250+ TOP MCQs on Method of Locating Instantaneous Centres in a Mechanism and Answers

Machine Kinematics Multiple Choice Questions on “Method of Locating Instantaneous Centres in a Mechanism”.

1. The direction of Corioli’s component of acceleration is the direction
a) of relative velocity vector for the two coincident points rotated by 90⁰ in the direction of the angular velocity of the rotation of the link
b) along the centripetal acceleration
c) along tangential acceleration
d) along perpendicular to angular velocity
Answer: a
Clarification: The direction of coriolis component of acceleration will not be changed in sign if both ω and v are reversed in direction. It is concluded that 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.

2. In a shaper mechanism, the Corioli’s component of acceleration will
a) not exist
b) exist
c) depend on position of crank
d) none of the mentioned
Answer: b
Clarification: None.

3. The magnitude of tangential acceleration is equal to
a) velocity² x crank radius
b) velocity²/ crank radius
c) (velocity/ crank radius)²
d) velocity x crank radius²
Answer: b
Clarification: The magnitude of tangential acceleration is equal to velocity²/ crank radius.
The magnitude of the Corioli’s component of acceleration of a slider moving at velocity V on a link rotating at angular speed ω is 2Vω.

4. Tangential acceleration direction is
a) along the angular velocity
b) opposite to angular velocity
c) perpendicular to angular velocity
d) all of the mentioned
Answer: d
Clarification: None.

5. The magnitude of the Corioli’s component of acceleration of a slider moving at velocity V on a link rotating at angular speed ω is
a) Vω
b) 2Vω
c) Vω/2
d) 2V/ω
Answer: b
Clarification: The magnitude of tangential acceleration is equal to velocity²/ crank radius.
The magnitude of the Corioli’s component of acceleration of a slider moving at velocity V on a link rotating at angular speed ω is 2Vω.

6. In a rotary engine the angular velocity of the cylinder center line is 25 rad/sec and the relative velocity of a point on the cylinder center line w.r.t. cylinder is 10 m/sec. Corioli’s acceleration will be
a) 500m/sec²
b) 250m/sec²
c) 1000m/sec²
d) 2000m/sec²
Answer: a
Clarification: Corioli’s component = 2Vω
= 2 x 10 x 25 = 500500m/sec².

7. Corioli’s component is encountered in
a) quick return mechanism of shaper
b) four bar chain mechanism
c) slider crank mechanism
d) all of the mentioned
Answer: a
Clarification: 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.

8. Klein’s construction gives a graphical construction for
a) slider-crank mechanism
b) velocity polygon
c) acceleration polygon
d) none of the mentioned
Answer: c
Clarification: Klein’s construction represents acceleration polygon.

9. The velocity of a slider with reference to a fixed point about which a bar is rotating and slider sliding on the bar will be
a) parallel to bar
b) perpendicular to bar
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: None.

10. Klien’s construction can be used to determine acceleration of various parts when the crank is at
a) inner dead center
b) outer dead center
c) right angles to the link of the stroke
d) all of the mentioned
Answer: d
Clarification: Klien’s construction can be used to determine acceleration in all the mentioned position.

11. The number of dead centers in a crank driven slider crank mechanism are
a) 0
b) 2
c) 4
d) 6
Answer: b
Clarification: None.

12. Corioli’s component acts
a) perpendicular to sliding surfaces
b) along sliding surfaces
c) both of the mentioned
d) all of the mentioned
Answer: a
Clarification: The coriolis component of acceleration is always perpendicular to the link.

13. The sense of Coriol’s component is such that it
a) leads the sliding velocity vector by 90⁰
b) lags the sliding velocity vector by 90⁰
c) is along the sliding velocity vector by 90⁰
d) leads the sliding velocity vector by 180⁰
Answer: a
Clarification: 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.

14. Klien’s construction can be used when
a) crank has a uniform angular velocity
b) crank has non-uniform velocity
c) crank has uniform angular acceleration
d) crank has uniform angular velocity and angular acceleration
Answer: a
Clarification: None.

15. Klein’s construction is useful to determine
a) velocity of various parts
b) acceleration of various parts
c) displacement of various parts
d) angular acceleration of various parts
Answer: b
Clarification: Klien’s construction can be used to determine acceleration.