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A piece of wire is bent in the shape of a parabola y= kx² (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk?
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A piece of wire is bent in the shape of a parabola y= kx² (y-axis vert...
tan θ = a/g
tan θ = dy/dx = 2kx
Hence,=> x = a/2gk
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A piece of wire is bent in the shape of a parabola y= kx² (y-axis vert...
Problem Analysis:
We have a wire bent in the shape of a parabola given by the equation y = kx². A bead of mass m is placed on the wire and it can slide without friction. Initially, when the wire is at rest, the bead stays at the lowest point of the parabola.
Given:
- Equation of the parabola: y = kx²
- Mass of the bead: m
- Acceleration of the wire: a
To Find:
The distance of the new equilibrium position of the bead from the y-axis.
Solution:
In order to determine the new equilibrium position of the bead, we need to consider the forces acting on it.
1. Weight of the Bead:
The weight of the bead acts vertically downwards and is given by the equation W = mg, where g is the acceleration due to gravity.
2. Normal Force:
The normal force on the bead acts perpendicular to the wire and is responsible for balancing the weight of the bead. At the lowest point of the parabola, the normal force is equal to the weight of the bead.
3. Centripetal Force:
When the wire is accelerated parallel to the x-axis, there is a centripetal force acting towards the center of the parabolic path. This force is responsible for keeping the bead in equilibrium.
4. Equating Forces:
At the new equilibrium position, the net force acting on the bead in the y-direction is zero. Therefore, we can equate the weight of the bead with the centripetal force to find the new equilibrium position.
5. Equating Weight and Centripetal Force:
Weight of the bead: mg
Centripetal force: m * (acceleration of the wire)
6. Equating the Equations:
mg = m * a
7. Canceling Mass:
g = a
8. Finding the Distance:
The distance of the new equilibrium position of the bead from the y-axis can be determined by substituting the value of g into the equation y = kx².
9. Substituting g into the Equation:
y = kx²
a = kx²
x = √(a/k)
10. Finding the Distance:
The distance of the new equilibrium position of the bead from the y-axis is √(a/k).
Answer:
Therefore, the correct option is A) a/gk, which is equivalent to √(a/k).
We have a wire bent in the shape of a parabola given by the equation y = kx². A bead of mass m is placed on the wire and it can slide without friction. Initially, when the wire is at rest, the bead stays at the lowest point of the parabola.
Given:
- Equation of the parabola: y = kx²
- Mass of the bead: m
- Acceleration of the wire: a
To Find:
The distance of the new equilibrium position of the bead from the y-axis.
Solution:
In order to determine the new equilibrium position of the bead, we need to consider the forces acting on it.
1. Weight of the Bead:
The weight of the bead acts vertically downwards and is given by the equation W = mg, where g is the acceleration due to gravity.
2. Normal Force:
The normal force on the bead acts perpendicular to the wire and is responsible for balancing the weight of the bead. At the lowest point of the parabola, the normal force is equal to the weight of the bead.
3. Centripetal Force:
When the wire is accelerated parallel to the x-axis, there is a centripetal force acting towards the center of the parabolic path. This force is responsible for keeping the bead in equilibrium.
4. Equating Forces:
At the new equilibrium position, the net force acting on the bead in the y-direction is zero. Therefore, we can equate the weight of the bead with the centripetal force to find the new equilibrium position.
5. Equating Weight and Centripetal Force:
Weight of the bead: mg
Centripetal force: m * (acceleration of the wire)
6. Equating the Equations:
mg = m * a
7. Canceling Mass:
g = a
8. Finding the Distance:
The distance of the new equilibrium position of the bead from the y-axis can be determined by substituting the value of g into the equation y = kx².
9. Substituting g into the Equation:
y = kx²
a = kx²
x = √(a/k)
10. Finding the Distance:
The distance of the new equilibrium position of the bead from the y-axis is √(a/k).
Answer:
Therefore, the correct option is A) a/gk, which is equivalent to √(a/k).
Community Answer
A piece of wire is bent in the shape of a parabola y= kx² (y-axis vert...
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A piece of wire is bent in the shape of a parabola y= kx² (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk?
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A piece of wire is bent in the shape of a parabola y= kx² (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk? 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 piece of wire is bent in the shape of a parabola y= kx² (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A piece of wire is bent in the shape of a parabola y= kx² (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk?.
A piece of wire is bent in the shape of a parabola y= kx² (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk? 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 piece of wire is bent in the shape of a parabola y= kx² (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A piece of wire is bent in the shape of a parabola y= kx² (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk?.
Solutions for A piece of wire is bent in the shape of a parabola y= kx² (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is A) a/gk B) a/2gk C) 2a/gk D) a/4gk? in English & in Hindi are available as part of our courses for Class 11.
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