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A small ring of mass m is attached at an end of a light string, the other end of which is tied to a small block B of mass 2m. The ring is free to move on a fixed smooth horizontal rod. Find the velocity of the ring when the string becomes vertical.?
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A small ring of mass m is attached at an end of a light string, the ot...
Problem:
A small ring of mass m is attached at an end of a light string, the other end of which is tied to a small block B of mass 2m. The ring is free to move on a fixed smooth horizontal rod. Find the velocity of the ring when the string becomes vertical.

Solution:
To solve this problem, we can use the principles of conservation of energy and conservation of linear momentum.

1. Initial Conditions:
- At the initial position, the ring and the block are at rest.
- The height of the ring from the ground is h.
- The length of the string is L.

2. When the string becomes vertical:
- At this point, the block B has fallen a distance h, and the string has become vertical.
- The ring has also moved horizontally a distance x.

3. Conservation of Energy:
- At the initial position, the total mechanical energy of the system is given by:
E1 = mgh
(where g is the acceleration due to gravity)
- At the final position, the total mechanical energy of the system is given by:
E2 = mgh + (1/2)mV^2 + (1/2)(2m)V^2
(where V is the velocity of the ring when the string becomes vertical)
- According to the conservation of energy, E1 = E2:
mgh = mgh + (1/2)mV^2 + (1/2)(2m)V^2

4. Conservation of Linear Momentum:
- At the initial position, the total linear momentum of the system is zero, as both the ring and the block are at rest.
- At the final position, the total linear momentum of the system is given by:
P2 = mV + 2mV
(where P2 is the total linear momentum of the system when the string becomes vertical)
- According to the conservation of linear momentum, P2 = 0:
mV + 2mV = 0

5. Solving the Equations:
- From the conservation of linear momentum equation, we get:
mV + 2mV = 0
3mV = 0
V = 0

- Substituting V = 0 in the conservation of energy equation, we get:
mgh = mgh + 0 + 0
mgh = mgh

6. Conclusion:
The velocity of the ring when the string becomes vertical is zero. This means that the ring comes to rest at the moment the string becomes vertical.

Summary:
- The velocity of the ring when the string becomes vertical is zero.
- This is because of the conservation of energy and conservation of linear momentum principles.
- The ring comes to rest at the moment the string becomes vertical.
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A small ring of mass m is attached at an end of a light string, the ot...
It's 18mg
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A small ring of mass m is attached at an end of a light string, the other end of which is tied to a small block B of mass 2m. The ring is free to move on a fixed smooth horizontal rod. Find the velocity of the ring when the string becomes vertical.?
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A small ring of mass m is attached at an end of a light string, the other end of which is tied to a small block B of mass 2m. The ring is free to move on a fixed smooth horizontal rod. Find the velocity of the ring when the string becomes vertical.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A small ring of mass m is attached at an end of a light string, the other end of which is tied to a small block B of mass 2m. The ring is free to move on a fixed smooth horizontal rod. Find the velocity of the ring when the string becomes vertical.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small ring of mass m is attached at an end of a light string, the other end of which is tied to a small block B of mass 2m. The ring is free to move on a fixed smooth horizontal rod. Find the velocity of the ring when the string becomes vertical.?.
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