Young's modulus and energy stored in stretched wire | NEB physics | Knowledge of Physics
Q.1.) What is Young's modulus of elasticity? Derive expression for it.
Ans:
Young's modulus:
If the restoring force developed in a body is normal to its cross-section area, the stress due to such force is called normal stress. Normal stress is produced in a body when there is longitudinal or volumetric strain. Young's modulus of elasticity of a material is defined as the ratio of normal stress and longitudinal(or linear) strain.
From fig.1, if F is the external force applied to the body of cross-section area A, and original length `l_0` with change in length(extension or compression) `\Delta l` produced on it, then
...................(1) We have,
Normal Stress = `\frac{F}{A}` ........(2) and
Longitudinal Strain = `\frac{l_1 - l_0}{l_0} = \frac{\Delta l}{l_0}` ........(3)
Using (2) and (3) in (1),
`Y = \frac{\frac{F}{A}}{\frac{\Delta l}{l_0}}`
= `\frac{F \times l_0}{A \times \Delta l}`
`\therefore` Y = `(\frac{F}{A}) (\frac{l_0}{\Delta l})` ..........(4)
Fig. Energy stored in stretched wire of original length `l_0`.
When `\frac{\Delta l}{l_0} = 1`, that is, extension or compression in wire is equal to its original length, and A =1, then from (4) we have
Y = F ............(5)
Therefore, Young's modulus of a body can also be defined as the force per unit area required to produce an extension or compression on it equal to the original length of the body.
Dimensional Formula of Young's Modulus:
Here, F is measured in Newton whose dimensional formula is `[MLT^-2]` which is same as that of pressure, A is measured in meter squared whose dimensional formula is `[L^2]` and both of `\Delta l` and `l` being measured in meter so `\frac{\Delta l}{l}` is dimensionless, that is, `[M^0L^0T^0]`.
So, `[Y] = \frac{[MLT^-2]}{[L^2]}[M^0L^0T^0]`
`\therefore` `[Y] = [ML^-1T^-2]` which is required dimensional formula for Young's modulus and it is same as the dimensional formula of pressure.
Q.2.) Derive the expression for the energy stored in stretched wire. Hence, prove that energy density is directly proportional to the product of stress and strain.
Solution:
Energy stored in stretched wire:
When a force F is applied normally to the wire of cross section area A, the wire gets stretched. When the wire is stretched, the stress is developed in it internally and resists the applied force. That is, the stress tries to prevent the applied force to further stretch the wire. To stretch the wire further, some work needs to done against the stress. The work done by the applied force in further stretching the wire is stored in wire as its potential energy.
In fig.1, a force F, which is applied normally to the cross-section area A of the wire of initial length `l_0`, stretches the wire until its length becomes `l_1`. Then extension produced on the wire is,
`\Delta l` = x = `L_1 - L_0 = x_1 - x_0`
Young's modulus is given as
`Y = \frac{F }{A} \frac{l}{\Delta l}`
`\therefore` `Y = \frac{A}{A} \frac{l_0}{x}` ----------(1)
Now, to stretch the wire further by small amount of length `dx`, the small amount of work that needs to be done by the force F is,
`dW = Fdx = \frac{A Y}{l_0} x dx`
The total work done in extending the wire from `x =x_0` to `x = x_1` as shown in above fig. q, is given by
`W = \int_{x_0}^{x_1}` dW
` = \int_{x_0}^{x_1} \frac{AY}{l_0} xdx`
` = \frac{AY}{l_0} \int_{x_0}^{x_1} xdx`
Using the technique of integration,
`\int_a^b x^n dx = [\frac{x^{n+1}}{n+1}]_a^b`
`= \frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1}`
`= \frac{b^{n+1}- a^{n+1}}{n+1}`
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`W = \frac{AY}{l_0} [\frac{x^{1+1}}{1+1}]_{x_0}^{x_1}`
or, `W = \frac{AY}{l_0} [\frac{x^2}{2}]_{x_0}^{x_1}`
or, `W = \frac{AY}{l_0} [\frac{x_1^2 - x_0^2}{2}]`
or, `W = \frac{AY}{2 l_0} [x_1^2 - x_0^2]` ----------(2)
Equation (2) represents the expression for the energy stored in a stretched wire.
Suppose the wire is initially at the origin, that is at `x_0= 0`, then work done on extending the wire from `x_0 = 0` to `x_1 = x`, that is, `\Delta l = x_1 - x_0 = x - 0 = x`, extension produced on the stretched wire, can be given as
`\therefore` `W = \frac{1}{2 l_0} AYx^2 ` ----------(3)
This equation (3) gives the energy stored in a stretched wire of length `l_0`, cross-setion A, material of Young's modulus Y and extension produced x.
Energy Density:
The energy density of a deformed body, or the stretched wire in this case, is the energy stored in the stretched wire per unit volume of the wire.
ED = `\frac{W}{V}` -------(4)
Where, W = work done on deforming the body or work done on stretching the wire
V = Volume of the body or wire.
Here, W is given by equation (3) and V is cross-section area multiplied by the original length of wire, that is, `V = Al_0`.
From equation (3),
`W = \frac{1}{2l_0} YA x^2`
or, `W = \frac{1}{2l_0} (Y A x) x`
or, `W = \frac{1}{2} (\frac{YAx}{l_0}) x `
or, `W = \frac{1}{2} F x`
Where, `F = \frac{YAx}{l_0}`, which is applied deforming force.
Again,
`W = \frac{1}{2} Fx`
or, `W = \frac{1}{2} Fx \frac{Al_0}{Al_0}`
or, `W = \frac{1}{2} Fx \frac{V}{Al_0}`
or, `\frac{W}{V} = \frac{1}{2} \frac{Fx}{Al_0}`
`\therefore` `\frac{W}{V} = \frac{1}{2} \frac{F}{A} \frac{x}{l_0}` -------(5)
This equation represents the energy density of stretched wire.
But, `\frac{F}{A}` = Stress(Normal Stress) and `\frac{x}{l_0}` = Strain(Longitudinal Strain)
So, `\frac{W}{V} = \frac{1}{2}` Stress `\times` Strain
`\therefore` `\frac{W}{V} = \frac{1}{2}` Stress `\times` Strain -------(6)
From (4) and (6),
ED = `\frac{1}{2}` Stress `\times` Strain -------(7)
Thus, equation (7) shows that energy density is proportional to the product of stress and strain developed on a stretched wire.
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