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 84 we obtain (7m1 /10 + 3m2 /4 + m3 /2)ẋ2 − (m1 + m2 + m3 )gx sin α = const . Differentiation finally yields the acceleration of the bars: 2(7m1 /10 + 3m2 /4 + m3 /2)ẋẍ − (m1 + m2 + m3 )g ẋ sin α = 0 Gr → a = ẍ = (m1 + m2 + m3 ) sin α g. 7m1 /5 + 3m2 /2 + m3 85 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies Example 3.33 The cylindrical E3.33 z y shaft shown in Fig. 3.56 has x an inhomogeneous mass density given by ρ = ρ0 (1 + αr). l Find the moments of inerFig. 3.56 tia Θx and Θy . Solution We determine the z mass moment of inertia Θx from r dr Z Z Θx = (y 2 + z 2 )dm = r2 dm . r y R y R With dm = ρ 2πr dr dx = 2πρ0 (r + αr2 )dr dx we obtain R4 Rl RR 3 R5 (r + αr4 )dr dx = 2πρ0 l +α 4 5 0 0 π 4 = ρ0 lR4 1 + αR . 2 5 Θx = 2πρ0 The mass moment of inertia Θy = Θz (symmetry) follows from Z Θy = (z 2 + x2 )dm with rdϕ z dm = ρr dϕ dr dx dr dϕ r r sin ϕ 2 = ρ0 (r + αr ) dϕ dr dx , Gr z = r sin ϕ . ϕ y 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 86 Using the symmetry we obtain π/2 Zl ZR Z Θy = 4ρ0 (r2 sin2 ϕ + x2 )(r + αr2 )dϕ dr dx 0 0 0 Z l ZR = πρ0 (r2 + 2x2 )(r + αr2 )dr dx 0 0 3 l2 2 2 R2 R = πρ0 R l + +α + Rl . 4 3 5 9 2 Note: For α = 0 the results reduce with m = ρ0 πR2 l to the mass moments of inertia of a homogeneous cylindrical shaft (note the term due to the parallel-axis theorem): Gr Θx = 1 mR2 , 2 Θy = 1 m 3R2 + 4l2 . 12 87 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies Determine the moment of inertia Θa of a homogeneous torus with a circular cross section and mass m. Example 3.34 a E3.34 R c a Fig. 3.57 Solution The mass moment of inertia Θa is determined from Z Θa = r2 dm . We can determine the geometrical relations dr a R c ϕ r = R + c sin ϕ , c sin ϕ dm = ρ 2πr2c cos ϕ dr from the figure. With r a dr = c cos ϕ dϕ we obtain dm = 4πρc2 (R + c sin ϕ) cos2 ϕ dϕ . Thus, Θa = 4πρc 2 +π/2 Z (R + c sin ϕ)3 cos2 ϕdϕ −π/2 = 4πρc 2 +π/2 Z −π/2 R3 cos2 ϕ + 3R2 c sin ϕ cos2 ϕ + 3Rc2 sin2 ϕ cos2 ϕ + c3 sin3 ϕ cos2 ϕ]dϕ Gr = 4πρc 2 π 3 3π R + Rc2 2 8 3 2 2 = 2π ρc R R + c . 4 2 2 2c cos ϕ 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 88 Note that the integrals of the odd functions over the even interval are zero. Using m = 2π 2 ρc2 R we finally get 3 Θ a = m R 2 + c2 . 4 Gr This result reduces to Θa = mR2 in the case of a thin ring (c ≪ R). 89 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies Example 3.35 A rope drum on a E3.35 rough surface is set into motion by pulling the rope with a constant force F0 . Determine the acceleration of point C assuming that the drum rolls (no slipping). What coefficient of static friction µ0 is necessary to ensure rolling? m, Θc r2 C r1 F0 α 111111111 000000000 µ0 Fig. 3.58 Solution The free-body diagram shows the forces that act on the drum. We apply the principles of linear and angular momentum: →: mac = F0 cos α − H , ↑ : 0 = N − mg + F0 sin α , vc y C : Θc ω̇ = r1 H − r2 F0 . ω r1 r2 mg F0 C α In addition we have the kinematic relation H vc = r1 ω → v̇c = ac = r1 ω̇ N between the velocity and the angular velocity (rolling without slip). We solve these four equations to obtain the acceleration and the forces: r2 cos α − F0 r1 ac = , Θc m 1+ 2 r1 m Gr Θc cos α r2 + r12 m r1 H = F0 , Θc 1+ 2 r1 m N = mg − F0 sin α . 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 90 In order to ensure rolling, the condition of static friction H ≤ µ0 N has to be satisfied. This leads to Θc cos α r2 + r12 m r1 . µ0 ≥ mg Θc 1+ 2 − sin α F0 r1 m Gr Note: For cos α > r2 /r1 we have ac > 0 (motion to the right), whereas for cos α < r2 /r1 we obtain ac < 0 (motion to the left). In the case of F0 sin α > mg the drum lifts off the ground. 91 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies Example 3.36 A homogeneous beam (mass M , length l) is initially in vertical position ① (Fig. 3.59). A small disturbance causes the beam to rotate about the frictionless support A (initial velocity equal to zero). In position ② it strikes a small sphere (mass m, radius r ≪ l). Assume the impact to be elastic (e = 1). Determine the angular velocities of the beam immediately before and after the impact and the velocity of the sphere after the impact. ① M l A ② m Fig. 3.59 Solution The angular velocity ω immediately before the impact follows from the conservation of energy (ΘA = M l2 /3): q M gl/2 = ΘA ω 2 /2 − M gl/2 → ω = 6g/l . Since the impact is assumed to be elastic, the conservation of the angular momentum ΘA ω = ΘA ω̄ + mlv̄ as well as the conservation of energy ΘA ω 2 /2 = ΘA ω̄ 2 /2 + mv̄ 2 /2 hold. Solving these two equations yields M − 3m ω, M + 3m Gr ω̄ = v̄ = 2lM ω. M + 3m E3.36 E3.37 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 92 A thin half-cylindrical shell of weight W = mg rolls without sliding on a flat surface (Fig. 3.60). Determine the angular velocity as a function of ϕ when the initial condition ϕ̇(ϕ = 0) = 0 is given. Example 3.37 W ϕ R 11111111 00000000 Fig. 3.60 Solution The position of the center of mass and the mass moment of inertia are given by Z π 1 2R (R sin α)Rdα = , c = Rπ 0 π 4 Θc = Θ0 − mc2 = mR2 − 2 mR2 π 4 = mR2 (1 − 2 ) . π Thus, xc = Rϕ + c cos ϕ = R(ϕ + 2R sin ϕ , π 2 ẋc = Rϕ̇(1 − sin ϕ) , π 2R ẏc = ϕ̇ cos ϕ . π R sin α 0 α c C 2 cos ϕ) , π R R dα Rϕ c cos ϕ c yc = c sin ϕ = y ϕ yc C x 11111111 00000000 Since we want to find the angular velocity as a function of the angle we apply the conservation of energy: T + V = T0 + V0 . With T0 = 0 , Gr T = V0 = 0 , 1 1 1 4 4 m(ẋ2c + ẏc2 ) + Θc ϕ̇2 = mR2 ϕ̇2 1 − sin ϕ + 2 sin2 ϕ 2 2 2 π π 1 4 4 2 + 2 cos2 ϕ + mR2 ϕ̇2 1 − 2 = mR2 ϕ̇2 1 − sin ϕ , π 2 π π 2 V = −mgyc = − mgR sin ϕ π 93 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies we obtain mR2 ϕ̇2 (1 − → 2 2 sin ϕ) − mgr sin ϕ = 0 π π s 2g sin ϕ ϕ̇(ϕ) = . R(π − 2 sin ϕ) Gr Note that the angular velocity attains its maximum for ϕ = 90◦ (lowest position of the center of mass): q ϕ̇max = 2g/R(π − 2) . E3.38 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 94 An elevator consists of a cabin (weight W = mg) which is connected through a rope (of negligible mass) with a rope drum and a band brake (coefficient of dynamic friction µ). Determine the necessary braking force F such that a cabin travelling downwards with velocity v0 stops after a distance h. Example 3.38 l r2 0 r1 µ F Θ0 v0 W Fig. 3.61 Solution First we determine the r2 brake momentum MB that acts on the drum. To this end we separate the drum and the lever. Equilibrium at the lever y A: ω0 S1 0 = −2r2 S2 + lF → S2 = Fl 2r2 Gr S2 A 2r2 l Fl . 2 Now we apply the work-energy theorem T1 − T0 = U . S2 S1 and the formula S1 = S2 e−µπ for belt friction (see Volume 1, Chapter 9) lead to MB = r2 S2 − r2 S1 = (1 − e−µπ ) W F 95 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies With the kinematic relations (constraints imposed by the rope) v0 = r1 ω0 , h = r1 ϕh the kinetic energies and the work are given by v2 Θ0 1 1 T1 = 0 , T0 = mv02 + Θ0 ω02 = 0 m + 2 , 2 2 2 r1 Z ϕh F lh U = mgh − MB dϕ = mgh − (1 − e−µπ ) . 2r1 0 Thus, we obtain Θ0 v2 F lh − 0 m + 2 = mgh − (1 − e−µπ ) . 2 r1 2r1 Gr Solving for the force F yields r1 v02 m 2gh Θ0 F = + . 1 + lh(1 − e−µπ ) mr12 v02 E3.39 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 96 Example 3.39 A homogeneous beam (mass m, length l) rotates about frictionless support A (Fig. 3.62) until it hits support B. The motion starts with zero initial velocity in the vertical l position. The coefficient of restitution e is given. m Calculate the impulsive forces at A and B. DetermiA B ne the distance a so that the impulsive force at A vanishes. a Calculate the change of the Fig. 3.62 kinetic energy. Solution First we calculate the angular velocity of the beam im- mediately before the impact with the aid of the conservation of energy (ΘA = ml2 /3): mgl/2 = ΘA ϕ̇2 /2 ϕ y x → ϕ̇2 = 3g/l . bH A bV A C a b B We then apply the principles of impulse and momentum: →: ↑: Gr y A: ¯C − ẋC ) = AbH , m(ẋ b, m(ẏ¯C − ẏC ) = AbV + B b . ΘA (ϕ̇¯ − ϕ̇) = −Ba 97 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies Here, the velocity of the center of mass and the angular velocity before the impact are given by q p ẋC = 0 , ẏC = − 3gl/2 , ϕ̇ = 3g/l . In order to obtain the velocities immediately after the impact we use the hypothesis e=− ẏ¯B ẏB → ϕ̇¯ = −eϕ̇ , which leads to ¯C = 0 , ẋ p ẏ¯C = e 3gl/2 , q ϕ̇¯ = −e 3g/l . Solving for the impulsive reactions yields bV = (1 + e)(3a − 2l)m √3gl , A 6a AbH = 0 , b = (1 + e)ml √3gl . B 3a The impulsive force at A vanishes if AbV = 0 → a = 2l/3 . The change of the kinetic energy is the difference ∆T = T1 − T0 of the energies before and after impact: Gr ∆T = 1 1 ΘA ϕ̄˙ 2 − ΘA ϕ̇2 2 2 → ∆T = −(1 − e2 )mgl/2 . E3.40 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 98 A homogeneous circular disk of weight W = mg is suspended from a pin-supported bar (of negligible mass). Initially the disk rotates with the angular velocity ω0 . a) Determine the amplitude of oscillation of the pendulum, if the bar suddenly prevents the disk from rotating. b) Calculate the energy loss ∆E. Example 3.40 l ϕ m ω0 r Fig. 3.63 Solution a) First we determine the angular velocity ω of the bar immediately after the rotation of the disk is prevented (note that the bar is still in the vertical position). Since there are no external moments acting with respect to A, the angular momentum is conserA ved. With ω ΘB = 1 2 mr , 2 ΘA = 1 2 1 mr + ml2 = m(r2 + 2l2 ) 2 2 Disk blocked we obtain Θ B ω0 = Θ A ω → ω= B ΘB r2 ω0 = 2 ω0 . ΘA r + 2l2 The maximum value ϕ1 of the angle ϕ (= amplitude of the oscillation) can be calculated from the conservation of energy: ϕ1 T1 +V1 = T0 +V0 . If we choose the potential energy V0 (initial position) to be equal to zero, we can write T1 = 0 , position 1 l(1 − cos ϕ1 ) position 0 Gr V1 = mgl(1 − cos ϕ1 ) , T0 = 1 mr2 ω02 r2 ΘA ω 2 = , 2 4 r2 + 2l2 V0 = 0 . 99 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies Then, conservation of energy leads to cos ϕ1 = 1 − r2 ω02 r2 . 2 4gl r + 2l2 b) The loss of energy ∆E is given by the difference ∆T of the kinetic energies before and after the blocking: ∆E = Gr = ΘB ω02 ΘA ω 2 mr2 ω02 m r4 ω02 − = − (r2 + 2l2 ) 2 2 2 4 4 (r + 2l2 )2 mr2 ω02 l2 . 2 r2 + 2l2 E3.41 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 100 M 11 00 0 1 00 10 11 A homogeneous angled bar of mass m is attached to a shaft with negligible mass. The rotation of the system is driven by the moment M0 . Determine the angular acceleration and the support reactions. Example 3.41 0 2l m ξ 2l 2l 111 000 The following moments Solution and products of inertia with respect to the body-fixed coordinate system ξ, η, ζ are needed: Fig. 3.64 M0 Aη Aξ (2l)2 m 2 20 Θζ = m + (2l)2 = ml2 , 3 3 3 9 l 1 m Θξζ = − 2l = − ml2 , 3 2 3 ω, ω̇ η ζ C ξc ξ Bη Bξ Θηζ = 0 . l η With the moments Mξ = 2lBη − 2lAη , Mη = 2lAξ − 2lBξ , Mζ = M0 the principle of angular momentum in component form yields y ξ : y η : Gr y ζ : ml2 3 ml2 2l(Aξ − Bξ ) = −ω 2 3 20 M0 = ml2 ω̇ 9 2l(Bη − Aη ) = −ω̇ mlω̇ , 6 mlω 2 Aξ − Bξ = − , 6 9M0 . ω̇ = 20ml2 → Bη − Aη = − → → 101 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies In order to be able to calculate the support reactions, we now have to formulate the principle of linear momentum. The center of mass moves on a circle. With the distance ξc = 4l/3 from the axis of rotation, we obtain the components of its acceleration as acξ = −ξc ω 2 and acη = ξc ω̇. Thus, −m ξc ω 2 = Aξ + Bξ , m ξc ω̇ = Aη + Bη which leads to 3 Aξ = − mlω 2 , 4 Aη = 27 M0 , 80 l 7 mlω 2 , 12 Bη = 21 M0 . 80 l Gr Bξ = − E3.42 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 102 A shaft (principal moments of inertia Θ1 , Θ2 , Θ3 ) rotates with constant angular velocity ω0 about its longitudinal axis. This axis undergoes a rotation α(t) about the z-axis of the fixed in 1 y space system x, y, z. 2 ω0 Calculate the moment which is 3, z exerted by the bearings on the shaft x α(t) for a) uniform rotation α = Ωt, Fig. 3.65 b) harmonic rotation α = α0 sin Ωt. Example 3.42 Solution We solve the problem with the aid of Euler’s equations Θ1 ω̇1 − (Θ2 − Θ3 )ω2 ω3 = M1 , Θ2 ω̇2 − (Θ3 − Θ1 )ω3 ω1 = M2 , Θ3 ω̇3 − (Θ1 − Θ2 )ω1 ω2 = M3 . With ω1 = ω0 , ω̇1 = 0 , ω2 = ω̇2 = 0 , ω3 = α̇ , ω̇3 = α̈ we obtain M1 = 0 , M2 = −(Θ3 − Θ1 )ω0 α̇ , M3 = Θ3 α̈ where α(t) represents an arbitrary rotation¡. a) In the special case of a uniform rotation α = Ωt we get α̇ = Ω , α̈ = 0 . Thus, Gr M1 = M3 = 0 , M2 = (Θ1 − Θ3 )ω0 Ω . 103 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies b) In the case of a harmonic rotation α = α0 sin Ωt we find α̇ = α0 Ω cos Ωt , α̈ = −α0 Ω2 sin Ωt and M1 = 0 , M2 = (Θ1 − Θ3 )ω0 Ω α0 cos Ωt , M3 = −Θ3 Ω2 α0 sin Ωt . Gr Note: Only a moment about the 2-axis acts in case a). It is caused by two opposite support reactions of equal magnitude (= couple) in the 3- and z-directions, respectively. E3.43 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 104 Example 3.43 A pin-supported rigid beam (mass m, length l) is initially at rest. At time t0 = 0 it starts to rotate due to an applied constant moment M0 . Determine the stress resultants (inter- M0 nal forces and moments) as functions of x for t > t0 . Neglect gravitational effects. l x ϕ A Fig. 3.66 Solution First we determine the angular acceleration and the an- gular velocity of the beam. The principle of angular momentum with respect to A yields x A: ΘA ϕ̈ = M0 → ϕ̈ = M0 . ΘA Integration with the initial condition ϕ̇(0) = 0 gives ϕ̇(t) = M0 t. ΘA Now we cut the beam at an arbitrary position and introduce the bending moment M and the shear force V into the free-body diagram (the normal force N will be considered later). The principle of angular momentum (with respect to the center of mass C̄) and the principle of linear momentum (in the y-direction) yield x C̄ : տ: Θ̄C̄ ϕ̈ = −M (x) − V (x) l−x , 2 m̄ ÿC̄ = V (x) . The acceleration ÿC̄ follows from the kinematics (circular motion): m̄, Θ̄C̄ V M y C̄ 1 (l 2 − x) x rC̄ = 12 (l + x) l+x ϕ̈ . 2 x m l2 Thus, with m̄ = (1 − ) m and ΘA = we obtain the shear l 3 force: x 2 l+x l + x M0 3 M0 V (x) = m̄ ϕ̈ = m̄ = 1− . 2 2 ΘA 2 l l Gr ÿC̄ = rC̄ ϕ̈ = 105 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies 1 m̄ (l − x)2 leads Introduction of the moment of inertia Θ̄C̄ = 12 to the bending moment: l−x − Θ̄C̄ ϕ̈ 2 x 2 3 1+ = − M0 1 − 4 l x 2 1 1+ = −M0 1 − l 2 M (x) = −V x 3 M0 x m l2 − 1− l 12 l ΘA x . l The normal force can be determined from the equation of motion in the x-direction: m̄ ր: m̄ ẍC̄ = −N (x) 2 where ẍC̄ = −rC̄ ϕ̇ is the centripetal acceleration. This leads to N y C̄ x rC̄ = 12 (l + x) 2 N (x) = m̄ rC̄ ϕ̇ x x+l = m (1 − ) l 2 9 M02 t2 = 2 m l3 1− M0 t ΘA x 2 l 2 . Note that the normal force increases with time t in contrast to the bending moment and the shear force. The stress resultant diagrams are presented below. M(x) V (x) + 3 M0 2 l M0 Gr N(x) + 9 M02 t2 2 ml3 E3.44 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 106 Example 3.44 The angled member A (weight W = mg) in Fig. 3.67 consists of two homogeneous bars. Derive the equation of motion for the member’s center of mass. g 2l l Fig. 3.67 Solution First, we locate the center of mass of the angled member. With the coordinate system as shown in the figure we obtain m l l xc = 3 2 = , m 6 m 2m 2l + l 3 = 4l. yc = 3 m 3 x C xc A a √ q 65 yc2 + x2c = l. 6 y ϕ 1 m 3 C Thus, the distance of the center of mass from point A is given by a= yc a 2 m 3 mg a sin ϕ The member rotates about a fixed axis that passes through A. We introduce the angle ϕ and apply the principle of angular momentum: Gr ΘA ϕ̈ = MA . 107 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies We measure ϕ from the position of equilibrium (C is vertically below A). With the mass moment of inertia ( " 2 #) (2l)2 7 m l2 l 2 2 ΘA = + (2l) + + m = ml2 3 12 2 3 3 3 we obtain x A: 7 ml2 ϕ̈ = −mga sin ϕ 3 → ϕ̈ + √ 65 g sin ϕ = 0 . 14 l Note: In the case of small oscillations (ϕ ≪ 1, sin ϕ ≈ ϕ) the equation of motion reduces to the differential equation √ 65 g ϕ̈ + ϕ=0 14 l Gr for harmonic vibrations. E3.45 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 108 Example 3.45 A bowling ball (mass m) is placed on a rough surface (coefficient of kinetic friction µ = 0.3) with velocity v0 = 5 m/s (Fig. 3.68). Initially, the ball does not rotate. What is the position xr of the ball when it stops sliding? Calculate the corresponding velocity vr . v0 11111111 00000000 µ Fig. 3.68 Solution When the ball is placed on the rough surface it slides. ω C r mg x R N The friction force R is opposed to the direction of the motion. Thus, the equations of motion are given by →: ↑ : y C : mẍ = −R , 0 = N − mg → N = mg , Θc ω̇ = rR . With Θc = 2mr2 /5 and the law of friction R = µN = µmg, the first and the third equation lead to (initial conditions v(t = 0) = v0 , x(t = 0) = 0, ω(t = 0) = 0) Gr v = ẋ = v0 − µgt , x = v0 t − 1 µg t2 , 2 ω= 5µg t. 2r 109 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies The ball rolls without sliding if the velocity of its center of mass is given by v = rω . This condition leads to the corresponding time tr : v0 − µgtr = 5 µ g tr 2 → Thus, 12v02 = 2.08 m , 49µg vr = v(tr ) = 5 m v0 = 3.57 . 7 s Gr xr = x(tr ) = tr = 2v0 = 0.49 s . 7µg E3.46 3 Dynamics of Rigid Bodies os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 110 Example 3.46 The double pendulum in Fig. 3.69 consists of two identical homogeneous bars (each mass m, length l). It is struck by a linear impulse Fb at point D. Determine the distance d of point D from the lower end of the pendulum so that the angular velocity ω2 of the lower bar is zero immediately after the impact. Calculate the impulsive forces at A and B. A g l l B D Fb d Fig. 3.69 Solution We separate the two bars and draw the free-body diagram. Note that there is no linear impulse in the vertical direction. The bars are at rest before the impact. The principles of linear and angular impulse and momentum are given by ① x A: ② →: x C2 : where ΘA = b, ΘA ω 1 = l B b, m v 2 = Fb − B l b lb ΘC2 ω 2 = B − d− F 2 2 1 2 ml , 3 ΘC2 = A ω1 ① l b B 1 l 2 d− Fb 1 l 2 b B v2 C2 ω2 ② 1 ml2 . 12 We desire the motion of bar ② to be a translation. Therefore, Gr ω2 = 0 , v2 = l ω1 . 111 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 3 Dynamics of Rigid Bodies This leads to 3 Fb ω1 = , 4 ml Fb Fb b= B 2 = 4 , ml 1+ ΘA ml2 l 5 ΘA d= = l. 2 8 ml2 1+ ΘA 2+ The impulsive force at A follows from the principle of linear im1 l ω1 : 2 pulse for bar ① with the kinematic relation v 1 = →: b+B b m v1 = A → b b= F . A 8 b A Gr b B ω1 v1 C1 1 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e Gr Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e Gr Chapter 4 4 E4.8 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 114 Example 4.8 A homogeneous disk (mass m, radius r) rolls without slipping on a rough surface (Fig. 4.9). Its center of mass C is connected with the wall by a spring (spring constant k). Derive the equation of motion ϕ using r a) Newton’s 2nd Law, b) dynamic equilibrium conditions. k m C A Fig. 4.9 Solution a) The free-body diagram shows the forces that act on the disk. x ϕ W F C H N We use the coordinates x and ϕ. Then the principles of linear and angular momentum yield ←: ↑: x C: mẍc = −F − H , 0=N −W → ΘC ϕ̈ = rH N =W , where ΘC = mr2 /2 . In addition, we have the kinematic relation xc = rϕ → ẍc = rϕ̈ , and the relation Gr F = kxc → F = krϕ 115 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 Principles of Mechanics for the force F in the spring. Solving these equations for the angle ϕ yields ϕ̈ + ω 2 ϕ = 0 where ω 2 = 2k . 3m b) If we apply the dynamic equilibrium conditions, the inertial force mẍc and the pseudo moment ΘC ϕ̈ (both acting in the negative coordinate directions) have to be drawn into the free-body diagram. ΘC ϕ̈ W mẍc F C B H N Dynamic moment equilibrium about point B then yields y B: ΘC ϕ̈ + rmẍc + rF = 0 . If we introduce the kinematic relation (see a)), the force F = krϕ and ΘC = mr2 /2 we again obtain ϕ̈ + ω 2 ϕ = 0 , ω2 = 2c . 3m Gr Note that we may choose point B to be the reference point for the moment equilibrium. This is advantageous since then the lever arm of the unknown force of static friction H is zero. The mass moment of inertia must be ΘC (not ΘB !). E4.9 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 116 Example 4.9 A cylinder (mass m, radius r) rolls without slipping on a circular path (radius R); see Fig. 4.10. Derive the equation of motion using dynamic equilibrium conditions. 11111111111111 00000000000000 00000000000000 11111111111111 00000000000000 11111111111111 ϕ 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 m R 00000000000000 11111111111111 00000000000000 11111111111111 r 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 Fig. 4.10 Solution We isolate the cylinder and introduce the coordinates ϕ and ψ (angle of rotation of the cylinder). With the tangential acceleration at = (R − r)ϕ̈ (in the positive ϕ-direction) of the center of mass C and the normal acceleration an = (R − r)ϕ̇2 (directed towards the center of the circular path), the inertial forces mat (opposite to at ) and man (opposite to an ) can be drawn on the free-body diagram. ΘC ψ̈ ϕ ψ m(R − r)ϕ̇2 C B H ϕ m(R − r)ϕ̈ N mg The pseudo moment ΘC ψ̈ acts in the negative ψ-direction. Moment equilibrium about point B yields the equation of motion: y B : ΘC ψ̈ + m(R − r)ϕ̈ + mgr sin ϕ = 0 where ΘC = mr2 /2 . The system has one degree of freedom. Therefore a relation exists between the two coordinates: vC = (R − r)ϕ̇ = rψ̇ → ψ̈ = (R/r − 1)ϕ̈ . This leads to 2g ϕ̈ + sin ϕ = 0 . 3(R − r) Gr Note that the moment equilibrium may be established with respect to point B. This is advantageous since the lever arms of the unknown force of static friction H and of the inertial force m(R − r)ϕ̇2 are zero. The mass moment of inertia, however, has to be taken with respect to the center of mass C. 117 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 Principles of Mechanics Example 4.10 Two blocks of weights W1 = m1 g and W2 = m2 g are suspended by a pin-supported rope drum (moment of inertia ΘA ) as shown in Fig. 4.11. Determine the angular acceleration of the drum and the force in rope ① using dynamic equilibrium conditions. Neglect the mass of the ropes. We first introduce the coordinates x1 and x2 describing the motion of the blocks. The inertial forces −mi ẍi point in the negative xi -directions (see the freebody diagram). In addition, we have to consider the pseudo moment −ΘA ϕ̈ which acts in the negative ϕdirection. Moment equilibrium about point A then yields E4.10 ϕ A 2 Fig. 4.11 ϕ ΘA ϕ̈ A x2 x1 m1 g m2 g m1 x¨1 m2 x¨2 −r1 m1 (g + ẍ1 ) + r2 m2 (g − ẍ2 ) − ΘA ϕ̈ = 0 . x1 = r1 ϕ → ẍ1 = r1 ϕ̈ , x2 = r2 ϕ → ẍ2 = r2 ϕ̈ we obtain the angular acceleration of the drum: Gr r2 m2 − r1 m1 g. r12 m1 + r22 m2 + ΘA m2 m1 Using the kinematic relations ϕ̈ = r1 1 Solution y A: ΘA r2 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 118 In order to determine the force in rope ① we cut the rope. Force equilibrium (see the free-body diagram) yields S1 ↑: S1 − m1 g − m1 ẍ1 = 0 or r2 (r1 + r2 )m2 + ΘA S1 = m1 (g + r1 ϕ̈) = m1 g 2 . r1 m1 + r22 m2 + ΘA x1 m1 g m1 x¨1 Gr Note: For r2 m2 > r1 m1 the drum rotates clockwise, for r2 m2 < r1 m1 it rotates counterclockwise. In the special case r2 m2 = r1 m1 the system is in static equilibrium (ϕ̈ = 0). 119 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 Principles of Mechanics Example 4.11 An angled arm (mass E4.11 b m) rotates with constant angular velocity Ω about point 0 (Fig. 4.12). Calculate the bending moment, shear force and normal force as functions of position using dynamic equilibrium conditions. Ω a 0 Fig. 4.12 Solution We introduce two coordinate systems xi , zi . Then, we make a cut at the arbitrary position x1 . The acceleration of the mass element dm at the position s (distance from the left end of the arm) is given by an = rΩ2 (pointing towards point 0; note that at = 0). Therefore, this element is subjected to the inertial force dmrΩ2 (see the free-body diagram). N(x2 = 0) x1 z1 ds dm(a − s)Ω2 ds N V s M(x1 = b) M N(x1 = b) ϕ x2 V V (x1 = b) r N Ω z2 V (x2 = 0) s dmrΩ2 M(x2 = 0) M Ω With dm = m ds = µds , a+b where µ = m/(a + b) is the mass per unit length, and with the geometrical relations cos ϕ = (b − s)/r , sin ϕ = a/r Gr we can determine the stress resultants through integration (note that M ′ = V ). Normal force (0 ≤ x1 ≤ b): Z Z x1 N (x1 ) = rΩ2 cos ϕdm = Ω2 (b − s)µds 0 = µΩ2 [bs − s2 /2]|x0 1 Shear force (0 ≤ x1 ≤ b): Z Z V (x1 ) = rΩ2 sin ϕdm = → x1 0 N (x1 ) = µΩ2 (bx1 − x21 /2) . Ω2 aµds → V (x1 ) = µΩ2 ax1 . 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 120 Bending moment (0 ≤ x1 ≤ b): Z x1 M (x1 ) = V (s)ds → M (x1 ) = µΩ2 ax21 /2 . 0 The matching conditions at the corner (x1 = b, x2 = 0) are given by N0 = N (x2 = 0) = V (x1 = b) = µΩ2 ab , V0 = V (x2 = 0) = −N (x1 = b) = −µΩ2 b2 /2 , M0 = M (x2 = 0) = M (x1 = b) = µΩ2 ab2 /2 . Now we make a cut at the position x2 . The mass element dm at the position s is subjected to the inertial force dm(a − s)Ω2 . This leads to the following stress resultants: Normal force, shear force, bending moment (0 ≤ x2 ≤ a): Rx N (x2 ) = N0 + 0 2 µΩ2 (a − s)ds → N (x2 ) = µΩ2 (ab + ax2 − x22 /2) , V (x2 ) = V0 → V (x2 ) = −µΩ2 b2 /2 , M (x2 ) = M0 + x2 V0 N + | V0 | N0 → M (x2 ) = µΩ2 b2 (a − x2 )/2 . V + N0 V0 M M0 + + Gr + − 121 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 Principles of Mechanics Example 4.12 A wheel (weight W1 = m1 g, moment of inertia ΘA ) on an inclined plane is connected to a block (weight W2 = m2 g) by a rope which is guided over an ideal pulley (Fig. 4.13). The wheel rolls on the plane without slipping. Determine the acceleration of the block applying d’Alembert’s principle. Neglect the masses of the rope and the pulley. m1 , ΘA r A m2 α Fig. 4.13 Since the constraint forces (force in the rope, static friction force) need not be determined, it is advantageous to apply d’Alembert’s principle. The motion is described by the coordinates xi and ϕ. The inertial forces mi ẍi and ϕ x1 the pseudo moment ΘA ϕ̈ (acting in ΘA ϕ̈ x2 the directions opposite to the chosen A α positive coordinate directions) are m1 x¨1 m2 g shown in the figure. D’Alembert’s m1 g principle (principle of virtual work) m2 x¨2 requires that the virtual work of all forces vanishes: Solution δU + δUI = 0 → −m1 ẍ1 δx1 − m1 g sin αδx1 − ΘA ϕ̈δϕ + m2 gδx2 − m2 ẍ2 δx2 = 0 . With the kinematic relations x1 = x2 = rϕ = x → δx1 = δx2 = rδϕ = δx ẍ = ẍ = rϕ̈ = ẍ 1 2 we obtain ΘA −m1 ẍ − m1 g sin α − 2 ẍ + m2 g − m2 ẍ δx = 0 . r Gr Since δx 6= 0, the expression in the brackets must vanish. Thus, ẍ = ẍ2 = g m2 − m1 sin α . ΘA m1 + m2 + 2 r Note that ẍ < 0 for m1 sin α > m2 . In this case, the wheel rolls down the inclined plane. E4.12 E4.13 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 122 Example 4.13 Two drums are connected by a rope and carry blocks of weights m1 g and m2 g (Fig. 4.14). Drum ① is driven by the moment M0 . Determine the acceleration of block ② using d’Alembert’s principle. Neglect the mass of the ropes. ΘA ΘB r2 M0 1 r1 A r2 B m2 m1 2 Fig. 4.14 We introduce the inertial forces mi ẍi and the pseudo moments ΘA ϕ̈1 , ΘB ϕ̈2 . They act in the directions opposite to the chosen positive coordinate directions. D’Alembert’s x1 principle requires Solution δU + δUI = 0 , ϕ1 ΘA ϕ¨1 M0 A ΘB ϕ¨2 B x2 m1 g m2 g m1 x¨1 m2 x¨2 which leads to −m1 (g + ẍ1 )δx1 + m2 (g − ẍ2 )δx2 +M0 δϕ1 − ΘA ϕ̈1 δϕ1 − ΘB ϕ̈2 δϕ2 = 0 . With the kinematic relations ẍ2 r1 x1 = r1 ϕ1 ϕ̈1 = ϕ̈2 = , ẍ1 = ẍ2 , r2 r2 x2 = r2 ϕ2 → r δx 2 1 , δx1 = δx2 δϕ1 = δϕ2 = r2 r2 ϕ1 = ϕ2 Gr ϕ2 123 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 Principles of Mechanics we obtain r1 r1 −m1 g + ẍ2 + m2 (g − ẍ2 ) r2 r2 ΘB ΘA M0 − 2 ẍ2 − 2 ẍ2 δx2 = 0 . + r2 r2 r2 Since δx2 = 6 0, the expression in the curly brackets must vanish. Thus, the acceleration of block ② is Gr m1 r1 M0 1− + m2 r2 r2 m2 g ẍ2 = g . ΘA ΘB m1 r1 2 + + 1+ m2 r2 m2 r22 m2 r22 E4.14 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 124 Example 4.14 The system shown in Fig. 4.15 consists of a block (mass M ), a homogeneous disk (mass m, radius r) and two springs (spring constant k). The block moves on a frictionless surface; the m disk rolls without slipping k on the block. A force F (t) r k F (t) acts on the block. M Derive the equations of motion using Lagrange’s forno friction malism. Fig. 4.15 Solution The system has x two degrees of freedom. We 00000000000000 11111111111111 ϕ 00000000000000 11111111111111 00000000000000 11111111111111 choose the displacement x C 00000000000000 11111111111111 F (t) 00000000000000 11111111111111 of the block and the angle 00000000000000 11111111111111 00000000000000 11111111111111 of rotation ϕ of the disk to 00000000000000 11111111111111 be the generalized coordinates. The two springs are unstressed for x = 0 and ϕ = 0. The kinetic energy and the potential energy of the springs, respectively, are 2 T = M ẋ2 /2 + ΘC ϕ̇2 /2 + mvC /2 , V = kx2 /2 + k(rϕ)2 /2 . With ΘC = mr2 /2 and the kinematic relation vC = ẋ − ϕ̇ we obtain the Lagrangian L =T −V → L = M ẋ2 /2 + mr2 ϕ̇2 /4 + m(ẋ − rϕ̇)2 /2 − kx2 /2 − kr2 ϕ2 /2 . Since the force F (t) is not given from a potential, we have to apply the Lagrangian equations in the form d ∂L ∂L d ∂L ∂L − = Qx , − = Qϕ . dt ∂ ẋ ∂x dt ∂ ϕ̇ ∂ϕ Gr The generalized forces Qx and Qϕ follow from the virtual work δU of the force F (t): δU = Qx δx + Qϕ δϕ = F (t)δx → Qx = F (t) , Qϕ = 0 . 125 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 Principles of Mechanics To set up the equations of motion, the following derivatives have to be calculated: ∂L = M ẋ + m(ẋ − rϕ̇) , ∂ ẋ d ∂L = M ẍ + m(ẍ − rϕ̈) , dt ∂ ẋ ∂L = mr2 ϕ̇/2 − mr(ẋ − rϕ̇) , ∂ ϕ̇ d ∂L = mr2 ϕ̈/2 − mr(ẍ − rϕ̈) , dt ∂ ϕ̇ ∂L ∂L = −kx , = −kr2 ϕ . ∂x ∂ϕ Thus, we obtain (M + m)ẍ − mrϕ̈ + kx = F (t) , Gr 3 −mẍ + mrϕ̈ + krϕ = 0 . 2 E4.15 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 126 Fig. 4.16 shows two blocks of mass m1 and m2 which can glide on a frictionless surface. They are coupled by springs (stiffnesses k1 k3 k2 m1 m2 k1 , k2 , k3 ). Derive the equations of Fig. 4.16 motion using the Lagrange formalism. Solution The system is conservative; it has two degrees of freedom. We introduce the two coordinates x1 and x2 which describe the positions of the two blocks. They are measured from the x1 x2 equilibrium positions of the k1 k3 k2 blocks. The kinetic and the m1 m2 potential energy, respectively, are given by 1 1 T = m1 ẋ21 + m2 ẋ22 , 2 2 1 1 1 2 V = k1 x1 + k2 x22 + k3 (x2 − x1 )2 . 2 2 2 Example 4.15 Thus, the Lagrangian of the system is L=T −V = 1 1 1 1 1 m1 ẋ21 + m2 ẋ22 − k1 x21 − k2 x22 − k3 (x2 − x1 )2 . 2 2 2 2 2 To set up the Lagrange equations d ∂L ∂L d ∂L ∂L − =0, − =0 dt ∂ ẋ1 ∂x1 dt ∂ ẋ2 ∂x2 Gr the following derivatives must be calculated: ∂L d ∂L ∂L = m1 ẋ1 , = m1 ẍ1 , = −k1 x1 + k3 (x2 − x1 ) , ∂ ẋ1 dt ∂ ẋ1 ∂x1 ∂L d ∂L ∂L = m2 ẋ2 , = m2 ẍ2 , = −k2 x2 − k3 (x2 − x1 ) . ∂ ẋ2 dt ∂ ẋ2 ∂x2 127 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 Principles of Mechanics Hence, we obtain m1 ẍ1 + k1 x1 − k3 (x2 − x1 ) = 0 → m1 ẍ1 + (k1 + k3 )x1 − k3 x2 = 0 , m2 ẍ2 + k2 x2 + k3 (x2 − x1 ) = 0 → m2 ẍ2 + (k2 + k3 )x2 − k3 x1 = 0 . Gr Note: The two coupled differential equations describe the coupled free vibrations of the two blocks. In the special case of k3 = 0 the system is decoupled and we obtain two independent equations of motion for two systems, each with one degree of freedom. E4.16 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 128 Example 4.16 Two simple pen- dulums (each mass m, length l) are connected by a spring (spring constant k, unstretched length b) as shown in Fig. 4.17. Derive the equations of motion using the Lagrange formalism. l g k m l m Fig. 4.17 Solution The system has two degrees of freedom. We choose the generalized coordinates ϕ1 and ϕ2 as shown in the figure. With the kinetic and the potential energy, respectively, 0 ϕ1 ϕ2 k m T = ml2 (ϕ̇1 − ϕ̇2 )2 /2 + ml2 (ϕ̇1 + ϕ̇2 )2 /2 → T = ml2 (ϕ̇21 + ϕ̇22 ) , m 2l sin ϕ2 V = −mgl cos(ϕ1 − ϕ2 ) − mgl cos(ϕ1 + ϕ2 ) + k(2l sin ϕ2 − b)2 /2 → V = −2mgl cos ϕ1 cos ϕ2 + k(2l sin ϕ2 − b)2 /2 (zero level: point 0 and unstressed spring) the Lagrangian becomes L=T −V → L = ml2 (ϕ̇21 + ϕ̇22 ) + 2mgl cos ϕ1 cos ϕ2 − k(2l sin ϕ2 − b)2 /2 . Gr To set up the Lagrange equations d ∂L ∂L d ∂L ∂L − =0, − =0 dt ∂ ϕ̇1 ∂ϕ1 dt ∂ ϕ̇2 ∂ϕ2 129 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 Principles of Mechanics the following derivatives are needed: ∂L d ∂L = 2ml2 ϕ¨1 , = 2ml2 ϕ̇1 , ∂ ϕ̇1 dt ∂ ϕ̇1 ∂L = −2mgl sin ϕ1 cos ϕ2 , ∂ϕ1 ∂L d ∂L 2 = 2ml2 ϕ¨2 , = 2ml ϕ˙2 , ∂ ϕ̇2 dt ∂ ϕ̇2 ∂L = −2mgl cos ϕ1 sin ϕ2 − k(2l sin ϕ2 − b)2l cos ϕ2 . ∂ϕ2 This leads to the equations of motion: lϕ¨1 + g sin ϕ1 cos ϕ2 = 0 , Gr mlϕ¨2 + mg cos ϕ1 sin ϕ2 + k(2l sin ϕ2 − b) cos ϕ2 = 0 . E4.17 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 130 A disk (weight m2 g, moment of inertia Θ2 ) glides along a frictionless homogeneous bar of weight m1 g (Fig. 4.18). Find the equations of motion using the Lagrange formalism. Example 4.17 x l m2 , Θ2 ϕ m1 Fig. 4.18 Solution The system is conservative; x it has two degrees of freedom. Its po0 sition is uniquely determined by the x cos ϕ l C1 cos ϕ distance x of point C1 from pin 0 and 2 by the angle ϕ (generalized coordinaC2 ϕ tes). With the kinetic and the potential energy, respectively, o 1 m1 l 2 2 n 1 1 T = ϕ̇ + m2 [(xϕ̇)2 + ẋ2 ] + Θ2 ϕ̇22 2 3 2 2 i 1 h m1 l 2 1 = + m2 x2 + Θ2 ϕ̇2 + m2 ẋ2 , 2 3 2 h i l l V = −m1 g cos ϕ − m2 gx cos ϕ = − m1 + m2 x g cos ϕ 2 2 (zero level at 0) and the Lagrangian L = T − V , we can write down the Lagrange equations d ∂L ∂L d ∂L ∂L =0, =0. − − dt ∂ ϕ̇ ∂ϕ dt ∂ ẋ ∂x Gr To this end, the following derivatives are needed: ∂L m1 l2 ∂L l = + m2 x2 + Θ2 ϕ̇ , = − m1 + m2 x g sin ϕ , ∂ ϕ̇ 3 ∂ϕ 2 2 d ∂L m1 l ∂L = + m2 x2 + Θ2 ϕ̈ + 2m2 xẋϕ̇ , = m2 ẋ , dt ∂ ϕ̇ 3 ∂ ẋ d ∂L ∂L = m2 ẍ , = m2 xϕ̇2 + m2 g cos ϕ . dt ∂ ẋ ∂x This leads to the coupled equations of motion: m l2 l 1 + m2 x2 + Θ2 ϕ̈ + 2m2 xẋϕ̇ + m1 + m2 x g sin ϕ = 0 , 3 2 m2 ẍ − m2 xϕ̇2 − m2 g cos ϕ = 0 → ẍ − xϕ̇2 − g cos ϕ = 0 . 131 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 Principles of Mechanics A thin half-cylindrical shell of weight W = mg rolls without sliding on a flat surface (Fig. 4.19). Derive the equation of motion using the Lagrange formalism. Solution The system is conservative; it has one degree of freedom. We introduce the coordinate ϕ as shown in the figure. With the distance a = 2r/π of the center of mass C from the center M of the shell we have Example 4.18 E4.18 r m Fig. 4.19 Θ C = Θ M − a2 m = r 2 m − a2 m rϕ M a x ϕ C y = (1 − 4/π 2 )mr2 , xc = rϕ − a sin ϕ yc = a cos ϕ → → ẋc = rϕ̇ − aϕ̇ cos ϕ , ẏc = −aϕ̇ sin ϕ . Thus, the potential energy V , the kinetic energy T , the Lagrangian and the pertinent derivatives are 2 mgr(1 − cos ϕ) , π h 2 1 1 1 2 T = m(ẋ2c + ẏc2 ) + ΘC ϕ̇2 = mr2 ϕ̇2 1 − cos ϕ 2 2 2 π 2 2 i 4 2 + sin ϕ + 1 − 2 = mr2 ϕ̇2 1 − cos ϕ , π π π h i 2 2 L = T − V = mr rϕ̇2 1 − cos ϕ − g(1 − cos ϕ) , π π h i ∂L 2 = mr 2rϕ̇ 1 − cos ϕ , ∂ ϕ̇ π V = mga(1 − cos ϕ) = Gr h 4 i d ∂L 2 = mr 2rϕ̈ 1 − cos ϕ + rϕ̇2 sin ϕ , dt ∂ ϕ̇ π π h2 i 2 ∂L = mr rϕ̇2 sin ϕ − g sin ϕ . ∂ϕ π π 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 132 Introduction into d ∂L ∂L − =0 dt ∂ ϕ̇ ∂ϕ yields the equation of motion: Gr ϕ̈(π − 2 cos ϕ) + ϕ̇2 sin ϕ + g sin ϕ = 0 . r 133 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 4 Principles of Mechanics A block (mass m1 ) can move horizontally on a smooth surface (Fig. 4.20). A simple pendulum (mass m2 ) is connected to the block by a pin. Find the equations of motion using the Lagrange formalism. Example 4.19 E4.19 m1 g l m2 Fig. 4.20 Solution The system is conserva- tive; it has two degrees of freedom. We use the generalized coordinates x and ϕ as shown in the figure. With the zero-level of the potential of the force m2 g chosen at the height of the mass m1 , we have x x m1 y l cos ϕ ϕ m2 l sin ϕ V = −m2 gl cos ϕ , 1 1 T = m1 ẋ2 + m2 [(ẋ + lϕ̇ cos ϕ)2 + (lϕ̇ sin ϕ)2 ] , 2 2 1 1 L = T − V = (m1 + m2 )ẋ2 + m2 lẋϕ̇ cos ϕ + m2 l2 ϕ̇2 + m2 gl cos ϕ . 2 2 Introduction of the derivatives ∂L = m2 lẋ cos ϕ + m2 l2 ϕ̇ , ∂ ϕ̇ ∂L = −m2 lẋϕ̇ sin ϕ − m2 gl sin ϕ , ∂ϕ d ∂L = m2 lẍ cos ϕ − m2 lẋϕ̇ sin ϕ + m2 l2 ϕ̈ , dt ∂ ϕ̇ Gr ∂L ∂L = (m1 + m2 )ẋ + m2 lϕ̇ cos ϕ , =0, ∂ ẋ ∂x d ∂L = (m1 + m2 )ẍ + m2 lϕ̈ cos ϕ − m2 lϕ̇2 sin ϕ dt ∂ ẋ 4 Principles of Mechanics os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 134 into the Lagrange equations d ∂L ∂L − =0, dt ∂ ϕ̇ ∂ϕ d ∂L ∂L − =0 dt ∂ ẋ ∂x yields the coupled equations of motion ẍ cos ϕ + lϕ̈ + g sin ϕ = 0 , Gr (m1 + m2 )ẍ + m2 lϕ̈ cos ϕ − m2 lϕ̇2 sin ϕ = 0 . Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e Gr Chapter 5 5 E5.11 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 136 Example 5.11 The system in Fig 5.30 consists of three bars and a beam (with negligible masses) and a block (mass m). Determine the circular frequency of the free vertical vibrations. EA EA l EA EI m l l Fig. 5.30 Solution The system consisting of the truss and the beam is equivalent to a system consisting of two springs in parallel (both springs undergo the same elongation when the block is displaced). To determine the spring constant kB of the beam, we subject the beam to the force FB = 1 which acts at the location of the block. This force produces the deflection (see Engineering Mechanics 2: Mechanics of Materials, Table 4.3) wB = 1 · (2l)3 . 48EI FB = 1 wB ③ ② ① wT FT = 1 Thus, the spring constant is given by kB = 1 48 EI 6 EI = = 3 . 3 wB (2l) l In order to find the spring constant kT of the truss, we apply the force FT = 1 at bar ①. This force causes the displacement (see Engineering Mechanics 2: Mechanics of Materials, Section 6.3) Gr wT = X S̄ 2 li i . EA 137 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations With S̄1 = 1 , S̄2 = S̄3 = − 1√ 2, 2 l1 = l , l2 = l3 = √ 2l we obtain √2 2 √ i √ 1 h 2 l wT = 1 ·l+2· 2 l = (1 + 2) EA 2 EA → kT = 1 EA √ = . wT (1 + 2)l Now, we replace the two springs in parallel by an equivalent single spring. Its spring constant k ∗ is given by k ∗ = kB + kT = 6 EI EA √ + . l3 (1 + 2)l Gr Thus, the eigenfrequency is s r k∗ 1 EAl2 1 √ . = 6 EI + ω= m l ml 1+ 2 E5.12 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 138 Example 5.12 The system in Fig. 5.31 consists r of a homogeneous drum (mass M , radius r), a block (mass m), a spring (spring constant k) and a string (with negligible mass). The support of the drum is frictionless. Assume that there is no slip between the string and the drum. Determine the natural frequency of the sys- k tem. M m Fig. 5.31 Solution We first draw the free-body dia- gram. The position of the block is given by the coordinate x, measured from the position with an unstrechted spring. Next, we write down the equations of motion for the block ↓: mẍ = mg − S1 and for the drum y 2 B : M r ϕ̈/2 = S1 r − S2 r . In addition, we need the kinematic relation ϕ B S2 S1 S2 S1 x mg ẋ = rϕ̇ and the equation S2 = kx for the restoring force in the spring. Solving yields the differential equation for harmonic oscillations: ẍ + 2k 2mg x= . M + 2m M + 2m Gr Thus, the natural frequency is r 2k ω= . M + 2m 139 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations Example 5.13 Two drums rotate in opposite directions as shown in Fig. 5.32. They support a homogeneous board of weight W (mass m). The coefficient of kinetic friction between the drums x W and the board is µ. Show that the board unµ dergoes a harmonic vibratiΩ Ω on and determine the natural frequency. a Fig. 5.32 Solution We separate the various parts of the system. The freebody diagram shows the forces acting if the board is displaced by an amount x (the friction forces act on the drums in the opposite directions of the rotations). x R2 W N2 R1 N1 R2 a/2 R1 a/2 Thus, the equation of motion in the x-direction of the board is given by mẍ = R2 − R1 . The normal forces follow from force equilibrium in the vertical direction and from moment equilibrium: a a +x −x N1 = W 2 , N2 = W 2 . a a With the law of kinetic friction R = µN , we obtain Gr R2 − R1 = µ(N2 − N1 ) = −µmg 2x . a E5.13 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 140 The equation of motion mẍ = −µmg 2x a thus leads to the differential equation for harmonic vibrations: g ẍ + 2µ x = 0 . a The natural frequency is given by r g ω = 2µ . a Gr Note that the natural frequency does not depend on the angular velocity Ω of the drums. 141 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations Example 5.14 A homogeneous bar (weight W = mg, length l) is submerged in a viscous fluid and undergoes vibrations about point A (Fig. 5.33). The drag force Fd acting A on every point of the bar is proportional to the local velocity (proportionality factor β). ϕ l a) Derive the equation of motion. Assume small amplitudes and neglect the buoyancy. b) Calculate the value Fig. 5.33 β = β ∗ for critical damping. Solution a) We consider an arbitrary element of length dx of the bar. It is A x subjected to the drag force dFd = β v(x) dx = βxϕ̇ dx . ϕ We restrict ourselves to small amplitudes (sin ϕ ≈ ϕ). In this case the principle of angular momentum yields l ΘA ϕ̈ = −mg ϕ − 2 x A: Zl dx mg dFd l lϕ sin ϕ ≈ 2 2 βx2 ϕ̇dx . 0 Evaluation of the integral and introduction of ΘA = ml2 /3 lead to the equation of motion ϕ̈ + βl 3 g ϕ̇ + ϕ=0 m 2 l → where βl , 2m Gr ξ= ω2 = 3g . 2l ϕ̈ + 2ξ ϕ̇ + ω 2 ϕ = 0 E5.14 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 142 b) Critical damping is characterized by ξ=ω or ζ=1. Thus, Gr β∗l = 2m r 3g 2l → m β = l ∗ r 6 g . l 143 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations Example 5.15 The pendulum of a clock consists of a homogeneous rod (mass m, length l) and a homogeneous disk (mass M , radius r) whose center is located at a distance a from point A (Fig. 5.34). Assume small amplitudes and determine the natural frequency of the corresponding oscillations. Choose m = M and r ≪ a and calculate the ratio a/l which yields the maximum eigenfrequency. 11 00 A a l M m 2r Fig. 5.34 Solution We introduce the angle ϕ as shown and apply the principle of angular momentum: x A: A ΘA ϕ̈ = − mgl/2 sin ϕ − M ga sin ϕ where ΘA = ml2 /3 + M (r2 + 2a2 )/2 . ϕ mg Mg If we assume small amplitudes (sin ϕ ≈ ϕ), we can linearize the equation of motion: ϕ̈ + (ml + 2M a)g ϕ=0. 2ΘA Hence, the eigenfrequency is obtained as s (ml + 2M a)g ω= . 2ΘA In the special case of m = M and r ≪ a the natural frequency simplifies to r 3l + 6a ω= g. 2l2 + 6a2 Gr The ratio a/l which yields the maximum eigenfrequency is found by setting the derivative dω/da equal to zero: ! r dω 1 7 = 0 → a/l = −1 . da 2 3 E5.15 E5.16 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 144 111 000 000 111 Example 5.16 The system in Fig 5.35 consists of a homogeneous pulley (mass M , radius r), a block (mass m) and a spring (spring constant k). Determine the equation of motion for the block and its solution for the initial conditions x(0) = 0, v(0) = v0 . Neglect the mass of the string and any lateral motion. k r M A 111 000 m Fig. 5.35 Solution We separate the pulley and the block and measure the displacements x of the block and xA of point A from the position of equilibrium. With this choice, we do not have to consider the weights M g and mg in the freebody diagram. Thus, the equations of motion are ① ↓ : ② ↓ : M ẍA = S1 + S2 − kxA , x A : ΘA ϕ̈ = rS1 − rS2 . mẍ = −S1 , ϕ S1 x , 2 2rϕ = x → ẍA = ẍ , 2 kxA S2 x 1 xA If we use the kinematic relations (Π=: b instantaneous center of rotation, see the figure) xA = 2 xA Π x ϕ̈ = ẍ 2r Gr and ΘA = M r2 /2 we can solve the equations of motion for x and obtain v u k k ẍ + x=0 → ω=u . t 3 3 4m + M 4m + M 2 2 145 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations The general solution of this differential equation is given by x(t) = A cos ωt + B sin ωt . The initial conditions x(0) = 0 → A=0, lead to Gr x(t) = v0 sin ωt . ω ẋ(0) = v0 → B= v0 ω E5.17 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 146 Example 5.17 A wheel (mass m, radius r) rolls without slipping on a circular path (Fig. 5.36). The mass of the rod (length l) can be neglected; the joints are g frictionless. Derive the equation of l motion and determine the natural frequency of small A ϕ oscillations. m r Fig. 5.36 Solution We apply conserva- tion of energy T + V = const ϕ to derive the equation of motion. The kinetic energy of the rolling wheel is given by (see the figure) vc C ψ T = mvc2 /2 + ΘC ψ̇ 2 /2 , the potential energy is (zero-level at the height of point A) V = −mgl cos ϕ . With the mass moment of inertia ΘC = mr2 /2 and the kinematic relations vc = lϕ̇ , vc = rψ̇ → lϕ̇ = rψ̇ the kinetic energy can be written as T = 3ml2 ϕ̇2 /4 . Introduction into the expression for conservation of energy gives 3lϕ̇2 /4 − g cos ϕ = const . Gr Differentiation yields 3 lϕ̇ϕ̈ + g ϕ̇ sin ϕ = 0 2 → ϕ̈ + 2g sin ϕ = 0 . 3l 147 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations If we restrict ourselves to small amplitudes (sin ϕ ≈ ϕ) this equation reduces to 2g ϕ̈ + ϕ = 0 . 3l Gr Thus, the natural frequency is obtained as r 2g ω= . 3l E5.18 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 148 Example 5.18 The simple pendulum in Fig. 5.37 is attached to a spring (spring constant k) and a dashpot (damping coefficient d). a) Determine the maximum value of the damping coefficient d so that the system undergoes vibrations. a Assume small amplitudes. b) Find the damping ratio ζ so that d the amplitude is reduced to 1/10 a of its initial value after 10 full cyk cles. Calculate the corresponding m period Td . 1010 10 1010 01 Fig. 5.37 Solution a) The equation of motion A follows from the principle of angular momentum (rotation about point A; small amplitudes: sin ϕ ≈ ϕ, cos ϕ ≈ 1): x A: ϕ Fd Fk ΘA ϕ̈ = −Fd a − Fk 2a − mg2aϕ . With mg ΘA = m (2a)2 , Fd = da ϕ̇ , Fk = k2a ϕ we obtain ϕ̈ + where Gr ξ= k d g ϕ̇ + + ϕ=0 4m m 2a d , 8m ω2 = k g + . m 2a → ϕ̈ + 2ξ ϕ̇ + ω 2 ϕ = 0 , 149 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations In order to have oscillations, the system must be underdamped: r k g d < + ζ <1 → ξ<ω → 8m m 2a r gm2 . → d < 8 km + 2a b) The necessary damping ratio follows with xn+10 = xn /10 from the logarithmic decrement: 2πζ xn 10 p = ln 10 = ln xn+10 1 − ζ2 v u 1 → ζ=u = 0.037 . t 20π 2 +1 ln 10 This leads to the period Gr 2π 2π Td = p ≈ = 2π 2 ω ω 1−ζ s 2am . 2ak + gm E5.19 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 150 Example 5.19 The structure in Fig. 5.38 consists of an elastic beam (flexural rigidity EI, axial rigidity EA → ∞, negligible mass) and three rigid bars (with negligible masses). The block (mass m) is suspended from a spring (spring constant k). Determine the eigenfrequency of the vertical oscillations of the block. 1010 1010 010 1 a a a a k m Fig. 5.38 Solution We reduce the structure to an equivalent system of a spring (spring constant k ∗ , see the figure) and a mass. To this end we first replace the beam and the bars by a spring with the spring constant kB . We can determine kB if we subject the beam to a force F which acts at the free end. This force produces the deflection w= 1111 0000 k∗ m F a3 EI (see Engineering Mechanics 2: Mechanics of Materials, Example 6.22). The spring constant is obtained from kB = F w → kB = EI . a3 Gr The displacement of the block is the sum of the elongations of the springs with spring constants k and kB . Therefore, the beam/bars and the given spring act as springs in series. Thus, the spring constant k ∗ of the equivalent system follows from 1 1 1 = + k∗ kB k → k∗ = kEI . ka3 + EI This yields the eigenfrequency s r k∗ kEI ω= → ω= . m (ka3 + EI)m 151 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations Example 5.20 A rod (length l, with negligible mass) is elastically supported at point A (Fig. 5.39). The rotational spring (spring constant kT ) is unstretched for ϕ = 0. The rod carries a point mass m at its free end. A kT Derive the equation of motion. g Determine the spring constant so ϕ that ϕ = π/6 is an equilibrium posim tion. Calculate the natural frequenl cy of small oscillations about this Fig. 5.39 equilibrium position. Solution To derive the equation of motion we apply the principle of angular momentum: y A: 0110 MA ϕ ΘA ϕ̈ = mgl cos ϕ − MA . mg With the mass moment of inertia ΘA = ml2 and the restoring moment MA = kT ϕ we obtain ϕ̈ = g kT cos ϕ − ϕ. l ml2 Since ϕ = ϕ0 = π/6 is a position of equilibrium, the condition ϕ̈(π/6) = √ 0 leads to the required spring constant (note that cos(π/6) = 3/2): √ √ 3 g π kT 3 3 − = 0 → kT = mgl . 2 l 6 ml2 π Now we consider small oscillations about this position of equilibrium. We assume that ϕ = ϕ0 + ψ with |ψ| ≪ 1 . Introduction into the equation of motion yields ψ̈ = g kT cos(ϕ0 + ψ) − (ϕ0 + ψ) . l ml2 Gr We use the trigonometric relation cos(ϕ0 + ψ) = cos ϕ0 cos √ ψ − sin ϕ0 sin ψ 3 1 → cos(ϕ0 + ψ) = cos ψ − sin ψ 2 2 E5.20 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 152 and linearize (cos ψ ≈ 1, sin ψ ≈ ψ) to obtain √ kT 3 g π kT 1g ψ+ − ψ̈ = − ψ − 2l ml2 2 l 6 ml2 √ ! 1 3 3 g → ψ̈ + + ψ=0. 2 π l Gr Thus, the natural frequency is given by v √ ! u u 1 3 3 g t + . ω= 2 π l 153 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations Example 5.21 A single story frame E5.21 consists of two rigid columns (with m negligible masses), a rigid beam of mass m and a spring-dashpot sysd tem as shown in Fig. 5.40. The ground is forced to vibration by an k earthquake; the acceleration üE = 45◦ b0 cos Ωt is known from measurements. üE Determine the maximum ampliFig. 5.40 tude of the steady state vibrations. Assume that the system is underdamped and that the vibrations have small amplitudes. 00000 11111 1111 0000 00000 0000 11111 1111 Solution We assume that the amplitudes of the vibrations are small. Then the elongation of the diagonal is obtained as √ 2 ∆= (x − uE ) . 2 The elongation produces the force x − uE m x F ∆ 45◦ uE 1111111111 0000000000 ˙ F = k∆+ d∆ in the spring-dashpot system (see the figure). The equation of motion of the beam is given by √ 2 →: mẍ = −F 2 d k → mẍ + (ẋ − u̇E ) + (x − uE ) = 0 . 2 2 Thus, the relative displacement y = x − uE is described by d k ẏ + y = m b0 cos Ωt 2 2 2ζ 1 → ÿ + ẏ + y = y0 cos Ωt , ω2 ω Gr mÿ + where ω2 = k , 2m ζ= d 2 r 1 , 2k m y0 = 2m b0 . k 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 154 Gr The maximum amplitude A is obtained for η = Ω/ω ≈ 1 (resonance!). In the case of small damping (ζ ≪ 1) we obtain s √ b 0 m3 y0 A = y0 Vmax ≈ =2 2 . 2ζ d k 155 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations The undamped system in Fig. 5.41 consists of a block (mass m = 4 kg) and a spring (spring constant k = 1 N/m). The block is subjected to a force F (t). The initial conditions x(0) = x0 = 1 m, ẋ(0) = 0 and the response t t−T ht − T i0 x(t) = x0 cos + 20 1 − cos 2t0 2t0 Example 5.22 x(t) to the excitation are given. Here, t0 = 1 s, T = 5 s and ht − T i0 = 0 for t < T and ht − T i0 = 1 for t > T. Calculate the force F (t). k m F (t) Fig. 5.41 Solution First we calculate the force F for t < T . Then ht−T i0 = 0 and the response is given by x(t) = x0 cos t . 2t0 The unknown force follows from the equation of motion: F = mẍ + kx . With ẋ = − x0 t sin 2t0 2t0 and ẍ = − x0 t cos 2 4t0 2t0 we obtain mx0 t F (t) = − 2 + kx0 cos . 4t0 2t0 Inserting the given numerical values of the parameters leads to Gr F (t) ≡ 0 for t<T . E5.22 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 156 Now we consider the case t > T . Then ht − T i0 = 1 and the response follows as t−T t + 20 1 − cos x(t) = x0 cos 2t0 2t0 x0 t t−T → ẋ = − sin + 20 sin 2t0 2t0 2t0 x0 t t−T . → ẍ = 2 − cos + 20 cos 4t0 2t0 2t0 Introduction into the equation of motion yields mx0 t F (t) = − 2 + kx0 cos 4t0 2t0 20mx0 t−T + − 20kx + 20kx0 0 cos 4t20 2t0 → F (t) ≡ 20 N for t > T . F (t) [N] 30 20 10 Gr 5 10 t[s] 157 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations Example 5.23 A simplified model of a car (mass m) is given by a spring-mass system (Fig. 5.42). The car drives with constant velocity v0 over an uneven surface in the form of a sine function (amplitude U0 , wavelength L). v0 a) Derive the equation of motion and determine the forcing m x U0 frequency Ω. c b) Find the amplitude of the vertical vibrations as a function of the velocity v0 . L c) Calculate the critical velocity Fig. 5.42 vc (resonance!). Solution a) We denote the verti- cal displacement of the car by x, the uneven surface is described by u. Then Newton’s law reads ↑: m x u mẍ = −k (x − u) . With the position of the car, s = v0 t, in the horizontal direction we obtain u = U0 cos u L U0 s 2πs 2πv0 t = U0 cos = U0 cos Ωt . L L Thus, m ẍ + k x = k U0 cos Ωt with Ω= 2πv0 . L b) We assume the solution of the equation of motion to be of the form of the right-hand side: x = x0 cos Ωt. This leads to the amplitude of the steady state vibrations: x0 = U0 U0 = Ω2 4π 2 v02 m 1− 2 1− ω L2 k Gr where ω 2 = k/m. c) Resonance occurs for Ω = ω: 4π 2 vc2 m =1 L2 k → L vc = 2π r k . m E5.23 E5.24 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 158 Example 5.24 A homogeneous wheel (mass m) is attached to a spring (spring constant k). The x wheel rolls without slipping on m a rough surface which moves k C r according to the function u = µ0 u u0 cosΩt (Fig. 5.43). a) Determine the amplitude of Fig. 5.43 the steady state vibrations. b) Calculate the coefficient µ0 of static friction which is necessary to prevent slipping. 00 11 00 11 00 11 00 11 Solution The equations of mo- tion for the wheel are given by x ↑ : →: y C : 0 = N − mg , (a) m ẍ = −k x + H , (b) ΘC ϕ̈ = −r H . mg kx C (c) ϕ H N With the kinematic relation ẍ = ü + rϕ̈ = −u0 Ω2 cos Ωt + rϕ̈ → x = u + rϕ r and ΘC = mr2 /2 we obtain from (b) and (c) the differential equation for forced vibrations: ẍ + 2 k 1 x = − u0 Ω2 cos Ωt . 3 m 3 a) We assume the solution to be of the form of the right-hand side: x = x0 cos Ωt. This leads to the amplitude of the steady state vibrations: u0 |x0 | = . 2 k 3 − 1 3 m Ω2 b) The condition |H|max ≤ µ0 N for static friction and (a), (b) yield the required coefficient of static friction: Gr |H|max µ0 ≥ = N → ẍ + µ0 ≥ k x m g u 0 Ω2 3g max = −x0 Ω2 + k −1 m Ω2 . 2 k − 1 3 m Ω2 g k x0 m 159 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations Example 5.25 A small homogeneous disk (mass m, radius r) is at- tached to a large homogeneous disk (mass M , radius R) as shown in Fig. 5.44. The torsion spring (spring constant kT ) is unstretg M A R ched in the position shown. kT Determine the eigenfrequency a r of the oscillations. Assume small amplitudes. m Fig. 5.44 Solution We apply the principle of angular momentum to derive the equation of motion: x A: kT ϕ ΘA ϕ̈ = −kT ϕ − mga sin ϕ . In the case of small amplitudes (sin ϕ ≈ ϕ) this equation reduces to ϕ̈ + ω 2 ϕ = 0 with ω2 = A ϕ mg kT + mga . ΘA Inserting the mass moment of inertia M R2 mr2 2 ΘA = + + ma 2 2 we can write the eigenfrequency in the form v u kT + mga u ω=u . t1 r2 M R2 + m( + a2 ) 2 2 Note that the problem can also be solved with the aid of the conservation of energy (we choose V = 0 for ϕ = 0): T + V = const → 1 1 ΘA ϕ̇2 + kT ϕ2 + mga(1 − cos ϕ) = const. 2 2 Differentiation yields Gr ΘA ϕ̇ϕ̈ + kT ϕϕ̇ + mga sin ϕ ϕ̇ = 0 . With sin ϕ ≈ ϕ and ϕ̇ 6= 0 we obtain the same result as above. E5.25 E5.26 5 Vibrations 1111 0000 000011 1111 00 00 11 00 11 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 160 Example 5.26 The systems ➀ and ➁ in Fig. 5.45 consist of two beams (negligible masses, flexural rigidity EI), a spring (spring constant k) and a box (mass m). Determine the spring constants k ∗ of the equivalent springs for the two systems. Solution We reduce both sys- tems to the equivalent simple systems of a mass and a spring. 00 11 00 11 00 11 00 11 00 11 00 11 EI k EI 1 m a EI 2a EI k m 11 00 00 11 2 Fig. 5.45 11 00 00 11 1 l , EI w In system ①, the three “springs” are attached to the mass. Therefore, they undergo the same deflection when the box is displaced: they act as springs in parallel. The equivalent spring constant k ∗ is the sum of the individual spring constants: X k∗ = ki . We obtain the spring constants kL and kR of the right and the left beam, respectively, if we subject the cantilevers to a force 1 at their free ends. The corresponding deflections are w= 1 · l3 3EI Gr (see Engineering Mechanics 2: Mechanics of Materials, Section 4.5) which leads to 161 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 5 Vibrations kL = 1 3EI , = wL (2a)3 kR = 1 3EI = 3 . wR a Thus, k ∗ = kL + kR + k = 27 EI 27EI + 8ka3 + k = . 8 a3 8a3 0000 1111 111 000 111 0000 111 000 000000 111 0001111 111 kL k∗ k̄ kR k k m m m Now we consider system ②. Here, the two beams act as springs in parallel with equivalent spring constant k̄. This then acts in series with given spring (spring constant k). Hence, 27 EI , 8 a3 1 1 8a3 1 1 + = = + k∗ 27EI k k̄ k k̄ = kL + kR = → k∗ = 27EIk = 27EI + 8ka3 27EI 8a3 + 27 EI k . Gr Note that the stiffness of system ② is smaller than the one of system ①. Therefore it vibrates with a smaller frequency. E5.27 5 Vibrations os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 162 Example 5.27 A homogeneous wheel (mass m, moment of inertia ΘC , radius r) rolls without slipping on a rough beam (mass M ). m, ΘC The beam moves without k C friction on roller supports r (Fig. 5.46). Determine the natural freM quency of the system. 1010 1010 111 000 Solution We separate the wheel and the beam and introduce the coordinates x1 , x2 and ϕ (see the figure). The coordinates are measured from the position of equilibrium. Then the equations of motion are ① →: y C : ② Fig. 5.46 x1 mg 1 kx1 C r ϕ H m ẍ1 = −kx1 − H , ΘC ϕ̈ = r H , 11 00 00 11 N H x2 A → : M ẍ2 = H . 2 If we use the kinematic relation x1 = x2 + rϕ → ẋ1 = ẋ2 + rϕ̇ and solve for x1 we obtain k ẍ1 + m+ M 1 + M r2 /ΘC x1 = 0 . Gr Thus, the natural frequency is given by v u k u . ω=u M t m+ 1 + M r2 /ΘC → ẍ1 = ẍ2 + rϕ̈ B Non-Inertial Reference Frames os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e Gr Chapter 6 6 E6.5 6 Non-Inertial Reference Frames 1111111 0000000 0000000 1111111 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 164 Example 6.5 Point A of the simple pendulum (mass m, length l) in Fig. 6.8 moves with a constant acceleration a0 to the right. Derive the equation of motion. a0 A g ϕ l m Fig. 6.8 Solution We introduce the ξ, η-co- ordinate system as shown in the figure. It is a translating coordinate system with point A as the origin. The equation of motion in the moving system is η eη A eξ ξ ϕ mar = F + Ff . S The (real) force F acting at the mass is given by m W = mg F = − S sin ϕ eξ + (S cos ϕ − mg)eη and the fictitious force Ff is Ff = − maf = − ma0 eξ . Note that the Coriolis force is zero since ω = 0. The components of the relative acceleration ar follow from the coordinates of the point mass in the moving system through differentiation: ξ = l sin ϕ, η = − l cos ϕ, η̇ = lϕ̇ sin ϕ, ξ¨ = lϕ̈ cos ϕ − lϕ̇2 sin ϕ, η̈ = lϕ̈ sin ϕ + lϕ̇2 cos ϕ . Gr ξ˙ = lϕ̇ cos ϕ, 165 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 6 Non-Inertial Reference Frames This yields the relative acceleration ar = ξ¨ eξ + η̈ eη = (lϕ̈ cos ϕ − lϕ̇2 sin ϕ)eξ +(lϕ̈ sin ϕ + lϕ̇2 cos ϕ)eη . Introduction into the equation of motion leads to the components of the equation of motion in the direction of the axes ξ and η: m (lϕ̈ cos ϕ − lϕ̇2 sin ϕ) = −S sin ϕ − ma0 , m (lϕ̈ sin ϕ + lϕ̇2 cos ϕ) = S cos ϕ − mg. These are two equations for the unknowns ϕ and S. Solving for ϕ yields the equation of motion lϕ̈ + g sin ϕ + a0 cos ϕ = 0 . Gr Note that the position ϕ0 = − arctan a0 /g is obtained for ϕ̈ = 0. The pendulum oscillates about this position for ϕ̈ 6= 0. E6.6 6 Non-Inertial Reference Frames os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 166 Example 6.6 The two disks in 11 00 1010 00 11 Fig. 6.9 rotate with constant angular velocities Ω and ω about their respective axes. Determine the absolute acceleration of point P at the instant shown. P ϕ r ω z y x 101010 101010 Ω Solution We describe the mo- a Fig. 6.9 tion of point P in a coordinate system x, y, z which is fixed to the large disk. The absolute acceleration of P is P r ϕ z aP = af + ar + ac , where the acceleration of the reference frame and the relative acceleration are given by 0 2 af = −(a + r cos ϕ)Ω , 0 y x 0 2 ar = −rω cos ϕ . −rω 2 sin ϕ We also write the angular velocity of the reference frame and the relative velocity as column vectors: 0 0 Ω= 0 , v r = −rω sin ϕ Gr Ω rω cos ϕ 167 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 6 Non-Inertial Reference Frames and we calculate the Coriolis acceleration: 2rωΩ sin ϕ . ac = 2Ω × v r → ac = 0 0 Combining yields 2rωΩ sin ϕ Gr 2 2 aP = −(a + r cos ϕ)Ω − rω cos ϕ . −rω 2 sin ϕ E6.7 6 Non-Inertial Reference Frames os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 168 Example 6.7 A horizontal circular platform (radius r) rotates with constant angular velocity Ω (Fig. 6.10). A block (mass m) is locked in a frictionless slot at a distance a from the center of the platform. At r a time t = 0 the block is released. Determine the velocity vr of the m block relative to the platform when it reaches the rim of the platform. Ω Fig. 6.10 Solution We describe the motion of the block in a coordinate system x, y which is fixed to the platform. The absolute acceleration of the block is given by x y a aB = af + ar + ac . Here, the acceleration of the reference frame, the relative acceleration and the Coriolis acceleration are " # " # " # −xΩ2 ẍ 0 af = , ar = , ac = . 0 0 2Ωẋ Thus, the absolute acceleration becomes " # ẍ − Ω2 x aB = . 2Ωẋ The equation of motion for the block is Gr maB = F , 169 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 6 Non-Inertial Reference Frames where the force F (which is exerted from the slot on the block) is " # 0 F = . Fy We now write down the x-component of the equation of motion: m(ẍ − Ω2 x) = 0 → ẍ − Ω2 x = 0 . The general solution of this differential equation is given by x(t) = A cosh Ωt + B sinh Ωt . With the initial conditions x(0) = a ẋ(0) = 0 → → A=a, B=0 we obtain x(t) = a cosh Ωt . When the block reaches the rim of the platform, the condition x(tR ) = r → cosh ΩtR = r/a Gr is satisfied. Thus the relative velocity is (note that cosh2 x − sinh2 x = 1) p ẋ(tR ) = aΩ sinh ΩtR → ẋ(tR ) = Ω r2 − a2 . E6.8 6 Non-Inertial Reference Frames os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 170 Example 6.8 A simple pendulum is attached to point 0 of a circular disk (Fig. 6.11). The disk rotates with a constant angular velocity Ω; the pendulum oscillates in the horizontal plane. Determine the circular frequency of the oscillations. Assume small amplitudes and neglect the weight of the mass. Ω r m l 0 00000 11111 00000 11111 00000 11111 Fig. 6.11 Solution We introduce a rotating ξ, η, ζ-coordinate system. Then Ω = Ωeζ , Ω̇ = 0 , a0 = r̈ 0 = −rΩ2 eξ , r 0P = l cos ϕ eξ + l sin ϕ eη . The relative velocity can be expressed by the relative angular velocity ϕ∗ (∗ : time derivative relative to the moving frame): vr = lϕ∗ → Ω vr η ϕ r 0 v r = −lϕ∗ sin ϕ eξ + lϕ∗ cos ϕ eη . Thus, the fictitious forces F f and F c are F f = − ma0 − mΩ × (Ω × r 0P ) = m(rΩ2 + lΩ2 cos ϕ)eξ + mΩ2 l sin ϕ eη , F c = − 2mΩ × v r = 2mΩlϕ∗ (eξ cos ϕ + eη sin ϕ) . Gr P r0P ξ 171 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 6 Non-Inertial Reference Frames With the tangential relative acceleration art = lϕ∗∗ the equation of motion in the tangential direction is obtained as տ: mlϕ∗∗ = m(lΩ2 + 2lϕ∗ Ω) sin ϕ cos ϕ − m[rΩ2 + (lΩ2 + 2lϕ∗ Ω) cos ϕ] sin ϕ mΩ2 l sin ϕ = − mrΩ2 sin ϕ . We assume small amplitudes (sin ϕ ≈ ϕ). This yields rΩ2 ϕ∗∗ + ϕ=0. l art η S 2mlϕ∗ Ω sin ϕ m(rΩ2 +lΩ2 cos ϕ) m ϕ 2mlϕ∗ Ω cos ϕ ξ Gr Hence, the circular frequency of the oscillations is q ω = r/l Ω . E6.9 6 Non-Inertial Reference Frames os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 172 Example 6.9 A drum rotates with angular velocity ω about point B (Fig. 6.12). Pin C is fixed to the drum; it moves in the slot of link AD. Determine the angular velocity A ωAD of link AD and the velocity vr of the pin relative to the link at the instant shown. D C 3l B 4l 111111111 000000000 ω Fig. 6.12 Solution We use the rotating co- ordinate system x, y as shown in the figure. The (absolute) velocity of pin C is given by " # cos β . v C = 3lω y − sin β x C a 3ℓ β A With the geometrical relations p a = 16l2 + 9l2 = 5l , sin β = 3/5 , 4ℓ cos β = 4/5 we can write " # 3lω 4 vC = . 5 −3 The velocity of the reference frame at point C and the velocity of pin C relative to the moving frame are " # " # 0 1 v f = β̇l , v r = vr . 5 0 With vC = vf + vr → " # " # " # 3lω 4 0 1 = β̇l + vr , 5 −3 5 0 Gr we finally obtain ωAD 9ω = β̇ = − , 25 " # 12 1 vr = ωl . 5 0 173 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 6 Non-Inertial Reference Frames Example 6.10 A point P moves along a circular path (radius r) on a platform with a constant relative velocity vr (Fig. 6.13). The platform rotates with a convr stant angular velocity ω about P point A. The eccentricity e is giω r ven. A e 0 Determine the relative, fixed frame-, Coriolis, and absolute accelerations of P . Fig. 6.13 Solution We introduce the coor- dinate system ξ, η, ζ as shown in the figure. Its origin is located at the center 0 of the platform; it rotates with the platform. Thus point P undergoes a circular motion relative to this system. With the magnitude ar = vr2 /r of the relative acceleration and its direction (from P to 0) we can write v2 ar = − r (eξ cos ϕ + eη sin ϕ) . r η vr ω A r 0P e 0 P ϕ ξ With ω = ωeζ , ω̇ = 0 , r0P = eξ r cos ϕ + eη r sin ϕ , v r = vr (−eξ sin ϕ + eη cos ϕ) , r̈ 0 = a0 = −e ω 2 eξ we obtain af = a0 + ω × (ω × r0P ) = −e ω 2 eξ + rω 2 [eζ × (eζ × eξ cos ϕ) + eζ × (eζ × eη sin ϕ)] = −(e + r cos ϕ)ω 2 eξ − rω 2 sin ϕeη , Gr ac = 2ω × v r = 2ωvr [eζ × (−eξ sin ϕ) + eζ × eη cos ϕ] = −2ωvr (eξ cos ϕ + eη sin ϕ) . E6.10 6 Non-Inertial Reference Frames os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 174 Thus, the absolute acceleration is found as a = af + ar + ac vr2 v2 + 2ωvr ) cos ϕ]eξ − [rω 2 + r + 2ωvr ] sin ϕ eη r r vr 2 vr 2 2 = −[eω + r(ω + ) cos ϕ]eξ − r(ω + ) sin ϕ eη . r r Gr = −[eω 2 + (rω 2 + 175 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 6 Non-Inertial Reference Frames Example 6.11 A circular ring (ra- dius r) rotates with constant angular velocity Ω about the xaxis (Fig. 6.14). A point mass m moves without friction inside the ring. Derive the equations of motion and determine the equilibrium positions of the point mass relative to the ring. E6.11 Ω 01101010 m ϕ x z y g r 11 00 01 Fig. 6.14 Solution We describe the motion of the point mass in the x, y, zcoordinate system (see the figure) which rotates with the ring. The absolute acceleration a of the point mass is given by a = af + ar + ac , where the acceleration of the reference frame and the relative acceleration are 0 −rϕ̇2 cos ϕ − rϕ̈ sin ϕ 2 2 af = −rΩ sin ϕ , ar = −rϕ̇ sin ϕ + rϕ̈ cos ϕ . 0 0 With the angular velocity of the ring and the relative velocity Ω −rϕ̇ sin ϕ Ω= 0 , v r = rϕ̇ cos ϕ 0 0 we can calculate the Coriolis acceleration 0 . ac = 2Ω × v r → ac = 0 2rΩϕ̇ cos ϕ Gr The equation of motion is given by ma = W + N , 6 Non-Inertial Reference Frames os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 176 where −mg W = 0 0 is the weight of the point mass and N x N = Ny with Ny /Nx = tan ϕ Nz is the force exerted from the ring on the point mass. Now, we can write down the components of the equation of motion: −m(rϕ̇2 cos ϕ + rϕ̈ sin ϕ) = −mg + Nx , −m(rϕ̇2 sin ϕ − rϕ̈ cos ϕ + rΩ2 sin ϕ) = Nx tan ϕ , 2mrΩϕ̇ cos ϕ = Nz . Positions of equilibrium relative to the ring are characterized by g ϕ̇ = 0 , ϕ̈ = 0 → rΩ2 + sin ϕ = 0 . cos ϕ This yields ϕ1 = 0 , ϕ2 = π , ϕ3,4 = π ± arccos Gr The positions ϕ3,4 exist only for Ω2 > g/r. g . rΩ2 177 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 6 Non-Inertial Reference Frames Example 6.12 A point P moves on a square plate along a circular path (radius r) with a constant relative velocity vr . The plate moves horizontally with the constant acceleration a0 (Fig. 6.15). Determine the magnitude of the absolute acceleration of P . 11111111111 00000000000 vr P r a0 1111111111 0000000000 0000000000 1111111111 Fig. 6.15 Solution The components of the relative velocity in the moving reference frame ξ, η are η vr P y fixed x vrξ = ξ ∗ = −vr sin ϕ , E6.12 r 0 ϕ ξ vrη = η ∗ = vr cos ϕ (∗: time derivative relative to the moving frame). Differentiation with respect to the moving frame leads to the components of the relative acceleration (note: rϕ∗ = vr ): vr2 cos ϕ , r v2 = −vr ϕ∗ sin ϕ = − r sin ϕ . r arξ = ξ ∗∗ = −vr ϕ∗ cos ϕ = − arη = η ∗∗ Since the reference frame undergoes a translation, the absolute acceleration is given by ax = a0 + arξ = a0 − ay = arη = − vr2 sin ϕ . r It has the magnitude r q Gr a= vr2 cos ϕ , r a2x + a2y = a20 + vr4 vr2 − 2a cos ϕ . 0 r2 r E6.13 6 Non-Inertial Reference Frames os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 178 Example 6.13 A crane starts to move from rest with a constant acceleration b0 along a straight track. At the same time, the jib begins to rotate with constant angubc 0 lar velocity ω, and the trolley on the jib begins to move towards point 0 with constant relative acceleration bc ω (Fig. 6.16). The initial positions of the b0 jib and of the trolley are given by ϕ0 and s0 . bc s Determine the absolute velocity and the absolute acceleration of the trolley ϕ as functions of the time t. b0 1111111 0000000 Fig. 6.16 Solution We use the fixed coordinate system x, y, z, where the x-axis coincides with the track. In addition, we introduce the rotating coor- y dinate system ξ, η, ζ, where the ξ-axis rotates with the jib. Then, the general equations for the absolute velocity z and the absolute acceleration are η ω 0 P r 0P ξ ϕ ζ x v = v 0 + ω × r0P + v r , a = a0 + ω̇ × r 0P + ω × (ω × r 0P ) + 2 ω × v r + ar . We measure the time t from the beginning of the motion. Then, making use of the given accelerations, angular velocity and initial conditions, we obtain a 0 = b 0 ex ω = ωeζ , ar = −bc eξ → v 0 = b 0 t ex , ω̇ = 0 , → v r = −bc t eξ → r 0P = (− 21 bc t2 + s0 )eξ , and ω × r 0P = ω(− 21 bc t2 + s0 )eη , Gr ω × (ω × r0P ) = −ω 2 (− 12 bc t2 2 ω × v r = −2 ω bc t eη , + s0 )eξ . 179 os En s, Ha gin ug ee er, rin g M Schr ö e Sp cha der, rin ge nics Wall ,G r2 3, 01 Dy ov 3 na idje mi cs e 6 Non-Inertial Reference Frames With the relation ex = eξ cos ϕ − eη sin ϕ, where ϕ = ϕ0 + ω t, we finally obtain v = [b0 t cos ϕ − bc t]eξ + [−b0 t sin ϕ + ω(− 21 bc t2 + s0 )]eη , Gr a = [b0 cos ϕ − ω 2 (− 12 bc t2 + s0 ) − bc ]eξ − [b0 sin ϕ + 2 ω bc t]eη .
