# Machine Kinematics Questions and Answers – Velocity of a point Link on a Link By Instanteneous Centre Method

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This set of Machine Kinematics online test focuses on “Velocity of a point Link on a Link By Instanteneous Centre Method”.

1. There are two points P and Q on a planar rigid body. The relative velocity between the two points
a) should always be along PQ
b) can be oriented along any direction
c) should always be perpendicular to PQ
d) should be along QP when the body undergoes pure translation

Explanation: Velocity of any point on a link with respect to another point (relative velocity) on the same link is always perpendicular to the line joining these points on the configuration (or space) diagram.
vQP = Relative velocity between P & Q
= vP − vQ always perpendicular to PQ.

2. For a four-bar linkage in toggle position, the value of mechanical advantage is
a) 0.0
b) 0.5
c) 1.0
d) ∞

Explanation: When the mechanism is toggle,then β = 00 and 1800.
So M.A = ∞.

3. The number of inversion for a slider crank mechanism is
a) 6
b) 5
c) 4
d) 3

Explanation: For a 4 bar slider crank mechanism, there are the number of links or inversions are 4. These different inversions are obtained by fixing different links once at a time for one inversion. Hence, the number of inversions for a slider crank mechanism is 4.
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4. Match the item in columns I and II
Column I Column II
Q. Instantaneous centre of velocity 2. Beam
S. Prime circle 4. Gear
a) P-4, Q-2, R-3, S-1
b) P-4, Q-3, R-2, S-1
c) P-3, Q-2, R-1, S-4
d) P-3, Q-4, R-1, S-2

Explanation: Column I Column II
Q. Instantaneous centre of velocity 3. Linkage
R. Section modulus 2. Beam
S. Prime circle 1. Cam
So correct pairs are, P-4, Q-3, R-2, S-1.

5. Match the items in columns I and II
Column I Column II
P. Higher Kinematic Pair 1. Grubler’s Equation
Q. Lower Kinemation Pair 2. Line contact
R. Quick Return Mechanism 3. Euler’s Equation
S. Mobility of a Linkage 4. Planar
5. Shaper
6. Surface contact
a) P-2, Q-6, R-4, S-3
b) P-6, Q-2, R-4, S-1
c) P-6, Q-2, R-5, S-3
d) P-2, Q-6, R-5, S-1

Explanation: In this question pair or mechanism is related to contact & machine related to it.
Column I Column II
P. Higher Kinematic Pair 2. Line Contact
Q. Lower Kinematic Pair 6. Surface Contact
R. Quick Return Mechanism 5. Shaper
S. Mobility of a Linkage 1. Grubler’s Equation

So correct pairs are, P-2, Q-6, R-5, S-1.

6. In a four-bar linkage, S denotes the shortest link length, L is the longest link length, P and Q are the lengths of other two links. At least one of the three moving links will rotate by 3600 if
a) S + L < P + Q
b) S + L > P + Q
c) S + P < L + Q
d) S + P > L + Q

Explanation: Here P,Q,R, & S are the lengths of the links.
According to Grashof’s law : “For a four bar mechanism, the sum of the shortest and longest link lengths should not be greater than the sum of remaining two link lengths, if there is to be continuous relative motion between the two links
S + L < P + Q.

7. The number of degrees of freedom of a planar linkage with 8 links and 9 simple revolute joints is
a) 1
b) 2
c) 3
d) 4

Explanation: Given l= 8, j= 9
We know that, Degree of freedom,
n =3(l − 1)−2j = 3(8 − 1)−2 x 9 = 3.

8. The lengths of the links of a 4-bar linkage with revolute pairs are p,q,r, and s units. given that p<q<r<s. Which of these links should be the fixed one, for obtaining a “double crank” mechanism ?

Explanation: Given p<q<r<s
“Double crank” mechanism occurs, when the shortest link is fixed. From the given pairs p is the shortest link. So, link of length p should be fixed.

9. 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

Explanation: Number of degrees of freedom = 2 & movability includes the six degrees of freedom of the device as a whole, as the ground link were not fixed. So, 4 degrees of freedom are constrained or arrested.

10. The minimum number of links in a single degree-of-freedom planar mechanism with both higher and lower kinematic pairs is
a) 2
b) 3
c) 4
d) 5

Explanation: From the Kutzbach criterion the degree of freedom,
n = 3(l − 1) − 2j − h
For single degree of Freedom (n = 1),
1 = 3(l − 1) − 2j − h
3l − 2j − 4 − h = 0 …(i)
The simplest possible mechanisms of single degree of freedom is four-bar mechanism. For this mechanism j = 4, h = 0
From equation (i), we have
3l − 2 x 4 − 4 − 0 = 0
or, l = 4.

11. The total number of instantaneous centres for a mechanism consisting of n links are
a) n/2
b) n
c) n – 1/2
d) n(n – 1)/2

Explanation: The number of instantaneous centres in a constrained kinematic chain is equal to the number of possible combinations of two links. The number of pairs of links or the number of instantaneous centres is the number of combinations of n links taken two at a time. Mathematically, number of instantaneous centres, N = n(n – 1)/2 where n = Number of links.

12. According to Aronhold Kennedy’s theorem, if three bodies move relatively to each other, their instantaneous centres will lie on a
a) straight line
b) parabolic curve
c) ellipse
d) none of the mentioned

Explanation: The Aronhold Kennedy’s theorem states that if three bodies move relatively to each other, they have three instantaneous centres and lie on a straight line.

13. In a mechanism, the fixed instantaneous centres are those centres which
a) remain in the same place for all configurations of the mechanism
b) vary with the configuration of the mechanism
c) moves as the mechanism moves, but joints are of permanent nature
d) none of the mentioned

Explanation: Fixed instantaneous centres remain in the same place for all configurations of the mechanism. The permanent instantaneous centres move when the mechanism moves, but the joints are of permanent nature.

14. The instantaneous centres which vary with the configuration of the mechanism, are called
a) permanent instantaneous centres
b) fixed instantaneous centres
c) neither fixed nor permanent instantaneous centres
d) none of the mentioned

Explanation: Fixed instantaneous centres remain in the same place for all configurations of the mechanism. The permanent instantaneous centres move when the mechanism moves, but the joints are of permanent nature. Neither fixed nor permanent instantaneous centres vary with the configuration of the mechanism.

15. When a slider moves on a fixed link having curved surface, their instantaneous centre lies
a) on their point of contact
b) at the centre of curvature
c) at the centre of circle
d) at the pin joint

Explanation: When the slider link moves on fixed link having constant radius of curvature, the instantaneous centre lies at the centre of curvature i.e. the centre of the circle, for all configuration of the links.

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