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A stone of mass 5 kg is placed at a distance 5m from the axis of rotaion on the rotating table ( angular velocity w with respect to earth) What would be the angular velocity w so that the stone starts slipping? {The frictional coefficient between stone and table is 1.5 }
- a)1.7 rad/sec
- b)1.5 rad/sec
- c)1.2 rad/sec
- d)None
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A stone of mass 5 kg is placed at a distance 5m from the axis of rotai...
If the stone starts slipping, then the centripetl force will be equal to the frictional force.
Most Upvoted Answer
A stone of mass 5 kg is placed at a distance 5m from the axis of rotai...
To solve this problem, we need to consider the forces acting on the stone and find the angular velocity at which the stone starts slipping.
1. Forces acting on the stone:
- Weight (mg): The weight of the stone acts vertically downwards.
- Normal force (N): The normal force acts perpendicular to the surface of the rotating table and balances the weight of the stone.
- Friction force (f): The friction force opposes the motion of the stone and prevents it from slipping.
2. Equilibrium condition:
For the stone to be in equilibrium and not slip, the friction force should be equal to the maximum static friction force. The maximum static friction force can be calculated using the formula: f_max = μN, where μ is the coefficient of friction.
3. Calculation:
- Weight of the stone (mg) = mass (m) × acceleration due to gravity (g)
= 5 kg × 9.8 m/s²
= 49 N
- Normal force (N) = Weight of the stone (mg) = 49 N
- Maximum static friction force (f_max) = coefficient of friction (μ) × Normal force (N)
= 1.5 × 49 N
= 73.5 N
4. Angular velocity at which the stone starts slipping:
The maximum static friction force provides the necessary centripetal force for the stone to remain in circular motion. The centripetal force can be calculated using the formula: F_c = m × r × ω², where r is the distance from the axis of rotation and ω is the angular velocity.
- Centripetal force (F_c) = Maximum static friction force (f_max)
- m × r × ω² = f_max
- 5 kg × 5 m × ω² = 73.5 N
- ω² = 73.5 N / (5 kg × 5 m)
- ω² = 2.94 rad²/s²
- ω = √(2.94 rad²/s²)
- ω ≈ 1.71 rad/s
5. Answer:
The angular velocity (ω) at which the stone starts slipping is approximately 1.71 rad/s. Therefore, the correct answer is option A (1.7 rad/s).
1. Forces acting on the stone:
- Weight (mg): The weight of the stone acts vertically downwards.
- Normal force (N): The normal force acts perpendicular to the surface of the rotating table and balances the weight of the stone.
- Friction force (f): The friction force opposes the motion of the stone and prevents it from slipping.
2. Equilibrium condition:
For the stone to be in equilibrium and not slip, the friction force should be equal to the maximum static friction force. The maximum static friction force can be calculated using the formula: f_max = μN, where μ is the coefficient of friction.
3. Calculation:
- Weight of the stone (mg) = mass (m) × acceleration due to gravity (g)
= 5 kg × 9.8 m/s²
= 49 N
- Normal force (N) = Weight of the stone (mg) = 49 N
- Maximum static friction force (f_max) = coefficient of friction (μ) × Normal force (N)
= 1.5 × 49 N
= 73.5 N
4. Angular velocity at which the stone starts slipping:
The maximum static friction force provides the necessary centripetal force for the stone to remain in circular motion. The centripetal force can be calculated using the formula: F_c = m × r × ω², where r is the distance from the axis of rotation and ω is the angular velocity.
- Centripetal force (F_c) = Maximum static friction force (f_max)
- m × r × ω² = f_max
- 5 kg × 5 m × ω² = 73.5 N
- ω² = 73.5 N / (5 kg × 5 m)
- ω² = 2.94 rad²/s²
- ω = √(2.94 rad²/s²)
- ω ≈ 1.71 rad/s
5. Answer:
The angular velocity (ω) at which the stone starts slipping is approximately 1.71 rad/s. Therefore, the correct answer is option A (1.7 rad/s).
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A stone of mass 5 kg is placed at a distance 5m from the axis of rotai...
If the stone starts slipping, then the centripetl force will be equal to the frictional force.
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A stone of mass 5 kg is placed at a distance 5m from the axis of rotaion on therotating table ( angular velocity w with respect to earth)What would be the angular velocity w so that the stone starts slipping?{The frictional coefficient between stone and table is 1.5 }a)1.7 rad/secb)1.5 rad/secc)1.2 rad/secd)NoneCorrect answer is option 'A'. Can you explain this answer?
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A stone of mass 5 kg is placed at a distance 5m from the axis of rotaion on therotating table ( angular velocity w with respect to earth)What would be the angular velocity w so that the stone starts slipping?{The frictional coefficient between stone and table is 1.5 }a)1.7 rad/secb)1.5 rad/secc)1.2 rad/secd)NoneCorrect answer is option 'A'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A stone of mass 5 kg is placed at a distance 5m from the axis of rotaion on therotating table ( angular velocity w with respect to earth)What would be the angular velocity w so that the stone starts slipping?{The frictional coefficient between stone and table is 1.5 }a)1.7 rad/secb)1.5 rad/secc)1.2 rad/secd)NoneCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone of mass 5 kg is placed at a distance 5m from the axis of rotaion on therotating table ( angular velocity w with respect to earth)What would be the angular velocity w so that the stone starts slipping?{The frictional coefficient between stone and table is 1.5 }a)1.7 rad/secb)1.5 rad/secc)1.2 rad/secd)NoneCorrect answer is option 'A'. Can you explain this answer?.
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