Machine Kinematics Multiple Choice Questions on “Single Slider Crank Chain and its Inversions – 1”.
1. The quick return mechanism which is an inversion of 4-bar linkage is
a) Drag link mechanism
b) Whitworth quick return mechanism
c) Crank and slotted lever mechanism
d) None of the mentioned
Answer: a
Clarification: Drag link mechanism is an inversion of 4-bar linkage, which is a crank mechanism with different crank lengths. It is made up of revolute pairs only.
2. Match list I with list II
List I List II
A. Pantograph 1. Scotch yoke mechanism
B. Single slider crank 2. Double lever mechanism
C. Double slider crank chain 3. Tchebicheff’s mechanism
D. Straight line motion mechanism 4. Double crank
5. Hand pump
a) A-4,B-3,C-5,D-1
b) A-2,B-5,C-1,D-3
c) A-2,B-1,C-5,D-3
d) A-4,B-5,C-2,D-1
Answer: b
Clarification: Pantograph is double lever mechanism. handpump is an inversion of single slider crank chain. Scotch yoke mechanism is an inversion of double slider crank chain.Tchebiceff’s mechanism is an approximate straight line motion mechanism.
3. Match list I with list II
List I List II
A. Scott-Russel 1. Intemittent mechanism motion
B. Geneva 2. Quick return mechanism motion
C. Offset slider crank 3. Simple motion harmonic mechanism
D. Scotch Yoke 4. Straight line mechanism motion
a) A-2,B-3,C-1,D-4
b) A-3,B-2,C-4,D-1
c) A-4,B-1,C-2,D-3
d) A-4,B-3,C-1,D-2
Answer: c
Clarification: Scott-Russel – Straight line mechanism motion
Geneva – Intemittent mechanism motion
Offset slider crank – Quick return mechanism motion
Scotch Yoke – Simple motion harmonic mechanism.
4. When a cylinder is located in a Vee-block, the number of degrees of freedom which are arrested is
a) 2
b) 4
c) 7
d) 8
Answer: b
Clarification: Before placement on Vee-block, cylinder has 6 degrees of freedom. After placement on Vee-block, the cylinder has only 2 degrees of freedom. Hence, the degrees of freedom which are arrested is 6 – 2 = 4.
5. Match list I with list II
Type of joint Motion constrained
A. Revolute 1. Three
B. Cylindrical 2. Five
C. Spherical 3. Four
4. Two
5. Zero
a) A-1,B-3,C-2
b) A-5,B-4,C-3
c) A-2,B-3,C-1
d) A-4,B-5,C-3
Answer: c
Clarification: For revolute pair, degree of freedom = 1 and constrained
DOF = 6 – 1 = 5
For cylindrical pair, dof =2 and constrained dof = 6 – 2 = 4
For spherical pair, dof = 3 and constrained dof = 6 – 3 = 3.
6. The number of binary links, number of binary joints and number of ternary joints in Peaucelliar mechanism is
a) 6,6,0
b) 8,2,4
c) 8,4,2
d) 8,8,0
Answer: b
Clarification: The Peaucelliar mechanism has 8 binary links, 2 binary joints nad 4 ternary joints.
7. The number of degree of freedom of a planer linkage with 8 links and 9 simple revolute joints is
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: L = 8 = number of links
P = 9 = number of simple revolute joints
F = 3(L – 1) – 2P
= 3(8 – 1) – 2 x 9
= 21 – 18
= 3.
8. The following list of statements is given:
1) Grashoff’s rule states that for a planar crank-rocker 4-bar mechanism, the sum of the shortest and longest link lengths cannot be less than the sum of the remaining two link lengths.
2) Inversions of a mechanism are created by fixing different links, one at a time.
3) Geneva mechanism is an intermittent motion device.
4) Grubler’s criterion assumes mobility of a planar mechanism to be one.
The number of correct statements in the above list is
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: Except statement 1, all other three statements are correct.
9. A mechanism has 8 links, out of which 5 are binary, 2 are ternary and 1 is quaternary. The number of instantaneous centres of rotation will be
a) 28
b) 56
c) 62
d) 66
Answer: d
Clarification: n = 5 + 4 + 3 = 12
Number of instantaneous centres, N = n(n – 1)/2
= 12 x (12 – 1)/2
= 66.
10. In a dynamically equivalent system, a uniformly distributed mass is divided into
a) Three point masses
b) Four point masses
c) Two point masses
d) Infinite point masses
Answer: c
Clarification: Dynamically equivalent system of a rigid body is made of two point masses.
11. A crank and slotted lever mechanism used in a shaper has a centre distance of 300 mm between the centre of oscillation of the slotted lever and the centre of rotation of the crank. The radius of the crank is 120 mm. Find the ratio of the time of cutting to the time of return stroke.
a) 1.62
b) 1.72
c) 1.82
d) 1.92
Answer: b
Clarification: Given : AC = 300 mm ; CB = 120 mm
sin∠CAB = sin (90°−α/ 2)
CB/AC = 120/300 = 0.4
∠CAB = 90°−α/ 2
or α / 2 = 90° – 23.6° = 66.4°
α = 2 × 66.4 = 132.8°
We know that
Time of cutting stroke/ Time of return stroke = 1.72.