理论力学英文版试卷A（附参考答案）

　XXX 工业大学

　20XX ～ 20XX 学年第 XX 学期期末考试试卷 学院 _________________

　班级 __________

　 姓名 __________

　 学号 ___________ 卷 《理论力学》课程试卷 B （附参考答案）

　Instructor: Yijian (Time: 2 Hours)

　Course Code:

　题号 1 2 3 4 5 6 总得分 题分 20 15 15 15 15 20

　得分

　1 ．Choose the correct answers with proper justification (10×2=20, 20%) 1 2 3 4 5 6 7 8 9 10 D D C A B B A C B C

　(1) For vector addition you have to use

　 law.

　A) Newton’s Second

　 B) the arithmetic

　 C) Pascal’s

　 D) the parallelogram (2) A particle moves along a horizontal path with its velocity varying with time as shown in the right figure. The average acceleration of the particle is _________. A) 0.4 m/s 2 →

　 B) 0.4 m/s 2 ← C) 1.6 m/s 2 →

　 D) 1.6 m/s 2 ← (3) If the position of a particle is defined by r = [(1.5t 2

　+ 1) i + (4t – 1) j] (m), its speed at t=1 s is

　 . A) 2 m/s

　 B) 3 m/s

　 C) 5 m/s

　 D) 7 m/s (4) As shown in the right figure, the speed of block B is

　 . A) 1 m/s

　B) 2 m/s C)

　4 m/s

　D) None of the above. (5) If the path function of a particle is s = 10 sin 2θ, the acceleration, a,

　is

　 . (Noting that θ=ωt, ω is the angular velocity at time t). A) 20 a sin2θ

　B) 20 a cos2θ-40ω 2 sin2θ C) 20 a cos2θ

　D) -40 a sin2θ

　(6) As shown in the right figure, the velocity of plane A with respect to plane B is

　 . A) (400 i + 520 j) km/hr B) (1220 i - 300 j) km/hr C) (-181 i - 300 j) km/hr D) (-1220 i + 300 j ) km/hr (7) If the pendulum is released from the horizontal position, the velocity of its bob in the vertical position is

　 . A) 3.8 m/s B) 6.9 m/s C) 14.7 m/s D) 21 m/s (8) If a slender bar rotates about end A, its angular momentum with respect to A is

　 . A) (1/12) m l 2 ω B) (1/6) m l 2 ω C) (1/3) m l 2 ω D) m l 2 ω

　 (9) The tangential acceleration of an object

　.

　A) represents the rate of change of the velocity vector’s direction.

　B) represents the rate of change in the magnitude of the velocity.

　C) is a function of the radius of curvature.

　 D) Both B and C. (10) Point A on the rod has a velocity of 8 m/s to the right. Where is the instantaneous center (IC) for the rod?

　 .

　A) Point A.

　B) Point B.

　 C) Point C.

　 D) Point D.

　3

　2. (15%) Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis

　Solution: Using the vector analysis, we have the x- and y-components of forces 600N and 800N as    o1o122600N cos30 300 3N 519.6N600N sin30 300N0800NxyxyFFFF     

　(2%) Hence, the force vectors are obtained as 1 1 12 2 2519.6 300800x yx yF FF F     F i j i jF i j j

　(4%) According to the vector addition rule, the resultant force is expressed in the vector form    1 2 1 2 1 2519.6 500R x x y yRx RyF F F FF F        F F F i ji j i j

　(5%) So, the magnitude and direction of the resultant force are obtained as        22 2 2o519.6 500 721.10N519.6arccos arccos 43.90721.1R Rx RyRxRF F FFF             

　(4%)

　3. (15%) The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles α, β, γ of the resultant force. Take x=20 m, y=15 m.

　Solution: Going from D to A, one must travel {-24k} m, then {20i} m and finally {15j} m. Thus, the unit vector for rope DA is

　 (2%)   22 220 15 24 m 20 15 2434.66 34.66 34.6620 15 24 mADAAr       i j k ru i j k

　    20 15 24400N 230.81 173.11 276.98 N34.66 34.66 34.66A A DAF         F u i j k i j k

　 Going from D to B, one must travel {-24k} m, then {-6i} m and finally {4j} m. Thus, the unit vector for rope DB is

　 (2%)     2 226 4 24 m 6 4 2425.06 25.06 25.066 4 24 mBDBBr          i j k ru i j k

　    6 4 24800N 191.54 127.69 766.16 N25.06 25.06 25.06B B DBF           F u i j k i j k

　 Going from D to C one must travel {-24k} m, then {-18j} m and finally {16i}m. Thus, the unit vector for rope DC is

　 (2%)     2 2216 18 24 m 8 9 1217 17 1716 18 24 mCDCCr       i j k ru i j k

　   8 9 12600N 282.35 317.65 423.53 N17 17 17C C DCF         F u i j k i j k

　 Hence, the resultant force is

　 (5%)   321.62 16.85 1466.67 NR A B C      F F F F i j k

　The magnitude and the coordinate direction angles of RF

　are

　(4%)    2 22 2 2 2oo321.62 16.85 1466.67 1501.7N321.62arccos arccos 77.631501.790.64arccos arccos 90.641501.7167.6arccos arccos 167.61501.7R Rx Ry RzRxRRyRRzRF F F FFFFFFF                                  o

　4. (15%) Determine the horizontal and vertical components of reaction at the pin A and the reaction on the beam at C.

　Solution: (1) Free-body diagram

　 (5%) We draw a free-body diagram for the overhanging beam as below:

　(2) Equations of equilibrium

　 (5%)

　Consider the counterclockwise moment of the force positive. According to the free-body diagram, we have x- and y-component equations and moment about point A of equilibrium as follows       ooo0, cos45 00, sin45 4kN 00, sin45 1.5m 4kN 3m 0x Ax Cy Ay CA CF F FF F FM F       

　After solving, we have

　 (5%)

　11.31kN, 8kN, 4kNAx Ay CF F F   

　The negetive sign indicates that the direction sence is oppisite to that shown in the free-body diagram.

　5. (15%) A projectile is launched from point A with the initial conditions shown in the figure. Determine the slant distance s which locates the point B of impact. Calculate the time of flight t.

　Solution: Establish a fixed x,y coordinate system (in the solution here, the origin of the coordinate system is placed at A). Apply the kinematic relations in x and y-directions. Motion in x-direction:

　(4%)  o0 0ocos400 120m/s cos40B A x AB A ABB ABx x v t x v tx t    

　Motion in y-direction:

　(5%)     2 o 20o 2 21 1120m/s sin402 210 120m/s sin40 9.81kg/s2B A y AB AB A AB ABB AB ABy y v t gt y t gty t t        Noting that

　 (2%)

　  o8 0 0 m t a n 2 0B By x   , o/sin20Bs y 

　After solving, we obtain the slant distance s and the flight time t as follows

　 (4%) 279.8m, 3.04ABs t t s   

　6. (20%) The rod AB has a mass of 10 kg. Piston B is attached to a spring of constant k = 800 N/m. The spring is un-stretched when 

　= 0°. Neglect the mass of the pistons. Determine the angular velocity of rod AB at 

　= 0° if the rod is released from rest when 

　= 30°.

　Solution: Initial position

　 (2%)

　Final position

　 (2%)

　 (1) Potential Energy:

　(5%) Let’s put the datum in line with the rod when 

　= 0°. Then, the gravitational potential energy and the elastic potential energy will be zero at position 2. => V 2

　= 0 Gravitational potential energy at position 1: - (10)( 9.81) ½ (0.4 sin 30) Elastic potential energy at position 1: ½ (800) (0.4 sin 30) 2

　So, V 1 = - 9.81 J + 16.0 J =6.19 J (2) Kinetic Energy:

　(5%) The rod is released from rest from position 1 (so v G1

　= 0,  1

　= 0). Therefore, T 1 = 0. At position 2, the angular velocity is  2 and the velocity at the center of mass is v G2 .

　Therefore, T 2 = ½ (10) (v G2 ) 2

　+ ½ (1/12)(10)(0.4 2 ) ( 2 ) 2

　At position 2, point A is the instantaneous center of rotation. Hence, v G2 = r  = 0.2  2 . Then, T 2 = 0.2  2 2

　+ 0.067  2 2 = 0.267  2 2

　(3) conservation of energy equation

　 (6%) Now apply the conservation of energy equation and solve for the unknown angular velocity,  2 . T 1 + V 1 = T 2

　+

　V 2

　0 + 6.19 = 0.267 2 2

　+ 0 =>  2 = 4.82 rad/s

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