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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  Test: Velocity & Acceleration Analysis - 3 - Mechanical Engineering MCQ

Test: Velocity & Acceleration Analysis - 3 - Mechanical Engineering MCQ


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10 Questions MCQ Test - Test: Velocity & Acceleration Analysis - 3

Test: Velocity & Acceleration Analysis - 3 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Test: Velocity & Acceleration Analysis - 3 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Velocity & Acceleration Analysis - 3 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Velocity & Acceleration Analysis - 3 below.
Solutions of Test: Velocity & Acceleration Analysis - 3 questions in English are available as part of our course for Mechanical Engineering & Test: Velocity & Acceleration Analysis - 3 solutions in Hindi for Mechanical Engineering course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt Test: Velocity & Acceleration Analysis - 3 | 10 questions in 30 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study for Mechanical Engineering Exam | Download free PDF with solutions
Test: Velocity & Acceleration Analysis - 3 - Question 1
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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

Detailed Solution for Test: Velocity & Acceleration Analysis - 3 - Question 1


V = rω

Test: Velocity & Acceleration Analysis - 3 - Question 2
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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

Detailed Solution for Test: Velocity & Acceleration Analysis - 3 - Question 2

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

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Test: Velocity & Acceleration Analysis - 3 - Question 3
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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

Detailed Solution for Test: Velocity & Acceleration Analysis - 3 - Question 3

Velocity of link 1 with respect to 2

∴ Coriolis component of acceleration

Test: Velocity & Acceleration Analysis - 3 - Question 4
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The number of links in a planer mechanism with revolute joints having 10 instantaneous centre is
Detailed Solution for Test: Velocity & Acceleration Analysis - 3 - Question 4

Number of instantaneous centre

for n = 5
we get Number of instantaneous centre = 10

Test: Velocity & Acceleration Analysis - 3 - Question 5
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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
Test: Velocity & Acceleration Analysis - 3 - Question 6
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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
Detailed Solution for Test: Velocity & Acceleration Analysis - 3 - Question 6


sin α = 2/4
α = 30°
∴ β = 60°

Test: Velocity & Acceleration Analysis - 3 - Question 7
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A slider sliding at 10 cm/s on a link which is rotating at 60 rpm is subjected to Coriolis acceleration of magnitude
Detailed Solution for Test: Velocity & Acceleration Analysis - 3 - Question 7

Coriolis acceleration

Test: Velocity & Acceleration Analysis - 3 - Question 8
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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?
Detailed Solution for Test: Velocity & Acceleration Analysis - 3 - Question 8

Since link 2 and link 3 are perpendicular each other
ω = 0
∴ fcor = 0

Test: Velocity & Acceleration Analysis - 3 - Question 9
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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 O1P is 125 mm, then the distance ‘of’ (in mm) between the crank centre to lever pivot centre point should be
Detailed Solution for Test: Velocity & Acceleration Analysis - 3 - Question 9



O1P = 125mm
Quick return mechanism


The extreme position of the crank (O1P) are shown in figure.
From right angle triangle O2O1P1, we find that

Test: Velocity & Acceleration Analysis - 3 - Question 10
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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
Detailed Solution for Test: Velocity & Acceleration Analysis - 3 - Question 10

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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