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A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)
Correct answer is '3.41'. Can you explain this answer?
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A cam operates a knife edge follower having a lift of 30 mm. The cam ...
Given information:
- Lift of the knife edge follower: 30 mm
- Follower motion: Simple Harmonic Motion (SHM) for 150° of cam rotation
- Cam rotation velocity: 120 rpm
To find:
- Maximum acceleration of the follower during the lift
Approach:
1. Calculate the time taken for the cam to rotate 150°.
2. Convert the rotation time to seconds.
3. Determine the angular velocity of the cam.
4. Calculate the maximum angular acceleration of the cam.
5. Find the maximum acceleration of the follower using the radius of the cam and the maximum angular acceleration.
Calculations:
1. Time taken for the cam to rotate 150°:
- Total angle of rotation in radians = 150° * (π/180) = 5π/6 radians
- Cam rotation time = 1/120 minutes (since the cam rotates at a uniform velocity of 120 rpm)
- Convert minutes to seconds: 1 minute = 60 seconds
- Cam rotation time = (1/120) * 60 = 0.5 seconds
2. Angular velocity of the cam:
- Angular velocity (ω) = Angle of rotation (θ) / Time taken (t)
- ω = (5π/6) / 0.5 = (5π/6) * (1/0.5) = 5π/3 radians/second
3. Maximum angular acceleration of the cam:
- Angular acceleration (α) = ω^2 * r
- r = Lift of the follower = 30 mm = 0.03 m
- α = (5π/3)^2 * 0.03 = (25π^2/9) * 0.03 = 25π^2/270 radians/second^2
4. Maximum acceleration of the follower:
- Maximum acceleration = α * r
- Maximum acceleration = (25π^2/270) * 0.03 = 3.405 m/s^2
Final answer:
The maximum acceleration of the follower during the lift is approximately 3.41 m/s^2.
- Lift of the knife edge follower: 30 mm
- Follower motion: Simple Harmonic Motion (SHM) for 150° of cam rotation
- Cam rotation velocity: 120 rpm
To find:
- Maximum acceleration of the follower during the lift
Approach:
1. Calculate the time taken for the cam to rotate 150°.
2. Convert the rotation time to seconds.
3. Determine the angular velocity of the cam.
4. Calculate the maximum angular acceleration of the cam.
5. Find the maximum acceleration of the follower using the radius of the cam and the maximum angular acceleration.
Calculations:
1. Time taken for the cam to rotate 150°:
- Total angle of rotation in radians = 150° * (π/180) = 5π/6 radians
- Cam rotation time = 1/120 minutes (since the cam rotates at a uniform velocity of 120 rpm)
- Convert minutes to seconds: 1 minute = 60 seconds
- Cam rotation time = (1/120) * 60 = 0.5 seconds
2. Angular velocity of the cam:
- Angular velocity (ω) = Angle of rotation (θ) / Time taken (t)
- ω = (5π/6) / 0.5 = (5π/6) * (1/0.5) = 5π/3 radians/second
3. Maximum angular acceleration of the cam:
- Angular acceleration (α) = ω^2 * r
- r = Lift of the follower = 30 mm = 0.03 m
- α = (5π/3)^2 * 0.03 = (25π^2/9) * 0.03 = 25π^2/270 radians/second^2
4. Maximum acceleration of the follower:
- Maximum acceleration = α * r
- Maximum acceleration = (25π^2/270) * 0.03 = 3.405 m/s^2
Final answer:
The maximum acceleration of the follower during the lift is approximately 3.41 m/s^2.
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A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)Correct answer is '3.41'. Can you explain this answer?
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A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)Correct answer is '3.41'. Can you explain this answer? 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 A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)Correct answer is '3.41'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)Correct answer is '3.41'. Can you explain this answer?.
A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)Correct answer is '3.41'. Can you explain this answer? 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 A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)Correct answer is '3.41'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)Correct answer is '3.41'. Can you explain this answer?.
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A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)Correct answer is '3.41'. Can you explain this answer?, a detailed solution for A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)Correct answer is '3.41'. Can you explain this answer? has been provided alongside types of A cam operates a knife edge follower having a lift of 30 mm. The cam raises the follower with SHM for 150° of the rotation. If the cam rotates at a uniform velocity of 120 rpm, then the maximum acceleration (in m/s2) of follower during the lift is . (Rounded up to two decimal places)Correct answer is '3.41'. Can you explain this answer? theory, EduRev gives you an
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