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A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. When the floor is suddenly removed, what is the minimum rotational speed o f the cylinder to enable the man to remain stuck to the wall with out falling (approximately)?
- a)5 rad/s
- b)10 rad/s
- c)15 rad/s
- d)20 rad/s
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A 70 kg man stands in contact against the inner wall of a hollow cylin...
Given parameters:
- Mass of man, m = 70 kg
- Radius of drum, r = 3 m
- Rotational speed of drum, ω = 200 rev/min = 20π rad/s
- Coefficient of friction between wall and clothing, μ = 0.15
To calculate:
- Minimum rotational speed of drum to enable man to remain stuck to wall without falling
Assumptions:
- The man remains in contact with the wall throughout the motion
- Friction is the only force acting on the man
Solution:
1. Calculate the gravitational force on the man:
- Gravitational acceleration, g = 9.8 m/s²
- Gravitational force, Fg = mg = 70 kg × 9.8 m/s² = 686 N
2. Calculate the maximum frictional force between the wall and the man:
- Maximum frictional force, Ff = μN
- Normal force, N = Fg (since the man is in contact with the wall)
- Maximum frictional force, Ff = μFg = 0.15 × 686 N = 103 N
3. Calculate the minimum required centripetal force on the man to keep him stuck to the wall:
- Centripetal force, Fc = mrω²
- Minimum required centripetal force, Fc,min = Ff (since friction is the only force acting on the man)
- Fc,min = mrω² = 70 kg × (3 m) × ω²
- ω² = Fc,min / (mr) = Ff / (mr) = 103 N / (70 kg × 3 m) = 1.55 m/s²
4. Calculate the minimum rotational speed required for the man to remain stuck to the wall:
- ω = √(Fc,min / (mr)) = √(1.55 m/s²) = 1.25 rad/s
5. Convert the answer to revolutions per minute (rpm):
- Minimum rotational speed, ω = 1.25 rad/s
- Minimum rotational speed, ω = 1.25 × (60 / 2π) rpm = 11.92 rpm ≈ 12 rpm
Answer:
The minimum rotational speed of the drum to enable the man to remain stuck to the wall without falling is approximately 12 rpm, which is closest to option A (5 rad/s) among the given options.
- Mass of man, m = 70 kg
- Radius of drum, r = 3 m
- Rotational speed of drum, ω = 200 rev/min = 20π rad/s
- Coefficient of friction between wall and clothing, μ = 0.15
To calculate:
- Minimum rotational speed of drum to enable man to remain stuck to wall without falling
Assumptions:
- The man remains in contact with the wall throughout the motion
- Friction is the only force acting on the man
Solution:
1. Calculate the gravitational force on the man:
- Gravitational acceleration, g = 9.8 m/s²
- Gravitational force, Fg = mg = 70 kg × 9.8 m/s² = 686 N
2. Calculate the maximum frictional force between the wall and the man:
- Maximum frictional force, Ff = μN
- Normal force, N = Fg (since the man is in contact with the wall)
- Maximum frictional force, Ff = μFg = 0.15 × 686 N = 103 N
3. Calculate the minimum required centripetal force on the man to keep him stuck to the wall:
- Centripetal force, Fc = mrω²
- Minimum required centripetal force, Fc,min = Ff (since friction is the only force acting on the man)
- Fc,min = mrω² = 70 kg × (3 m) × ω²
- ω² = Fc,min / (mr) = Ff / (mr) = 103 N / (70 kg × 3 m) = 1.55 m/s²
4. Calculate the minimum rotational speed required for the man to remain stuck to the wall:
- ω = √(Fc,min / (mr)) = √(1.55 m/s²) = 1.25 rad/s
5. Convert the answer to revolutions per minute (rpm):
- Minimum rotational speed, ω = 1.25 rad/s
- Minimum rotational speed, ω = 1.25 × (60 / 2π) rpm = 11.92 rpm ≈ 12 rpm
Answer:
The minimum rotational speed of the drum to enable the man to remain stuck to the wall without falling is approximately 12 rpm, which is closest to option A (5 rad/s) among the given options.
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Community Answer
A 70 kg man stands in contact against the inner wall of a hollow cylin...
Given:
- Mass of the man (m) = 70 kg
- Radius of the cylindrical drum (r) = 3 m
- Rotational speed of the drum (ω) = 200 rev/min
- Coefficient of friction between the wall and the man's clothing (μ) = 0.15
To find:
- Minimum rotational speed of the cylinder to enable the man to remain stuck to the wall without falling
Solution:
1. Determine the gravitational force on the man:
- Gravitational force (Fg) = mg = 70 kg x 9.8 m/s² = 686 N
2. Determine the centrifugal force on the man:
- Centrifugal force (Fc) = mrω² = 70 kg x (3 m) x (200 rev/min x 2π/60 s)² = 6591.8 N
3. Determine the force of friction between the man's clothing and the wall:
- Force of friction (Ff) = μN, where N is the normal force on the man
- At the minimum rotational speed, the force of friction is equal and opposite to the gravitational force: Ff = Fg
- Therefore, μN = mg
- N = mg/μ = 686 N / 0.15 = 4573.3 N
- Ff = μN = 0.15 x 4573.3 N = 686 N
4. Determine the net force on the man:
- Net force (Fnet) = Fc - Ff
- At the minimum rotational speed, the net force is zero: Fnet = 0
- Therefore, Fc = Ff
- mrω² = μN
- ω² = μN/mr = 0.15 x 4573.3 N / (70 kg x 3 m) = 3.69 rad/s²
- ω = √(3.69) = 1.92 rad/s
5. Convert the answer to RPM:
- ω = 1.92 rad/s x 60 s/2π = 18.3 rev/min
6. Round the answer to the nearest option:
- The minimum rotational speed is approximately 5 rad/s (Option A).
Therefore, the correct answer is option 'A' (5 rad/s).
- Mass of the man (m) = 70 kg
- Radius of the cylindrical drum (r) = 3 m
- Rotational speed of the drum (ω) = 200 rev/min
- Coefficient of friction between the wall and the man's clothing (μ) = 0.15
To find:
- Minimum rotational speed of the cylinder to enable the man to remain stuck to the wall without falling
Solution:
1. Determine the gravitational force on the man:
- Gravitational force (Fg) = mg = 70 kg x 9.8 m/s² = 686 N
2. Determine the centrifugal force on the man:
- Centrifugal force (Fc) = mrω² = 70 kg x (3 m) x (200 rev/min x 2π/60 s)² = 6591.8 N
3. Determine the force of friction between the man's clothing and the wall:
- Force of friction (Ff) = μN, where N is the normal force on the man
- At the minimum rotational speed, the force of friction is equal and opposite to the gravitational force: Ff = Fg
- Therefore, μN = mg
- N = mg/μ = 686 N / 0.15 = 4573.3 N
- Ff = μN = 0.15 x 4573.3 N = 686 N
4. Determine the net force on the man:
- Net force (Fnet) = Fc - Ff
- At the minimum rotational speed, the net force is zero: Fnet = 0
- Therefore, Fc = Ff
- mrω² = μN
- ω² = μN/mr = 0.15 x 4573.3 N / (70 kg x 3 m) = 3.69 rad/s²
- ω = √(3.69) = 1.92 rad/s
5. Convert the answer to RPM:
- ω = 1.92 rad/s x 60 s/2π = 18.3 rev/min
6. Round the answer to the nearest option:
- The minimum rotational speed is approximately 5 rad/s (Option A).
Therefore, the correct answer is option 'A' (5 rad/s).
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A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. When the floor is suddenly removed, what is the minimum rotational speed o f the cylinder to enable the man to remain stuck to the wall with out falling (approximately)?a)5 rad/sb)10 rad/sc)15 rad/sd)20 rad/sCorrect answer is option 'A'. Can you explain this answer?
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