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This mock test of Test: Velocity & Acceleration Analysis - 3 for Mechanical Engineering helps you for every Mechanical Engineering entrance exam.
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QUESTION: 1

Consider a four-bar mechanism shown in the given figure. The driving link DA is rotation uniformly at a speed of 100 rpm clockwise

Q. The velocity of A will be

Solution:

V = rω

QUESTION: 2

The directions of Coriolis component of acceleration, 2 ωV, of the slider A with respect to the coincident point B is shown in figure 1,2, 3 and 4. Directions shown by figures

Solution:

f_{cor} = 2Vω

2 and 4 are wrong according to above equation. These should be as show below.

QUESTION: 3

In the figure shown, the relative velocity of link 1 with respect of link 2 is 12 m/sec. Link 2 rotates at a constant speed of 120 rpm. The magnitude of Coriolis component of acceleration of link 1 is

Solution:

Velocity of link 1 with respect to 2

∴ Coriolis component of acceleration

QUESTION: 4

The number of links in a planer mechanism with revolute joints having 10 instantaneous centre is

Solution:

Number of instantaneous centre

for n = 5

we get Number of instantaneous centre = 10

QUESTION: 5

The figure below shows a planer mechanism with single degree of freedom. The instant center 24 for the given configuration in located at a position

Solution:

QUESTION: 6

Figure shows a quick return mechanism. The crank OA rotates clockwise uniformly.

OA = 2 cm; OO' = 4 cm

Q. The ratio of time for forward motion to that for return motion is

Solution:

sin α = 2/4

α = 30°

∴ β = 60°

QUESTION: 7

A slider sliding at 10 cm/s on a link which is rotating at 60 rpm is subjected to Coriolis acceleration of magnitude

Solution:

Coriolis acceleration

QUESTION: 8

In the figure given the link 2 rotates at an angular velocity of 2 rad/s. What is the magnitude of Coriolis acceleration experienced by the link 4?

Solution:

Since link 2 and link 3 are perpendicular each other

ω = 0

∴ f_{cor }= 0

QUESTION: 9

A simple quick return mechanism is shown in the figure. The forward to return ratio of the quick return mechanism is 2 : 1. If the radius of the crank O_{1}P is 125 mm, then the distance ‘of’ (in mm) between the crank centre to lever pivot centre point should be

Solution:

O_{1}P = 125mm

Quick return mechanism

The extreme position of the crank (O_{1}P) are shown in figure.

From right angle triangle O_{2}O_{1}P_{1}, we find that

QUESTION: 10

For the configuration shown, the angular velocity of link AB is 10 rad/s counter clockwise. The magnitude of the relative sliding velocity (in ms^{-1}) of slider B with respect to rigid link CD is

Solution:

As AB is perpendicular to slider for given condition, so sliding velocity of slider equals to velocity of link AB

= ω x AB = 10 x 0.25

= 2.5 m/s

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