# Machine Kinematics Questions and Answers – Linear Acceleration

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This set of Machine Kinematics Multiple Choice Questions & Answers (MCQs) focuses on “Linear Acceleration”.

1. The acceleration of a particle at any instant has two components, radial component and tangential component. These two components will be
a) parallel to each other
b) perpendicular to each other
c) inclined at 450
d) opposite to each other

Explanation: Both the components will be perpendicular to each other.

2. The centre of gravity of a coupler link in a four bar mechanism will experience
a) no acceleration
b) only linear acceleration
c) only angular acceleration
d) both linear and angular acceleration

Explanation: None

3. When a point moves along a straight line, its acceleration will have
b) tangential component only
c) coriolis component only
d) radial and tangential components both

Explanation: The tangential component, is parallel to the velocity of the particle at the given instant.
The centripetal or radial component, is perpendicular to the velocity of the particle at the given instant.
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4. When a point at the end of a link moves with constant angular velocity, its acceleration will have
b) tangential component only
c) coriolis component only
d) radial and tangential components both

Explanation: The centripetal or radial component, is perpendicular to the velocity of the particle at the given instant.
The tangential component, is parallel to the velocity of the particle at the given instant.

5. In a shaper mechanism, the coriolis component of acceleration does not exists.
a) True
b) False

Explanation: In a shaper mechanism, the coriolis component of acceleration exists.

6. The tangential component of acceleration of the slider with respect to the coincident point on the link is called coriolis component of acceleration.
a) True
b) False

Explanation: When a point on one link is sliding along another rotating link, such as in quick return motion mechanism, then the coriolis component of the acceleration must be calculated.

7. The coriolis component of acceleration acts
a) along the sliding surface
b) perpendicular to the sliding surface
c) at 450 to the sliding surface
d) parallel to the sliding surface

Explanation: None

8. The coriolis component of acceleration is taken into account for
a) slider crank mechanism
b) four bar chain mechanism
c) quick return motion mechanism
d) all of the mentioned

Explanation: When a point on one link is sliding along another rotating link, such as in quick return motion mechanism, then the coriolis component of the acceleration must be calculated.

9. The coriolis component of acceleration depends upon
a) velocity of slider
b) angular velocity of the link
c) all of the mentioned
d) none of the mentioned

Explanation: None

10. A body in motion will be subjected to coriolis acceleration when that body is
a) in plane rotation with variable velocity
b) in plane translation with variable velocity
c) in plane motion which is a resultant of plane translation and rotation
d) restrained to rotate while sliding over another body

Explanation: When a point on one link is sliding along another rotating link, such as in quick return motion mechanism, then the coriolis component of the acceleration must be calculated.

11. A slider moves at a velocity v on a link revolving at ωrad/s. The coriolis component of acceleration is
a) ωv
b) 2ωv
c) ω2v
d) 2ωv2

Explanation: None

12. The coriolis component of acceleration leads the sliding velocity by
a) 450
b) 900
c) 1350
d) 1800

Explanation: The direction of coriolis component of acceleration is obtained by rotating v, at 90°, about its origin in the same direction as that of ω.

13. The sense of coriolis component 2ωv is same as that of the relative velocity vector v rotated at
a) 450 in the direction of rotation of the link containing the path
b) 450 in the direction opposite to the rotation of the link containing the path
c) 900 in the direction of rotation of the link containing the path
d) 1800 in the direction opposite to the rotation of the link containing the path

Explanation: The direction of coriolis component of acceleration is obtained by rotating v, at 90°, about its origin in the same direction as that of ω.

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