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In the figure shown, the relative velocity of link 1 with respect to link 2 is 12 m/sec. Link 2 rotates at a constant speed of 120 rpm. The magnitude of Carioles component of acceleration of link 1 is (a) 302m/s2 (b) 604 m/s2 (c) 906m/s2 (d) 1208 m/s2?
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In the figure shown, the relative velocity of link 1 with respect to l...
Analysis:
To solve this problem, we need to determine the magnitude of the Coriolis component of acceleration of link 1. The Coriolis component of acceleration arises due to the relative velocity between two bodies in motion. In this case, we have link 1 and link 2, with a given relative velocity of 12 m/sec. Link 2 is rotating at a constant speed of 120 rpm.
Formula:
The formula to calculate the Coriolis component of acceleration is given by:
a_c = 2 * ω * v_r
where a_c is the Coriolis component of acceleration, ω is the angular velocity, and v_r is the relative velocity.
Calculation:
Given that link 2 is rotating at a constant speed of 120 rpm, we can calculate the angular velocity (ω) using the formula:
ω = (2 * π * n) / 60
where n is the rotational speed in rpm.
Substituting the given values, we have:
ω = (2 * π * 120) / 60 = 4π rad/sec
The relative velocity (v_r) is given as 12 m/sec.
Using the formula for the Coriolis component of acceleration, we can calculate the magnitude:
a_c = 2 * ω * v_r
= 2 * 4π * 12
= 96π m/s^2
To find the numerical value of the acceleration, we can approximate π as 3.14:
a_c ≈ 96 * 3.14
≈ 301.44 m/s^2
Therefore, the magnitude of the Coriolis component of acceleration of link 1 is approximately 301.44 m/s^2.
Conclusion:
The magnitude of the Coriolis component of acceleration of link 1 is approximately 301.44 m/s^2. Hence, the correct option is (a) 302 m/s^2.
To solve this problem, we need to determine the magnitude of the Coriolis component of acceleration of link 1. The Coriolis component of acceleration arises due to the relative velocity between two bodies in motion. In this case, we have link 1 and link 2, with a given relative velocity of 12 m/sec. Link 2 is rotating at a constant speed of 120 rpm.
Formula:
The formula to calculate the Coriolis component of acceleration is given by:
a_c = 2 * ω * v_r
where a_c is the Coriolis component of acceleration, ω is the angular velocity, and v_r is the relative velocity.
Calculation:
Given that link 2 is rotating at a constant speed of 120 rpm, we can calculate the angular velocity (ω) using the formula:
ω = (2 * π * n) / 60
where n is the rotational speed in rpm.
Substituting the given values, we have:
ω = (2 * π * 120) / 60 = 4π rad/sec
The relative velocity (v_r) is given as 12 m/sec.
Using the formula for the Coriolis component of acceleration, we can calculate the magnitude:
a_c = 2 * ω * v_r
= 2 * 4π * 12
= 96π m/s^2
To find the numerical value of the acceleration, we can approximate π as 3.14:
a_c ≈ 96 * 3.14
≈ 301.44 m/s^2
Therefore, the magnitude of the Coriolis component of acceleration of link 1 is approximately 301.44 m/s^2.
Conclusion:
The magnitude of the Coriolis component of acceleration of link 1 is approximately 301.44 m/s^2. Hence, the correct option is (a) 302 m/s^2.
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In the figure shown, the relative velocity of link 1 with respect to link 2 is 12 m/sec. Link 2 rotates at a constant speed of 120 rpm. The magnitude of Carioles component of acceleration of link 1 is (a) 302m/s2 (b) 604 m/s2 (c) 906m/s2 (d) 1208 m/s2?
Question Description
In the figure shown, the relative velocity of link 1 with respect to link 2 is 12 m/sec. Link 2 rotates at a constant speed of 120 rpm. The magnitude of Carioles component of acceleration of link 1 is (a) 302m/s2 (b) 604 m/s2 (c) 906m/s2 (d) 1208 m/s2? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In the figure shown, the relative velocity of link 1 with respect to link 2 is 12 m/sec. Link 2 rotates at a constant speed of 120 rpm. The magnitude of Carioles component of acceleration of link 1 is (a) 302m/s2 (b) 604 m/s2 (c) 906m/s2 (d) 1208 m/s2? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the figure shown, the relative velocity of link 1 with respect to link 2 is 12 m/sec. Link 2 rotates at a constant speed of 120 rpm. The magnitude of Carioles component of acceleration of link 1 is (a) 302m/s2 (b) 604 m/s2 (c) 906m/s2 (d) 1208 m/s2?.
In the figure shown, the relative velocity of link 1 with respect to link 2 is 12 m/sec. Link 2 rotates at a constant speed of 120 rpm. The magnitude of Carioles component of acceleration of link 1 is (a) 302m/s2 (b) 604 m/s2 (c) 906m/s2 (d) 1208 m/s2? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In the figure shown, the relative velocity of link 1 with respect to link 2 is 12 m/sec. Link 2 rotates at a constant speed of 120 rpm. The magnitude of Carioles component of acceleration of link 1 is (a) 302m/s2 (b) 604 m/s2 (c) 906m/s2 (d) 1208 m/s2? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the figure shown, the relative velocity of link 1 with respect to link 2 is 12 m/sec. Link 2 rotates at a constant speed of 120 rpm. The magnitude of Carioles component of acceleration of link 1 is (a) 302m/s2 (b) 604 m/s2 (c) 906m/s2 (d) 1208 m/s2?.
Solutions for In the figure shown, the relative velocity of link 1 with respect to link 2 is 12 m/sec. Link 2 rotates at a constant speed of 120 rpm. The magnitude of Carioles component of acceleration of link 1 is (a) 302m/s2 (b) 604 m/s2 (c) 906m/s2 (d) 1208 m/s2? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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