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A 1 kg stone at end of a 1 m long shring is whirled on vertical circle with a constant speed of 4 m per second the answer in the string is 6 Newton then the stone age?
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A 1 kg stone at end of a 1 m long shring is whirled on vertical circle...
Understanding the Problem
In this scenario, we have a stone of mass 1 kg being whirled in a vertical circle with a radius of 1 m at a constant speed of 4 m/s. The tension in the string is provided as 6 Newtons.
Forces Acting on the Stone
- Weight of the Stone: The gravitational force acting downwards is given by:
- Weight (W) = mass (m) × gravity (g)
- W = 1 kg × 9.81 m/s² = 9.81 N
- Centripetal Force Requirement: For circular motion, a centripetal force is required to keep the stone moving in a circle. This force is provided by the tension in the string and the weight of the stone.
Applying Newton's Second Law
At the top of the circle, the forces acting on the stone can be expressed as:
- Tension (T) + Weight (W) = Centripetal Force (F_c)
Where:
- T = 6 N (tension in the string)
- W = 9.81 N (weight of the stone)
- F_c = m × (v²/r) = 1 kg × (4 m/s)² / 1 m = 16 N
Calculating Tension
At the top of the circle:
- 6 N + 9.81 N = 16 N
This indicates that the system is in equilibrium, and the tension is sufficient to provide the required centripetal force needed for circular motion.
Conclusion
In summary, the tension in the string (6 N) along with the weight of the stone (9.81 N) balances the centripetal force needed (16 N) for maintaining the circular motion of the stone. Thus, the stone is effectively moving in a vertical circle under the influence of gravity and tension in the string.
In this scenario, we have a stone of mass 1 kg being whirled in a vertical circle with a radius of 1 m at a constant speed of 4 m/s. The tension in the string is provided as 6 Newtons.
Forces Acting on the Stone
- Weight of the Stone: The gravitational force acting downwards is given by:
- Weight (W) = mass (m) × gravity (g)
- W = 1 kg × 9.81 m/s² = 9.81 N
- Centripetal Force Requirement: For circular motion, a centripetal force is required to keep the stone moving in a circle. This force is provided by the tension in the string and the weight of the stone.
Applying Newton's Second Law
At the top of the circle, the forces acting on the stone can be expressed as:
- Tension (T) + Weight (W) = Centripetal Force (F_c)
Where:
- T = 6 N (tension in the string)
- W = 9.81 N (weight of the stone)
- F_c = m × (v²/r) = 1 kg × (4 m/s)² / 1 m = 16 N
Calculating Tension
At the top of the circle:
- 6 N + 9.81 N = 16 N
This indicates that the system is in equilibrium, and the tension is sufficient to provide the required centripetal force needed for circular motion.
Conclusion
In summary, the tension in the string (6 N) along with the weight of the stone (9.81 N) balances the centripetal force needed (16 N) for maintaining the circular motion of the stone. Thus, the stone is effectively moving in a vertical circle under the influence of gravity and tension in the string.
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A 1 kg stone at end of a 1 m long shring is whirled on vertical circle with a constant speed of 4 m per second the answer in the string is 6 Newton then the stone age?
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A 1 kg stone at end of a 1 m long shring is whirled on vertical circle with a constant speed of 4 m per second the answer in the string is 6 Newton then the stone age? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 1 kg stone at end of a 1 m long shring is whirled on vertical circle with a constant speed of 4 m per second the answer in the string is 6 Newton then the stone age? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 1 kg stone at end of a 1 m long shring is whirled on vertical circle with a constant speed of 4 m per second the answer in the string is 6 Newton then the stone age?.
A 1 kg stone at end of a 1 m long shring is whirled on vertical circle with a constant speed of 4 m per second the answer in the string is 6 Newton then the stone age? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 1 kg stone at end of a 1 m long shring is whirled on vertical circle with a constant speed of 4 m per second the answer in the string is 6 Newton then the stone age? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 1 kg stone at end of a 1 m long shring is whirled on vertical circle with a constant speed of 4 m per second the answer in the string is 6 Newton then the stone age?.
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