A piece of wire is bent in shape of a parabola y 4x2 OX- axis is taken as vertical) with a bead of mass 20 g on it. The bead can slide on the wire without friction. It stays at the lowermost point of the parabola when the wire is at rest. The wire is now accelerated parallel to x- axis with a constant acceleration 4 m/s2. The distance of new equilibrium position of the bead from the axis of parabola, is
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A piece of wire is bent in the shape of a parabola y=kx
2
(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
Hard
JEE Advanced
Solution
verified
Verified by Toppr
Correct option is B)
Step 1: Choosing the Frame Of Reference and Drawing the FBD
Solving problem from the frame of wire which is accelerating towards right.
So, Pseudo force on mass m will act on left direction.
Pseudo force = mass of bead × acceleration of Observer
=ma
Step 2: Equilibrium conditon
At equilibrium position, the acceleration of block will be zero in the chosen frame.
Therefore, Balancing the forces in x and y direction, we get:
Nsinθ=ma ....(1)
Ncosθ=mg ....(2)
Dividing these two equations, we get
tanθ=
g
a
....(3)
search-icon-header
Search for questions & chapters
search-icon-image
Class 11
>>Physics
>>Laws of Motion
>>Equilibrium of Bodies
>>A piece of wire is bent in the shape of
Question
Bookmark
A piece of wire is bent in the shape of a parabola y=kx
2
(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
Hard
JEE Advanced
Solution
verified
Verified by Toppr
Correct option is B)
Step 1: Choosing the Frame Of Reference and Drawing the FBD
Solving problem from the frame of wire which is accelerating towards right.
So, Pseudo force on mass m will act on left direction.
Pseudo force = mass of bead × acceleration of Observer
=ma
Step 2: Equilibrium conditon
At equilibrium position, the acceleration of block will be zero in the chosen frame.
Therefore, Balancing the forces in x and y direction, we get:
Nsinθ=ma ....(1)
Ncosθ=mg ....(2)
Dividing these two equations, we get
tanθ=
g
a
....(3)
Step 2: Slope of cuve
Equation of curve, y=kx
2
From figure, we see that tangent to the curve at equilibrium point also makes the angle $$\theta$ with the x axis.
Therefore, Slope of curve =tanθ
⇒
dx
dy
=tanθ
⇒ 2kx=tanθ
⇒ 2kx=
g
a
(Using Equation 3)
∴ x=
2gk
a
hope it's helpful to you
The distance of the bead from the axis of the parabola at the equilibrium position is 8.72 m.
Explanation:
Let m be the mass of bead, N is the normal force, a is the acceleration of wire and g be acceleration due to gravity.
First of all, free body diagram is made. The free body diagram is shown in the figure attached.
Here the bead experience following forces:
- Pseudo force due to acceleration of bead
- Normal force due to contact
- Weight of bead
At equilibrium all forces must be balanced. So,
ma = N cos θ (1)
mg = N sin θ (2)
Dividing (2) by (1),
tan θ = g/a
It is given that a = 4 m/s^2, Taking g = 10 m/s^2.
tan θ = 10/2 = 2.5 (3)
By differentiating the equation of parabola
y = 4x^2
dy/dx = 8x (4)
From figure, slope of the tangent is
π/2 + tan ^-1 2.5 (5)
The slope of the tangent is the first derivative of the equation of parabola. So, equating (4) and (5)
8x = π/2 + tan ^-1 2.5 (6)
By solving (6) we get the value of x = 8.72 m.
Therefore, the distance of bead from X axis of the parabola is 8.72 m.
