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A light string ABCDE whose midpoint is C passes through smooth rings B and D, which are fixed in a horizontal plane distant 2a apart. To each of the points A,C,E is attached a mass m. Initially C is held at rest at O(midpoint BD) and is then set free. What is the distance OC when C comes to instantaneous rest?
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Problem: A light string ABCDE whose midpoint is C passes through smooth rings B and D, which are fixed in a horizontal plane distant 2a apart. To each of the points A,C,E is attached a mass m. Initially C is held at rest at O(midpoint BD) and is then set free. What is the distance OC when C comes to instantaneous rest?
Solution:
To solve this problem, we need to understand the concept of potential energy and kinetic energy. The total energy of the system is the sum of potential energy and kinetic energy. At the instant when C comes to rest, all its kinetic energy is converted into potential energy.
Step 1: Find the initial potential energy of the system
The initial potential energy of the system is given by U = 3mg(a/2) = 3mga/2, where m is the mass of each point A, C, and E.
Step 2: Find the initial kinetic energy of the system
The initial kinetic energy of the system is zero because C is held at rest.
Step 3: Find the potential energy of the system when C comes to rest
When C comes to rest, all its kinetic energy is converted into potential energy. Let the distance OC be x. At this instant, the potential energy of the system is U' = 2mg(a-x) + mgx = 2mga + mgx - 2mgx = 2mga - mgx.
Step 4: Equate the initial and final potential energies
Since energy is conserved, we can equate the initial potential energy and the final potential energy. Therefore, 3mga/2 = 2mga - mgx, which gives us x = a/2.
Answer: The distance OC when C comes to instantaneous rest is a/2.
Solution:
To solve this problem, we need to understand the concept of potential energy and kinetic energy. The total energy of the system is the sum of potential energy and kinetic energy. At the instant when C comes to rest, all its kinetic energy is converted into potential energy.
Step 1: Find the initial potential energy of the system
The initial potential energy of the system is given by U = 3mg(a/2) = 3mga/2, where m is the mass of each point A, C, and E.
Step 2: Find the initial kinetic energy of the system
The initial kinetic energy of the system is zero because C is held at rest.
Step 3: Find the potential energy of the system when C comes to rest
When C comes to rest, all its kinetic energy is converted into potential energy. Let the distance OC be x. At this instant, the potential energy of the system is U' = 2mg(a-x) + mgx = 2mga + mgx - 2mgx = 2mga - mgx.
Step 4: Equate the initial and final potential energies
Since energy is conserved, we can equate the initial potential energy and the final potential energy. Therefore, 3mga/2 = 2mga - mgx, which gives us x = a/2.
Answer: The distance OC when C comes to instantaneous rest is a/2.
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A light string ABCDE whose midpoint is C passes through smooth rings B and D, which are fixed in a horizontal plane distant 2a apart. To each of the points A,C,E is attached a mass m. Initially C is held at rest at O(midpoint BD) and is then set free. What is the distance OC when C comes to instantaneous rest?
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