Machine Kinematics

Method of Locating Instantaneous Centres in a Mechanism

Question 1
Marks : +2 | -2
Pass Ratio : 100%
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
Vω
2Vω
Vω/2
2V/ω
Explanation:
The magnitude of tangential acceleration is equal to velocity2/ crank radius.
Question 2
Marks : +2 | -2
Pass Ratio : 100%
Klien’s construction can be used to determine acceleration of various parts when the crank is at
inner dead center
outer dead center
right angles to the link of the stroke
all of the mentioned
Explanation:
Klien’s construction can be used to determine acceleration in all the mentioned position.
Question 3
Marks : +2 | -2
Pass Ratio : 100%
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
parallel to bar
perpendicular to bar
both of the mentioned
none of the mentioned
Explanation:
None.
Question 4
Marks : +2 | -2
Pass Ratio : 100%
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
500m/sec2
250m/sec2
1000m/sec2
2000m/sec2
Explanation:
Corioli’s component = 2Vω
Question 5
Marks : +2 | -2
Pass Ratio : 100%
In a shaper mechanism, the Corioli’s component of acceleration will
not exist
exist
depend on position of crank
none of the mentioned
Explanation:
None.
Question 6
Marks : +2 | -2
Pass Ratio : 100%
The direction of Corioli’s component of acceleration is the direction
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
along the centripetal acceleration
along tangential acceleration
along perpendicular to angular velocity
Explanation:
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.
Question 7
Marks : +2 | -2
Pass Ratio : 100%
Corioli’s component is encountered in
quick return mechanism of shaper
four bar chain mechanism
slider crank mechanism
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.
Question 8
Marks : +2 | -2
Pass Ratio : 100%
The magnitude of tangential acceleration is equal to
velocity2 x crank radius
velocity2/ crank radius
(velocity/ crank radius)2
velocity x crank radius2
Explanation:
The magnitude of tangential acceleration is equal to velocity2/ crank radius.
Question 9
Marks : +2 | -2
Pass Ratio : 100%
Tangential acceleration direction is
along the angular velocity
opposite to angular velocity
perpendicular to angular velocity
all of the mentioned
Explanation:
None.
Question 10
Marks : +2 | -2
Pass Ratio : 100%
Klein’s construction gives a graphical construction for
slider-crank mechanism
velocity polygon
acceleration polygon
none of the mentioned
Explanation:
Klein’s construction represents acceleration polygon.