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A cylindrical disc of mass m = 1 kg and radius r = 0.15 m was spinning at ω = 5 rad/s when it was placed on a flat horizontal surface and released (refer to the figure). Gravity g acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive.
Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).
Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).
Correct answer is '0.25'. Can you explain this answer?
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Most Upvoted Answer
A cylindrical disc of mass m = 1 kg and radius r = 0.15 m was spinning...
Given data :
Mass of cylindrical disc (m) = 1 kg
Radius of disc (r) = 0.15 m
Initial angular velocity of disc (ω0 )= 5 rad/s
Gravitational acceleration = g (m/s2)
Let us horizontal velocity of disc = v m/s
We know, Newton’s second law for linear motion F = ma … (i)
And Newton’s second law for angular motion
Mass of cylindrical disc (m) = 1 kg
Radius of disc (r) = 0.15 m
Initial angular velocity of disc (ω0 )= 5 rad/s
Gravitational acceleration = g (m/s2)
Let us horizontal velocity of disc = v m/s
We know, Newton’s second law for linear motion F = ma … (i)
And Newton’s second law for angular motion
Newton’s 1st law of motion, for linear motion
v = u+ at [ u = 0 , because when disc touches the ground initial velocity will be zero]
Newton’s 1st law of motion for angular motion
v = 0.25 m/s
Hence, the correct answer is 0.25.
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A cylindrical disc of mass m = 1 kg and radius r = 0.15 m was spinning at ω = 5 rad/s when it was placed on a flat horizontal surface and released (refer to the figure). Gravity g acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive.Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).Correct answer is '0.25'. Can you explain this answer?
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A cylindrical disc of mass m = 1 kg and radius r = 0.15 m was spinning at ω = 5 rad/s when it was placed on a flat horizontal surface and released (refer to the figure). Gravity g acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive.Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).Correct answer is '0.25'. 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 cylindrical disc of mass m = 1 kg and radius r = 0.15 m was spinning at ω = 5 rad/s when it was placed on a flat horizontal surface and released (refer to the figure). Gravity g acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive.Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).Correct answer is '0.25'. 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 cylindrical disc of mass m = 1 kg and radius r = 0.15 m was spinning at ω = 5 rad/s when it was placed on a flat horizontal surface and released (refer to the figure). Gravity g acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive.Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).Correct answer is '0.25'. Can you explain this answer?.
A cylindrical disc of mass m = 1 kg and radius r = 0.15 m was spinning at ω = 5 rad/s when it was placed on a flat horizontal surface and released (refer to the figure). Gravity g acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive.Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).Correct answer is '0.25'. 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 cylindrical disc of mass m = 1 kg and radius r = 0.15 m was spinning at ω = 5 rad/s when it was placed on a flat horizontal surface and released (refer to the figure). Gravity g acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive.Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).Correct answer is '0.25'. 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 cylindrical disc of mass m = 1 kg and radius r = 0.15 m was spinning at ω = 5 rad/s when it was placed on a flat horizontal surface and released (refer to the figure). Gravity g acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive.Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).Correct answer is '0.25'. Can you explain this answer?.
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