Engineering Questions with Answers - Multiple Choice Questions
Machine Kinematics Questions and Answers – Four Bar Chain or Quadric Cycle Chain
1 - Question
In a crank and slotted lever quick-return motion, the distance between the fixed centres is 150 mm and the length of the driving crank is 75mm. the ratio of the time taken on the cutting and return strokes is
a) 1.5
b) 2.0
c) 2.2
d) 2.93
View Answer
Answer: b
Explanation: OA = 150mm
AB = 75 mm
sin (90 – α/2) = AB/OA
cosα/2 = 1/2
therefore, α = 1200
β = 3600 – 1200 = 2400
Quick return ratio = β/α = 240/120 = 2
2 - Question
A helical coil spring of stiffness k is cut to two equal halves and then these are connected in parallel to support a vibrating mass m. The angular frequency of vibration, ωn is
a) √k/m
b) √2k/m
c) √4k/m
d) √k/4m
View Answer
Answer: c
Explanation: stiffness ∞ 1/length
keq = 2k + 2k = 4k
ω = √4k/m.
3 - Question
Consider the following statements:
In a slider-crank mechanism, the slider is at its dead centre position when the
(i) slider velocity is zero
(ii) slider velocity is maximum
(iii) slider acceleration is zero
(iv) slider acceleration is maximum
Which of the above statements are correct?
a) 1 and 4
b) 1 and 3
c) 2 and 3
d) 2 and 4
View Answer
Answer: a
Explanation: At dead centre velocity is zero, because instantaneous acceleration is maximum.
4 - Question
Which one of the following mechanisms is an inversion of double slider-crank chain?
a) Elliptic trammels
b) Beam engine
c) Oscillating cylinder engine
d) Coupling rod of a locomotive
View Answer
Answer: a
Explanation: Elliptic trammels are inversion of double slider crank chain.
5 - Question
The number of instantaneous centres of rotation for a 10-link kinematic chain is
a) 36
b) 90
c) 120
d) 45
View Answer
Answer: d
Explanation: Instantaneous centre = n(n – 1)/2 = 45.
6 - Question
A slider moves with uniform velocity on a revolving link of length r with angular velocity ω. The Coriolis acceleration component of a point on the slider relative to a coincident point on the link is equal to
a) ω parallel to the link
b) 2ω perpendicular to the link
c) ω perpendicular to the link
d) 2ω parallel to the link
View Answer
Answer: b
Explanation: coriolis components of acceleration for slider is perpendicular to the link.
7 - Question
In a crank and slotted lever type quick return mechanism, the link moves with an angular velocity of 20 rad/s, while the slider moves with the linear velocity of 1.5 m/s. The magnitude and direction of Coriolis component of acceleration with respect to angular velocity are
a) 30 m/s2 and direction is such as to rotate slider velocity in the same sense as the angular velocity
b) 30 m/s2 and direction is such as to rotate slider velocity in the opposite sense as the angular velocity
c) 60 m/s2 and direction is such as to rotate slider velocity in the same sense as the angular velocity
d) 60 m/s2 and direction is such as to rotate slider velocity in the opposite sense as the angular velocity.
View Answer
Answer: c
Explanation: Magnitude of Coriolis acceleration = 2Vω
= 2 x 1.5 x 20
= 60 m/s2
Rotate the link in direction of rotation and see the direction of linear velocity that will be direction of Coriolis acceleration.
8 - Question
Which of the following are associated with Ackerman steering mechanism used in automobiles?
1. Has both sliding and turning pairs
2. Less friction and hence long life
3. Mechanically correct in all positions
4. Mathematically not accurate except in three positions.
5. Has only turning pairs 6. Controls movement of two front wheels
a) 2, 4, 5 and 6
b) 1, 2, 3 and 6
c) 2, 3, 5 and 6
d) 1, 2, 3 and 5
View Answer
Answer: a
Explanation: Ackerman steering mechanism has no sliding mechanism & not suitable for all the positions.
9 - Question
The crankshaft of reciprocating engine having a 20 cm crank and 100 cm connecting rod rotates at 210 r.p.m. When the crank angle is 45o, the velocity of piston is nearly
a) 1.8m/s
b) 1.9m/s
c) 3.5m/s
d) 19m/s
View Answer
Answer: c
Explanation: Vp = ωr(sinϴ + sin2ϴ/2n)
ω = 2пN/60 = 21.98rad/sec
r = 6.2m
n = l/r = 100/20 = 5
therefore, Vp = 21.980 x 0.2(sin45o + sin90o/2 x 5)
= 3.5m/sec.