瘁事锯派伯另认钩仑拉报为敞
chapter 1 Introduction
c1p1.Q
1、多选题：
What is the main content of “Theory of Machines and Mechanisms”?
A: Composition of mechanisms
B: Design of commonly used mechanisms
C: Analysis and design of the combined mechanisms
D: Kinetics of mechanisms
答案: Composition of mechanisms;
Design of commonly used mechanisms;
Analysis and design of the combined mechanisms;
Kinetics of mechanisms
2、多选题：
What should students pay attention to in the study of ‘Theory of Machines and Mechanisms’?
A: Combining theory with practice
B: Combining kinematic diagrams with actual machines/mechanisms
C: Combining kinematic diagrams with moving machines/mechanisms
D: Combining intuitive thinking and logical thinking
答案: Combining theory with practice;
Combining kinematic diagrams with actual machines/mechanisms;
Combining kinematic diagrams with moving machines/mechanisms;
Combining intuitive thinking and logical thinking
3、判断题：
The study objects of the course ‘Theory or Machines and Mechanisms’ are machines and mechanisms.
A: 正确
B: 错误
答案: 正确
c1p2.Q
1、多选题：
What are the main features of machines?
A: Assembly of man-made components
B: Determinate relative motions among different components
C: Can conduct work/energy conversion
D: Well designed shapes and structures
答案: Assembly of man-made components;
Determinate relative motions among different components;
Can conduct work/energy conversion
c1p3.Q
1、多选题：
What are the main features of mechanisms?
A: Assemblage of man-made components
B: Determinate relative motions among different components
C: Can conduct work/energy conversion
D: Well designed shapes and structures
答案: Assemblage of man-made components;
Determinate relative motions among different components
2、判断题：
The definition of mechanisms is ‘assembly of man-made components to realize determinate motion/force transmission among different components’.
A: 正确
B: 错误
答案: 正确
Chapter 1 Quiz
1、多选题：
What are the main features of machines?
A: Assembly of man-made components
B: Determinate relative motions among different components
C: Can conduct work/energy conversion
D: Well designed shapes and structures
答案: Assembly of man-made components;
Determinate relative motions among different components;
Can conduct work/energy conversion
2、多选题：
What are the main features of mechanisms?
A: Assembly of man-made components
B: Determinate relative motions among different components
C: Can conduct work/energy conversion
D: Well designed shapes and structures
答案: Assembly of man-made components;
Determinate relative motions among different components
3、判断题：
The study objects of ‘Theory of Machines and Mechanisms’ are mechanisms and machines.
A: 正确
B: 错误
答案: 正确
Chapter 10 Balance of machinery
4.Q
1、单选题：
For a balance level G100, the product of allowable offset [e] and angular velocity ω is
A: 100m/s
B: 100000m/s
C: 100mm/s
D: 10m/s
答案: 100m/s
2、多选题：
The methods to evaluate the allowable unbalance of a rotor include
A: Allowable mass-radius product
B: Allowable offset
C: Inertial forces
D: Inertial moment
答案: Allowable mass-radius product;
Allowable offset
5.Q
1、单选题：
The purpose of balance for mechanisms is to eliminate
A: Dynamic pressure of mechanisms applied to the frame
B: Forces of mechanisms applied to the frame
C: Friction
D: Friction torque
答案: Dynamic pressure of mechanisms applied to the frame
2、单选题：
Complete balance of inertial forces of a mechanism is to make the overall center of the masses in the mechanism
A: Stationary
B: Move at a constant velocity
C: Move linearly at a constant speed
D: Move along a circular trajectory at a constant speed
答案: Stationary
3、多选题：
Methods to balance the inertial force of the mechanism include
A: Symmetrical design of mechanisms
B: Adding masses
C: Use test method
D: None of the above
答案: Symmetrical design of mechanisms;
Adding masses
4、多选题：
What need to be performed to have a fully balanced mechanism?
A: Balance inertial force
B: Balance inertial moment
C: Balance total force
D: Balance total moment
答案: Balance inertial force;
Balance inertial moment
c10p1.Q
1、单选题：
What is the purpose of machinery balance?
A: To reduce or eliminate the unbalanced forces acted on links
B: To reduce the weight of the mechanism
C: To increase the strength of the mechanism
D: None of the above
答案: To reduce or eliminate the unbalanced forces acted on links
2、多选题：
What are the disadvantages of unbalanced machinery?
A: Lower efficiency
B: Lower strength
C: Lower reliability
D: More vulnerable to fatigue
答案: Lower efficiency;
Lower strength;
Lower reliability;
More vulnerable to fatigue
3、多选题：
Types of machinery balance include
A: Balance of rigid rotors
B: Balance of flexible rotors
C: Balance of the mechanisms
D: None of the above
答案: Balance of rigid rotors;
Balance of flexible rotors;
Balance of the mechanisms
4、多选题：
Methods of machinery balance for rigid rotors include
A: Theoretical calculation method
B: Testing method
C: Virtual prototype method
D: Virtual reality method
答案: Theoretical calculation method;
Testing method
c10p3.Q
1、单选题：
Static balance of rigid motor requires
A: Balance of inertial force
B: Balance of inertial moment
C: Balance of inertial force and inertial moment
D: None of the above
答案: Balance of inertial force
2、单选题：
Only the rigid motor is statically balanced, provided that it is theoretically assumed that all unbalanced masses on the rigid rotor are distributed in
A: The same plane
B: Different parallel planes
C: Different arbitrary planes
D: Not sure
答案: The same plane
3、单选题：
The condition for a rigid rotor to reach static balance is
A: The sum of the product mass and diameter is zero
B: The vector sum of the mass diameter product is zero
C: Balanced force
D: The rotor does not exert force on the frame
答案: The vector sum of the mass diameter product is zero
4、单选题：
Dynamic balance of rigid motor require
A: Balance of inertial force
B: Balance of inertial moment
C: Balance of inertial force and inertial moment
D: Not sure
答案: Balance of inertial force and inertial moment
5、单选题：
When conducting dynamic balancing on a rigid rotor, how many balanced planes should be selected at least?
A: 1
B: 2
C: 3
D: 4
答案: 2
6、多选题：
The condition for a rotor to achieve dynamic balance is
A: Equivalent inertial force is zero
B: The equivalent inertial moments produced by the inertial forces is zero
C: The rotor does not apply force on the frame
D: Forces are in their equilibrium condition
答案: Equivalent inertial force is zero;
The equivalent inertial moments produced by the inertial forces is zero
Chapter 10 Quiz
1、单选题：
What is the purpose to conduct balancing for mechanical systems?
A: To reduce or eliminate the unbalanced forces acted on links
B: To reduce the weight of the mechanism
C: To increase the strength of the mechanism
D: Other reasons
答案: To reduce or eliminate the unbalanced forces acted on links
2、单选题：
For a rigid rotor that needs static balancing, theoretically, the concentrated masses on it are distributed in ( ) plane perpendicular to the axis of rotation.
A: 1
B: 2
C: 3
D: 4
答案: 1
3、单选题：
The static balancing of a rigid rotor only needs to balance _____.
A: Inertial forces
B: Moments
C: Volumes
D: Other answer
答案: Inertial forces
4、单选题：
In a disc rotor shown in the figure, there are four eccentric masses located in the same plane of rotation, and their size and distance to the rotation center are m1=5kg, m2=7kg, m3=8kg, m4=10kg; r1=r4=10cm, r2=20cm, r3=15cm. Assume the distance of the added mass to the rotational center rb=15cm. Please find mb and θb for the added mass.
A: mb=4.32kg, θb=149.24°
B: mb=5.37kg, θb=119.73°
C: mb=7.35, θb=29.53°
D: mb=8.45kg, θb=89.37°
答案: mb=5.37kg, θb=119.73°
5、单选题：
To conduct dynamic balancing to a rotor, in addition to the inertial force, () must also be balanced.
A: Inertial forces
B: Moments
C: Masses
D: Other answer
答案: Moments
6、单选题：
As shown in the picture, there is a general rotor. The mass of the rotor is 15kg and rotation speed n=3000r/min. Try to determine the allowable unbalanced mass-radius product in the two equilibrium planes [mr]I and [mr]II .
A: [mr]I=200.63 g·mm, [mr]II=100.32 g·mm
B: [mr]I=200.63 kg·mm, [mr]II=100.32 kg·mm
C: [mr]I=200.63 g·cm, [mr]II=100.32 g·cm
D: [mr]I=200.63 g·m, [mr]II=100.32 g·m
答案: [mr]I=200.63 g·mm, [mr]II=100.32 g·mm
Chapter 11 Kinetics of mechanical systems
c11p1.Q
1、单选题：
What are the general characteristics of motion for driving links in a machine at the steady running stage?
A: Periodic fluctuation
B: Non-periodic fluctuation
C: Keep changing between periodic and non-periodic fluctuation
D: Remains constant
答案: Periodic fluctuation
2、单选题：
What is the relationship between the input work Wd and output work Wr in one cycle of motion during the steady running period when the work Wf done by friction is ignored?
A: Wd = Wr
B: Wd < Wr
C: Wd > Wr;
D: Not sure
答案: Wd = Wr
3、多选题：
What is the main purpose of kinetics analysis of mechanical systems?
A: Determine the actual motions of links under external forces
B: Control the periodic speed fluctuation
C: Force analysis of machines
D: Solve the magnitude of the mechanical driving force
答案: Determine the actual motions of links under external forces;
Control the periodic speed fluctuation
4、多选题：
What are the basic periods of machine running?
A: Warm-up
B: Start-up
C: Steady running
D: Stopping
答案: Start-up;
Steady running;
Stopping
5、多选题：
What may be the types of the driving force acting on the machine?
A: Constant
B: Function of locations
C: Function of velocity
D: Function of time
答案: Constant;
Function of locations;
Function of velocity
6、多选题：
What may be the types of the resistance acting on the machine?
A: Constant
B: Function of location
C: Function of velocity
D: Function of time
答案: Constant;
Function of location;
Function of velocity;
Function of time
c11p2.Q
1、单选题：
When building up an equivalent kinetic models for a mechanical system, what is the principle we can use to obtained the equivalent moment of inertia?
A: The kinetic energy remains unchanged
B: The power remains unchanged
C: The force remains unchanged
D: The moment of inertia remains unchanged
答案: The kinetic energy remains unchanged
2、单选题：
When building up an equivalent kinetic models for a mechanical system, what is the principle we can use to obtained the equivalent torque?
A: The kinetic energy remains unchanged
B: The power remains unchanged
C: The force remains unchanged
D: The moment of inertia remains unchanged
答案: The power remains unchanged
3、单选题：
Under which of the following conditions one can calculate the angular acceleration ε in an equivalent kinetic model by using the expression ε = Me/Je?
A: The machine is at the starting moment
B: The machine is in the steady-running period
C: The moment of inertia Je is a function of location
D: The equivalent torque Me is constant
答案: The machine is at the starting moment
4、单选题：
Under which of the following conditions one can calculate the angular acceleration ε in an equivalent kinetic model by using the expression ε = Me/Je?
A: The moment of inertia Je is constant
B: The moment of inertia Je is a function of location
C: The equivalent torque Me is constant
D: The equivalent torque Me is a function of location
答案: The moment of inertia Je is constant
5、判断题：
he equivalent weight (or moment of inertia) in the equivalent kinetic model for a mechanical system is calculated by using the principle of same kinetic energy. Therefore, they cannot be calculated until the actual motion of the mechanical system is determined.
A: 正确
B: 错误
答案: 错误
c11p3.Q
1、单选题：
A: Equal to each other
B: Mostly the same
C: Does not equal to each other
D: Not sure
答案: Equal to each other
2、单选题：
The figure shows the diagram of the equivalent driving torque Med and the equivalent resistance torque Mer of a machine whose moment of inertia Je is constant. At which angular position, the angular velocity of the equivalent link reaches its maximal value ωmax.
A: 0
B: π/2
C: 3/2π
D: 2π
答案: 3/2π
3、单选题：
The figure shows the diagram of the equivalent driving torque Med and the equivalent resistance torque Mer of a machine whose moment of inertia Je is constant. At which angular position, the angular velocity of the equivalent link reaches its minimal value ωmin.
A: 0
B: π/2
C: 3/2π
D: 2π
答案: π/2
4、判断题：
A: 正确
B: 错误
答案: 正确
c11p4.Q
1、单选题：
What causes periodic speed fluctuation in machine operation?
A: There are links which do reciprocating motion in machines, and the inertia forces are difficult to balance
B: Uneven weight distribution of the rotational links in machines
C: Each instantaneous driving power and resistance power are not equal to each other. However their average values are equal to each other in a cycle
D: Improper placement of the moving parts in machine
答案: Each instantaneous driving power and resistance power are not equal to each other. However their average values are equal to each other in a cycle
2、单选题：
Which parameter is the most important when applying the following equation to calculate the moment of inertia of a flywheel for periodic speed fluctuation control?
A: maximum fluctuation of kinetic energy
B: average angular speed
C: coefficient of the speed fluctuation
D: equivalent moment of inertia of the original mechanical system
答案: maximum fluctuation of kinetic energy
3、判断题：
After a flywheel is mounted in a machine, the speed fluctuation can be completely eliminated.
A: 正确
B: 错误
答案: 错误
Chapter 11 Quiz
1、单选题：
When the equivalent link is a rotating link with fixed axis, its equivalent moment of inertia Je is obtained according to the principle of ___, while the equivalent torque Me is obtained according to the principle of ___.
A: Same kinetic energy; Same kinetic energy
B: Same kinetic energy; Same power
C: Same power; Same kinetic energy
D: Same power; Same power
答案: Same kinetic energy; Same power
2、单选题：
As shown in following figure, the CNC vertical milling table driven by a servo motor. The mass of the table and workpiece is m4 = 350kg; the lead of ball screw 3 is l = 6mm, and the moment of inertia J3 = 1.2 × 10^(-3)kg•m^2; the number of teeth of gear 1, 2 is z1 = 25, z2 = 45, and the moment of inertia of gear 1, 2 is J1 = 7.2 × 10^(-4)kg•m^2, J2 = 1.92 × 10^(-3)kg•m^2. When selecting a servo motor, the allowed load moment of inertia must be greater than the equivalent load moment of inertia converted to the motor shaft. Then the equivalent moment of inertia converted by the mechanism to the motor shaft is Je1 = ___ kg*m^2.
A: 1.65 × 10^(-3)
B: 1.73 × 10^(-3)
C: 1.78 × 10^(-3)
D: 1.82 × 10^(-3)
答案: 1.78 × 10^(-3)
3、单选题：
As the mechanism shown in figure, the resistance torque acting on link 3 is M3 = 10Nm, the moment of inertia of link 3 to axis C is J3 = 0.016kg•m^2, and the mass and the moment of inertia of other links are ignored. If take crank 1 as equivalent link, given crank length lAB = 100mm. In the instantaneous moment shown in figure, φ1 = 90°, φ3 = 30°, find the equivalent moment of inertia Je1 and equivalent resistance torque Mer1 at this instantaneous moment.
A: Je1 = 0.001kg•m^2, Mer1 = 2.5Nm
B: Je1 = 0.002kg•m^2, Mer1 = 3.5Nm
C: Je1 = 0.003kg•m^2, Mer1 = 4.5Nm
D: Je1 = 0.004kg•m^2, Mer1 = 5.5Nm
答案: Je1 = 0.001kg•m^2, Mer1 = 2.5Nm
4、单选题：
As the winch shown in figure (a), if the weight of workpiece is G = 1000N, the radius of wheel is r = 0.2m, the number of teeth of each gear in reducer is z1 = 17, z2 = 64, z2’ = 32, z3 = 85. The moment of inertia of each wheel around its axis is J1 = 0.1 kg•m^2, J2 = 0.3kg•m^2, J2’ = 0.2 kg•m^2, J3 = 1 kg•m^2. When the workpiece drops at v = 1m/s, the driving torque is suddenly interrupted and the braking torque is applied on the shaft of wheel 1 at Mf = 40Nm. Find how much time t does the work piece need to be stopped and in this period of time the height h of the workpiece moves down.
A: t = 0.265s, h = 0.033m
B: t = 0.365s, h = 0.133m
C: t = 0.465s, h = 0.233m
D: t = 0.565s, h = 0.333m
答案: t = 0.465s, h = 0.233m
5、单选题：
In engineering practice, if the ω does not fluctuate much, the formula for calculating the average angular speed of the shaft is ( ).
A: 
B: 
C: 
D: 
答案:
6、单选题：
The coefficient of the speed fluctuation defined as ( ), is commonly used to evaluate the speed fluctuation of the shaft in mechanical system.
A: 
B: 
C: 
D: 
答案:
7、单选题：
As the mechanism shown in figure (a), the driving torque acting on gear 1’s shaft is constant, and the fluctuation of the resistance torque M2 acting on the gear 2’s shaft is shown in diagram (b), the rotation speed of gear 2’s shaft is n2 = 60r/min, the moment of inertia of gear 2 is J2 = 29.2kg•m^2, the mass and moment of inertia of gear 1 and other links are ignored. Find:
(1) To make sure the coefficient of the speed fluctuation is δ = 0.04, the moment of inertia Jf2 of flywheel should be installed on gear 2’s shaft;
(2) If z1 = 22, z2 = 85, the moment of inertia Jf1 of flywheel should be installed on gear 1’s shaft
A: Jf2 = 523.049kg•m^2, Jf1 = 35.039kg•m^2
B: Jf2 = 623.049kg•m^2, Jf1 = 41.744kg•m^2
C: Jf2 = 723.049kg•m^2, Jf1 = 48.437kg•m^2
D: Jf2 = 823.049kg•m^2, Jf1 = 55.143kg•m^2
答案: Jf2 = 723.049kg•m^2, Jf1 = 48.437kg•m^2
8、多选题：
What are the three periods in the running of a mechanical system?
A: Start-up
B: Steady running
C: Stopping
D: Constant speed
答案: Start-up;
Steady running;
Stopping
9、填空题：
The figure shows a crank-slider mechanism used in engine, the length of crank is l1 = 100mm, φ1 = 90°; the mass of slider 3 is m3 = 10kg, and the mass and moment of inertia of other links are ignored. The driving force acting on slider is F3 = 1000N, the work resistance torque acting on crank is M1 = 90Nm. The angular acceleration at the beginning of the crank rotation is ε10 = ____rad/s^2.
答案: 100
Chapter 2 Composition of mechanisms
c2p1.Q
1、单选题：
What is the difference between a link and a part?
A: A part is a basic component of a machine from a point view of manufacturing. A link is a basic component of a mechanism from a point view of motion transmission. It can be a part or an assemblage of several parts.
B: Links are lines, and parts are entities.
C: Links are virtual, and parts actually exist.
D: Other differences.
答案: A part is a basic component of a machine from a point view of manufacturing. A link is a basic component of a mechanism from a point view of motion transmission. It can be a part or an assemblage of several parts.
2、单选题：
How many frames can exist in a mechanism?
A: 1
B: 2
C: 3
D: Not sure. Maybe several.
答案:
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