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A small bead of mass m can slide on a frictionless fixed ring of radius r. With the help of two identical strings of force constant k, the bead is connected to two nails A and B each on dia meter at a distance 0.5r from centre O of the ring. Relaxed length of each string is negligible as compared to the radius of the ring. The bead is given a small velocity at point C. What can you predict about subsequent motion of the bead before any of the string strikes a nail?
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A small bead of mass m can slide on a frictionless fixed ring of radiu...
The bead will move in a centripetal circular path towards one of the nails, speeding up quickly as it gets closer. The strings will not become taut unless the bead happens to reach one of the nails, but the chances of that are very small.
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A small bead of mass m can slide on a frictionless fixed ring of radiu...
Motion of the Bead on the Fixed Ring

The motion of the bead on the fixed ring can be analyzed by considering the forces acting on the bead at different positions. Let's break down the analysis into different stages:

Stage 1: Initial Velocity at Point C
- At the beginning, the bead is given a small velocity at point C.
- As there is no friction and the bead is on a frictionless surface, it will continue to move in a circular path along the ring.

Stage 2: Tension in the Strings
- As the bead moves along the ring, the tension in the strings connected to nails A and B comes into play.
- The tension in each string will always be directed towards the center of the ring (point O).
- The tension in the strings will provide the necessary centripetal force to keep the bead in circular motion.

Stage 3: Length of the Strings
- As the bead moves, the strings connected to nails A and B will start to unwind from the nails.
- The length of each string will increase as the bead moves away from the respective nail.
- However, the relaxed length of each string is negligible compared to the radius of the ring, so the change in string length can be ignored for simplicity.

Stage 4: Velocity and Acceleration
- As the bead moves, its velocity will change due to the tension in the strings.
- The centripetal force provided by the tension in the strings will cause the bead to accelerate towards the center of the ring.
- The acceleration of the bead can be calculated using the equation: a = v^2 / r, where v is the velocity of the bead and r is the radius of the ring.

Stage 5: Bead's Path
- The bead will continue to move along the ring, following a circular path.
- The path of the bead will depend on the initial velocity given at point C and the tension in the strings.
- The bead will keep moving until one of the strings strikes its respective nail.
- At that point, the tension in that string will become zero, and the motion of the bead will be affected.

Predicting the Subsequent Motion
- Since the relaxed length of each string is negligible compared to the radius of the ring, the bead will continue to move in circular motion until one of the strings strikes its respective nail.
- The direction of the bead's motion will depend on the initial velocity given at point C.
- If the initial velocity is such that the bead is moving in the clockwise direction, then the bead will continue to move in the same direction until one of the strings strikes its respective nail.
- If the initial velocity is such that the bead is moving in the counterclockwise direction, then the bead will continue to move in the same direction until one of the strings strikes its respective nail.
- Once a string strikes its respective nail, the tension in that string becomes zero, causing the bead to move in a straight line tangent to the point of contact.
- The subsequent motion of the bead after one of the strings strikes its respective nail will depend on the initial velocity, the tension in the remaining string, and the angle at which the bead strikes the nail.

Conclusion
- The bead will continue to move in circular motion on the fixed ring until one
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A small bead of mass m can slide on a frictionless fixed ring of radius r. With the help of two identical strings of force constant k, the bead is connected to two nails A and B each on dia meter at a distance 0.5r from centre O of the ring. Relaxed length of each string is negligible as compared to the radius of the ring. The bead is given a small velocity at point C. What can you predict about subsequent motion of the bead before any of the string strikes a nail?
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A small bead of mass m can slide on a frictionless fixed ring of radius r. With the help of two identical strings of force constant k, the bead is connected to two nails A and B each on dia meter at a distance 0.5r from centre O of the ring. Relaxed length of each string is negligible as compared to the radius of the ring. The bead is given a small velocity at point C. What can you predict about subsequent motion of the bead before any of the string strikes a nail? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A small bead of mass m can slide on a frictionless fixed ring of radius r. With the help of two identical strings of force constant k, the bead is connected to two nails A and B each on dia meter at a distance 0.5r from centre O of the ring. Relaxed length of each string is negligible as compared to the radius of the ring. The bead is given a small velocity at point C. What can you predict about subsequent motion of the bead before any of the string strikes a nail? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small bead of mass m can slide on a frictionless fixed ring of radius r. With the help of two identical strings of force constant k, the bead is connected to two nails A and B each on dia meter at a distance 0.5r from centre O of the ring. Relaxed length of each string is negligible as compared to the radius of the ring. The bead is given a small velocity at point C. What can you predict about subsequent motion of the bead before any of the string strikes a nail?.
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