Dynamics - Midterm Exam Type 1

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Candace James
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1 Dynaics - Midter Exa Type 1 1. Two particles of ass and 2 slide on two vertical sooth uides. They are connected to each other and to the ceilin by three sprins of equal stiffness and of zero unstretched lenth. Gravity is actin downwards, as shown. The syste is released fro rest at the position x 1 = x 2 = 0 at tie t 0. Denote with R 1 and R 2 the reaction forces exerted by the left and the riht uides, respectively, on the particles. Which of the followin stateents ust be true for t > t 0? (a) R 1 = R 2 = 0 L 0 x 1 x 2 2 Sooth uides (b) R 1 < R 2 (c) R 1 = R 2 = (x 2 x 1 ) 2 + L 2 0 (d) R 1 > R 2 (e) R 1 = R 2 Solution: The forces exerted on each ass by the uides are only in the horizontal direction. The other force present in the horizontal direction is the sprin force provided by the sprin connectin the two asses, which is equal and opposite on each ass. Therefore, if one applies the linear oentu principle on each ass in the horizontal direction (alon which no otion occurs), it ust hold that R 1 = R 2. 1

2 2. A particle of ass oves in the 3D space spanned by the unit vectors [e x e y e z ]. It is subject to the forces F 1 = aze z, with a > 0, and F 2. At tie t = t 1, the particle is at rest at point A = (1, 1, 1) and at tie t = t 2, the particle reaches point B = (5, 2, 3) with velocity v 2 = 3v 2 e x. What is the wor W F 2 AB done by the force F 2 between points A and B? (a) W F 2 AB = 9 2 v 2 a (b) W F 2 AB = B A F 1 ds (c) W F 2 AB = 15 2 v (d) W F 2 AB = 0 (e) W F 2 AB = 9 2 v a Solution: By applyin the wor-enery principle, one can write B A (F 1 +F 2 ) ds = T B T A B A azdz+w F 2 AB = 9 2 v2 2 W F 2 AB = 9 2 v2 2 +4a 2

3 3. A particle of ass is connected to a slider A by a assless rod of lenth L. The slider oves on a sooth, vertical uide, with its position is iven by x A (t). Gravity acts downwards, as shown. Denote with θ the anle that the bar encloses with the vertical axis. What is the correct equation of otion of the syste? A x y (a) L θ + sin θ = 0 (b) θ + ẍ A θ = 0 L (c) L θ + ( ẍ A ) sin θ = 0 (d) θ + cos θ + ẍ A sin θ = 0 (e) L θ + ( + ẍ A ) sin θ = 0 Solution: The equation of otion can be found by applyin the anular oentu principle with respect to point A, which writes Ḣ A + v A P = M A, ( ) where the linear oentu is iven by P = ṙ = d dt [ (xa + L cos θ)e x + L sin θe y ] = [(ẋa L sin θ θ)e x +L cos θ θe y ], the velocity of the ovin hine A is siply v A = x A e x, the anular oentu H A is H A = r AB P = (L cos θe x + L sin θe y ) [(ẋ A L sin θ θ)e x + L cos θ θe y ], and the oent of the external forces is M A = r AB e x = (L cos θe x + L sin θe y ) e x. Then, by coputin all the ters of Eqn. (*), one finds L θ + ( ẍ A ) sin θ = 0. 3

4 4. A dis rolls without slippin on a horizontal plane. A sooth uide of lenth 2R is riidly connected to the dis, as shown. A particle of ass is then free to slide in the uide, with its position alon the uide denoted by s. The anle θ easures the orientation of the uide relative to the horizontal direction. A coordinate frae [e s e θ ] is attached to the center of the dis and rotates with it. What is the total velocity of the particle expressed in the [e s e θ ] frae? e! s R e s (a) v = (ṡ R θ cos θ)e s + (R θ sin θ + s θ)e θ (b) v = (s + R cos θ) θe s + (R sin θ + ṡ)e θ (c) v = ṡe s + R θe θ (d) v = (ṡ + R θ sin θ)e s + (R θ cos θ + s θ)e θ (e) v = R θ cos θe s + s θe θ Solution: The position r of the particle with respect to a fixed frae [e x e y ] is iven by r = ( Rθ + s cos θ)e x + s sin θe y. e y r e x e! s R e s The velocity is then iven by ṙ = ( R θ + ṡ cos θ s sin θ θ)e x + (ṡ sin θ + s cos θ θ)e y. We now need to project ṙ on e s and e θ, which are respectively iven by e s = cos θe x + sin θe y e θ = sin θe x + cos θe y. By perforin the scalar products, we obtain ṙ e s = ṡ R θ cos θ ṙ eθ = R θ sin θ + s θ 4

5 5. A circular uide rotates about its center, with its rotation anle denoted by θ(t). A particle of ass is subject to a eneral force F as it oves alon the uide. If the otion of the particle is observed in the frae [y 1 y 2 ] attached to the rotatin uide, which of the followin stateents ust be true? (a) The Coriolis force is parallel to the Euler force. y 2 F y 1 s (t)! x 2 R x 1 (b) The Coriolis force is parallel to the centrifual force. (c) The Coriolis force is orthoonal to the centrifual force. (d) The Coriolis force is the null vector. (e) The anitude of the Coriolis force is equal to the anitude of the centrifual force. Solution: Since the relative velocity is tanent to the circular uide, the Coriolis force is radial, as the centrifual force. The Coriolis and centrifual forces are therefore parallel. 5

6 6. A particle of ass is perceived to ove alon a radial, rectilinear trajectory by an observer rotatin at constant anular velocity Ω > 0. Which of the followin stateents ust be true? (a) No active force is present. y 2 x 2 Perceived trajectory in [y1 y2] frae t y1 x 1 (b) An active force has to balance the Euler force. (c) An active force has to balance the Coriolis force. (d) No inertial forces are present. (e) An active force has to balance the centrifual force. Solution: Since the anular velocity is constant, no Euler force is present. The centrifual force is actin alon a radial direction, while the Coriolis force is noral to the perceived trajectory. The Coriolis force, therefore, would curve the trajectory of the particle perceived by the rotatin observer. As the trajectory in the ovin frae is rectilinear, there ust be an active force that balance the Coriolis force. In practice, the particle could be forced to ove on a rotatin, rectilinear uide. The reaction force provided by the uide is balancin the Coriolis force, and is then responsible of pushin the particle into the rotational otion. 6

7 7. Two particles of equal ass and neliible diensions are connected by, L 0 a sprin of stiffness. They are released at the sae heiht fro rest s s x on a sooth conical wede. At the tie of release, the sprin is at its unstretched lenth L 0. The inclination of 4 4 the wede is π, as shown; denotes the 4 constant of ravity, actin downwards. Because of syetry, the particles are at equal heihts at all ties. Denote with x the vertical direction, and with s the position of each particle alon the wede, as shown. What is the acceleration s when the syste reaches its iniu heiht? (a) s = 2 (b) s = 0 2 (c) s = 2 (d) s = (e) s = 2 Solution: First, we need to deterine the axiu drop x of the syste. we can use the conservation of echanical enery between the initial condition and the confiuration when the axiu drop is achieved, with zero velocity, as 0 = 1 2 L2 2x = 2x 2 2x x =, where, fro the eoetry of the proble, one can easily see that L = 2x. At the confiuration found, we can then write the linear oentu principle for one particle in the s direction, as s = L cos π 4 + sin π 4, which leads to s =

8 8. The wor done by the internal forces of an arbitrary syste of particles is zero when... (a)...no exernal forces are applied to the syste. (b)...the syste is subject to potential forces only. ( c)...the echanical enery is conserved. (d)...the linear and anular oentu w.r.t the center of ass are conserved. (e)...the particles are connected by riid lins only. Solution: The wor done by internal forces on a syste of particles is zero only if the particles are connected by riid lins only. 8

9 9. A particle of ass is attached by an undaped sprin of stiffness to the floor, subject to ravity. When the sprin is unstretched, the distance between the particle and the ceilin is h. The particle then is released with zero velocity (v 0 = 0) fro a defored confiuration, denoted by s. What is the axiu rane of s such that the particle does not reach the ceilin? (a) s (b) h < s < h ( < 2 (c) h 2 < s < h (d) h 2 < s < h2 (e) s < h h ) 2 + h2 + 2 ceilin Solution: The syste is conservative. Therefore, we can write the conservation of echanical enery between two relevant confiurations, naely: 1. stretched sprin, still (i.e. no velocity) confiuration and 2. particle touchin the ceilin with zero velocity. In particular, we can require that the enery of an arbitrary stretched confiuration is lower than the enery related to the contact with ceilin, as 1 2 s2 + s < 1 2 h2 + h. This is a second order alebraic equation that can be solved for s, ivin 2 h < s < h. h 9

10 10. A particle of ass oves with velocity v(t) on a straiht line. Which of the followin stateents ust be true? (a) No torques are actin on the particle. (b) No forces are actin on the particle. (c) The resultant force actin on the particle is parallel to the velocity. (d) The resultant force actin on the particle is noral to the velocity. (e) The anular oentu of the particle with respect to an arbitrary point is conserved. Solution: Since the trajectory is straiht, accordin to the linear oentu principle no force ust act alon a noral direction. Bein the otion alon the trajectory arbitrary, an acceleration v(t) iht exist. Therefore, the resultant force ust be parallel to the trajectory. 10

11 11. A particle of ass is constrained to ove alon a circular path in the counterclocwise direction with velocity v constant in anitude. An observer is travelin alon the sae path with a velocity v (t) of the sae anitude ( v (t) = v ) in the clocwise direction. Denote with T the tie required for the ass to coplete a full cycle. How any ties in the tie interval [0, T ] is the tie derivative of anular oentu Ḣ(t) of the particle with respect to the observer (t) equal to zero? (a) 1 (b) 4 (c) 2 (d) 3 (e) For all ties (Ḣ 0) Solution: We require that Ḣ = M v P =. v (t) v (t) This happens when M = 0 and when v P = 0. Since the particle is ovin on a circular path with constant anular velocity (i.e. v is constant), the resultant applied force is radial. Therefore, M = 0 when either and the particle overlap, or when they are placed on the sae diaater. This happens 4 ties in a coplete loop. Liewise, in these positions, v P = 0, as v v 0. The correct answer is thus 4. 11

12 12. A particle of ass 1 slides down a sooth surface. Initially, the particle is placed at heiht h 1 with zero initial velocity. What is the axiu heiht h 2 the ass 2 reaches after the ipact with 1, if the coefficient of restitution is e and 2 is at rest before ipact? h 1 h2 (a) h 2 = 2 1 h 1 1 (b) h 2 = h ( ) 2 (1 + 2 )(1 e) (c) h 2 = h 1 2 ( ) 2 1 (1 + e) (d) h 2 = h (e) h 2 = 1 ( ) 1 (1+e) h 1 Solution: Before ipact, the conservation of echanical enery holds T 0 + V 0 = T 1 + V 1 1 h 1 = 1 2 1v 2 1 v 1 = 2h 1, where v 1 is the velocity of the first particle before ipact. At ipact, we can apply the conservation of linear oentu for the whole syste, and the definition of the coefficient of restitution e: 1 v v 2 = 1 v v 2+, e = v 2+ v 1+ v 1 v 2, where v 2 = 0. The velocity v 2+ of the second particle after ipact is then v 2+ = 1(1 + e) v 1. After ipact, the conservation of echanical enery can be applied aain, as T 2 + V 2 = T 3 + V 3 1 ( ) 2 2 1v (1 + e) = 2 h 2 h 2 = h

13 13. A syste consists of four particles, positioned as depicted. The center of ass of the syste is at the oriin. Which of the followin stateents ust be true? (a) 1 = 4 and 2 = 3 (b) 1 = 3 and 2 = 4 (c) 4 = (d) 1 = 2 and 3 = 4 e y e x (e) 1 = 2 = 3 = 4 Solution: By tain into account the syetry of the proble, one can deduce that 1 = 3 and 2 = 4 ust hold for the center of ass to be at the oriin. Alternatively, one can apply the definition of the center of ass r C = 1 M 4 i r i, i=1 which results in the syste of equations = = 0 which is solvable only if 1 = 3 and 2 = 4. 13

14 14. A particle of ass is attached to a assless dis D 2, which rotates in the plane of the fiure with a positive, nonconstant anular velocity Ω 2 (t) > 0. The dis D 2 itself is hined at P to a bier dis D 1 that spins at a constant anular velocity Ω 1. Which of the followin diaras represents correctly the inertial forces actin on the particle, in the frae of reference [e y1 e y2 ] fixed to the dis D 1? Solution: The correct answer is (b). Since the rotatin frae spins at constant anular velocity, no Euler force is present. Moreover, the centrifual force is always directed alon the relative position y. These two observations are sufficient to exclude all the other possible answers. 14

The bloc is towed by a force F alon a sooth, horizontal surface. What is the anitude F such that the velocity of the bloc reains constant?

15 15. A bloc of ass is attached to a chain of ass per unit lenth ρ (the ass of a chain of lenth l is iven by chain = ρl). At t = t 0, all of the chain is at rest and the bloc has an initial velocity v 0 pointin to the riht, as shown. The bloc is towed by a force F alon a sooth, horizontal surface. What is the anitude F such that the velocity of the bloc reains constant? (Inore ravity) v 0 F v 0 F (a) F = 1 ρ 2 l 2 v 0 2 (b) F = ( + ρl)v 0 t t 2 0 (c) F = 0 (d) F = ρv0 2 1 (e) F = ρlv 0 t t 0 Solution: The linear oentu principle for variable ass writes d (Mv) = F Ṁv + M v = F. dt In our case, M = + ρl Ṁ = ρ l = ρv 0 and v = 0, therefore F = Ṁv 0 = ρv

16 16. A particle of ass oves in the vertical plane with the coordinates (x, y) indicatin its position. nly potential forces act on the particle that derive fro the potential V (x, y) = 1 2 (x x 0) 2 + (y y 0 ). The initial position of the particle is denoted by (x 0, y 0 ). Which of the followin diaras is a correct echanical odel of the syste? y x (a) y x (b) y x (c) y x (d) y x (e) Solution: By applyin the linear oentu principle for a sinle particle and the definition of potential force, one obtains: a = V { ẍ = (x x 0 ) ÿ = { ẍ + (x x 0 ) = 0 ÿ = which are the equations of otion of the confiuration illustrated in fiure (a). 16, (1)

17 17. Two blocs of equal ass are connected by a riid assless rod as they slide alon an inclined plane. Let α denote the inclination anle and assue that there is no friction between bloc A and the plane, while there is friction between bloc B and the plane, with friction coefficient µ = 1/ 3. What is the inclination anle α for which the two asses ove at constant velocity? ( ) 1 (a) α = arccos 4 3 ( π ) (b) α = arcsin 6 (c) α = π 2 ( ) 1 (d) α = arctan 2 3 ( ) 2 (e) α = arctan 9 3 µ B Solution: The principle of linear oentu for a syste of particles in the direction of the incline (w.r.t. the plane) ives A 2a t,cm = 2 sin α µn B where 2a t,cm is the acceleration of the center of ass in the direction of the incline. By applyin the linear oentu principle to each ass in the direction noral to the incline, one easily finds that N A = N B = cos α. Since the two asses are riidly connected, the acceleration of the center of ass is equal to the acceleration of each bloc. These latter are null if the riht hand side of the equation of otion in the tanential direction is zero and thus the anle α can be coputed as 2 sin α µ cos α = 0 tan α = µ 2 α = arctan

18. A particle of ass is connected to the floor of a ovin platfor by a sprin. The syste lays on the vertical plane and the platfor underoes a constant acceleration a, with a > 0.

18 18. A particle of ass is connected to the floor of a ovin platfor by a sprin. The syste lays on the vertical plane and the platfor underoes a constant acceleration a, with a > 0. Denote as [y 1 y 2 ] the reference frae fixed to the platfor, while [x 1 x 2 ] is the inertial frae fixed to the round. When y 2 = 0, the sprin is unstretched. What is the correct equation of otion w.r.t. the ovin frae [y 1 y 2 ], in the y 2 direction? y 2 x 2 x 1 y 1 a Movin platfor (a) ÿ 2 = ( a) + y 2 (b) ÿ 2 = ( a) y 2 (c) ÿ 2 = 0 (d) ÿ 2 = ( + a) y 2 (e) ÿ 2 = ẍ 2 Solution: The equation of otion in the inertial reference frae [x 1 x 2 ] is ẍ 2 = l ; l = x 2 (x 2 y 2 ) = y 2 ẍ 2 = y 2 and the coordination transforation between [x 1 x 2 ] and [y 1 y 2 ] is: x 2 = y at2 ẋ 2 = ẏ 2 + at ẍ 2 = ÿ 2 + a. Substitutin this latter result into the first equation of otion leads to the equation of otion in the relative reference frae: ÿ 2 = ( + a) y 2 The platfor acceleration adds to the ravitational acceleration. 18

19. A syste of five asses is shown in the fiure, with all asses located in the vertical plane. The asses are connected by sprins and riid assless lins as shown.

Thus, the nuber of derees of freedo is 2 5 7 = 3. Another, ore intuitive way to realize this, is by splittin the overall syste into saller subsystes.

19 19. A syste of five asses is shown in the fiure, with all asses located in the vertical plane. The asses are connected by sprins and riid assless lins as shown. How any derees of freedo does this syste have? (a) 5 (b) 1 (c) 2 (d) 4 (e) 3 L L 2 L 2 L 2 Solution: The syste is coposed by 5 particles in a 2D doain and by 7 independent constraints (i.e. riid lins). Thus, the nuber of derees of freedo is = 3. Another, ore intuitive way to realize this, is by splittin the overall syste into saller subsystes. First, the couple of asses lined by a horizontal lin has one deree of freedo. Fixin this deree, the sae reasonin is valid for the couple of asses 2 and the botto-left one of ass. Then, the last bottoriht ass has a further deree of freedo since it oves independently fro the forer syste (i.e. it consist of a pendulu). Therefore, the overall syste has 3 derees of freedo. Note that sprins and specific asses values do not affect at all the nuber of derees of freedo. 3L 2 3 L L 19

20 20. A ball of ass is launched in the vertical plane with an initial velocity vector v 0, inclided by an anle α with respect to the x axis, and subject to ravity. After the ball reaches the axiun heiht y ax,0, it falls to the round and underoes a collision with a coefficient of restitution e = 0.5. What is the ratio between the axiu heiht y ax,1 (after one bounce) and y ax,0? (a) y ax,1 y ax,0 = 2 y v 0 y ax,0 (b) y ax,1 y ax,0 = 2 5 (c) y ax,1 y ax,0 = y ax,0 v 2 0 y ax,1... x (d) y ax,1 y ax,0 = 1 4 (e) y ax,1 y ax,0 = 1 16 Solution: The total echanical enery of the particle when the particle is leavin the round (i.e. tae-off, to) and when it reaches the axiu heiht h before the first bounce (index 0) are respectively: E to,0 = 1 2 v 2 = 1 2 v2 x, v2 y,0 ; E h,0 = 1 2 v2 x,0 + y ax,0 where v x,0 = v cos(α) and v y,0 = v sin(α) and due to the conservation of the total echanical enery one obtains: y ax,0 = v2 y,0 2 ( ). When the particle is re-approachin the round (riht before the first bounce), its velocity vector is v = (v x,0, v y,0 ) always due to enery conservation and to particle ineatics. Due to the non perfect bounce (i.e. collision with coefficient of restitution) we now that: v x,1 = v x,0 ; v y,1 = ev y,0 where the subscript 1 denotes velocities after the first bounce. Reasonin aain with enery, one obtain the analoue of forula ( ) but now with index 1. This results in y ax,1 = 1 a y ax,0 4 a Note that it is possible to eneralize the result to n arbitrary bounces. It is easy to see that y ax,n = v2 y,n 2 = v2 y,n 1 2 e 2 = v2 y,0 2 e2n = y ax,0 e 2n yax,n y ax,0 e = 1/2, so the desired ratio is 1/4. 20 = e 2n. In our case, n = 1 and

21 Solution Version n e 2. e 3. c 4. a 5. b 6. c 7. c 8. e 9. c 10. c 11. b 12. d 13. b 14. d 15. d 16. d 17. e 18. d Solutions Pae 1

Vector Spaces in Physics 8/6/2015. Chapter 4. Practical Examples.

Vector Spaces in Physics 8/6/2015. Chapter 4. Practical Examples. Vector Spaces in Physics 8/6/15 Chapter 4. Practical Exaples. In this chapter we will discuss solutions to two physics probles where we ae use of techniques discussed in this boo. In both cases there are