250+ TOP MCQs on Acceleration in the Slider Crank Mechanism & Coriolis Component and Answers

Machine Kinematics Objective Questions & Answers on “Acceleration in the Slider Crank Mechanism & Coriolis Component”.

1. A point is moving at the end of the link rotating with constant angular velocity ω, what will be the value of tangential component of acceleration?
a) 0
b) ω²R
c) Infinite
d) ω²R/2
Answer: a
Clarification: Any point at the end of the link which is moving with a constant angular velocity has no component as tangential acceleration, it only possesses radial component of acceleration.

2. The tangential component of acceleration is maximum when the link rotates with a constant angular velocity.
a) True
b) False
Answer: b
Clarification: The tangential component of acceleration is zero when the link is rotating with a constant angular velocity, hence the given statement is false.

3. A point is moving at the end of the link rotating with constant angular velocity ω, what will be the value of radial component of acceleration?
a) 0
b) ω²R
c) Infinite
d) ω²R/2
Answer: b
Clarification: Any point at the end of the link which is moving with a constant angular velocity has no component as tangential acceleration, it only possess radial component of acceleration whose value is given by ω²R, where R is the distance of the point from the reference point.

4. In a slider crank mechanism, the crank rotates with a constant angular velocity of 300 rpm, Length of crank is 150mm, and the length of the connecting rod is 600mm. Determine linear velocity of the midpoint of the connecting rod in m/s. Crank angle = 45° from IDC.
a) 4.1
b) 4.4
c) 4.8
d) 5.2
Answer: a
Clarification: Let the Intersection point of crank and connecting rod be termed as ‘x’
velocity of x = ωxL(crank)
= 4.713 m/s
We draw a corresponding velocity diagram for the same,
where this velocity is perpendicular to the crank, this diagram when drawn to scale turns out to be a triangle.
From the reference point we can measure the velocity of midpoint of connecting rod
V = 4.1 m/s.

5. In a slider crank mechanism, the crank rotates with a constant angular velocity of 300 rpm, Length of crank is 150mm, and the length of the connecting rod is 600mm. Determine acceleration of the midpoint of the connecting rod in m/s². Crank angle = 45° from IDC.
a) 117
b) 144
c) 148
d) 252
Answer: a
Clarification: Let the Intersection point of crank and connecting rod be termed as ‘x’
velocity of x = ωxL(crank)
= 4.713 m/s
We draw a corresponding velocity diagram for the same,
where this velocity is perpendicular to the crank, this diagram when drawn to scale turns out to be a triangle.
From the reference point we can measure the velocity of midpoint of connecting rod V = 4.1 m/s
From the acceleration diagram of the same , we find that acceleration of midpoint of the connecting rod comes out to be
117m/s².

6. What will be the shape of the velocity diagram of the slider crank mechanism if there are three links including the slider.
a) Triangle
b) Parallelogram
c) Square
d) Trapezium
Answer: a
Clarification: When there are two links and a slider in a slider crank mechanism, there are in total 3 links. In this case the shape of the velocity polygon is a triangle.

7. If the normal component of the acceleration is doubled, what will be the effect on the radial component?
a) Doubled
b) Halved
c) Remains same
d) Becomes 4 times
Answer: a
Clarification: The component which is normal to the motion is known as the normal component which comes from the angular velocity, this is also known as the radial component. Hence radial and normal components are the same.

8. If the body is not rotating with a constant angular velocity then there are both radial and tangential component of acceleration.
a) True
b) False
Answer: a
Clarification: The radial component always exists as long as the body is rotating with some angular velocity, however the tangential components acts only if the angular velocity is not constant. In this case both the components will act on the given link.

9. In the given figure, the direction of radial velocity vector and angular velocity is given what will be the direction of coriolis force?

a) Along the radial velocity vector
b) Opposite to radial velocity vector
c) Perpendicular to radial velocity vector towards right
d) Perpendicular to radial velocity vector towards left
Answer: d
Clarification: The direction of the coriolis component of acceleration is obtained by rotating the radial velocity vector by 90 degrees in the direction of the angular velocity.

10. Coriolis component of acceleration exists when there is relative motion between two points from the ground frame.
a) True
b) False
Answer: a
Clarification: When there is relative motion between two points from the ground frame, Pseudo force exists.

11. Calculate the coriolis component of acceleration in m/s² from the following data:
ω = 12 rad/s
v = 2 m/s
R = 1 m
a) 24
b) 12
c) 36
d) 6
Answer: a
Clarification: The coriolis component of acceleration is given by
2ωV
inserting the values we get
a = 24 m/s².

12. Which component of acceleration is parallel to the given link?
a) Radial
b) Tangential
c) Coriolis
d) Pseudo
Answer: a
Clarification: The radial component also known as the normal component is parallel to the link and always exists as long as there is angular velocity.

13. Which of the following mechanism will have coriolis component?
a) Quick return motion mechanisms
b) Slider crank mechanism
c) Four bar chains
d) Gnome engine
Answer: a
Clarification: Quick return motion mechanism have sliders attached to the link, hence out of the given options quick return motion mechanisms will have coriolis component of acceleration.

14. Which component of acceleration is parallel to the velocity of given link?
a) Radial
b) Tangential
c) Coriolis
d) Pseudo
Answer: b
Clarification: The radial component also known as the normal component is parallel to the link and always exists as long as there is angular velocity. However the tangential component acts in a direction parallel to the velocity of the link.

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