ELASTICITY

ELASTICITY

This can be defined as the ability of a material (elastic material) to return or regain its original shape/ size after deformation/ after the removal of stress/ force/ after it has been compressed.

TERMS USED IN ELASTICITY

1. Elastic Limit: this can be defined as the maximum load which a body can withstand and still retain its original size/ shape after the load/ stress has been removed. It can also be defined as a point on a stress-strain graph beyond which Hooke’s law is violated or disobeyed.
2. Yield Point: This can be defined as that point beyond which a stretched material fails to return to its original position.
3. Maximum Load: This can be defined as that load beyond which a material fails to stand any further addition of load.
4. Breaking Point: This can be defined as the point at which material breaks away from its original position after being stretched beyond its yield point and the material cannot withstand any further load.

HOOKE’S LAW

It states that provided the elastic or proportional limit is not exceeded, the force applied to an elastic material is directly proportional to the extension produced.

Mathematically,

                        F α e

                        F = Ke

                        Where K= Elastic Constant/ Force Constant/ Stiffness

                                    F= Applied Force

                                    e = Extension/ Compression

Example 1: A spring of natural length 3m is extended by 0.01m by a force of 4N. What will be its length when the applied force is 12N?

Solution:

Given: l₀ = 3m; e = 0.01m; F =4N

Required: l =?

Analysis:

            F = Ke

            K = F/e

            K = 4/0.01 = 400Nm^-1.

When F = 12N, K = 400Nm^-1;

            e = F/K = 12/ 400

              e = 0.03m

            Therefore,

                              l = l₀ + e   

                                l = 3 + 0.03 = 3.03m.

YOUNG’S MODULUS OF ELASTICITY

Suppose a force F is applied to a wire of original length l₀ and cross-sectional area which extend its length by extension, we can also state Hooke’s Law as;

                                   F/A α e/l₀

                       Where F/A = Tensile Stress; e/l₀ = Tensile Strain

Hence, Hooke’s law is also stated as Stress is proportional to Strain, i.e.

                             Stress α Strain

                        Stress = ϵ Strain,

                    Where ϵ = Young Modulus

ENERGY STORED OR WORKDONE ON SPRINGS AND ELASTIC STRINGS

Work is done when an elastic material is stretched or compressed. If the force stretching or compressing the material is F, then the workdone is given by;

                                     Work = Average Force x Extension/ Compression

                                Work = ½ (F_i + F_f) e

                                But F_i = 0N, F_f = F

                                W = ½ Fe

                                From Hooke’s law,

                                 F = Ke

                               Therefore,

                               W = ½ Ke²

ELASTIC POTENTIAL ENERGY

Elastic potential energy is the ability of the stretched or compressed elastic material to do work. This is energy stored in the material as a result of stretched or compressed state. Elastic potential energy is also given by;

                   W = ½ Fe

                W = ½ Ke²

Example 2: A spiral spring is compressed by 0.02m. calculate the energy stored in the spring if the force constant is 400Nm^-1.

Solution

Given: e = 0.02m; K = 400Nm^-1.

Required: E =?

Analysis:

            E = ½ Ke²

            E = ½ x 400 x 0.02 x 0.02

            E = 0.8J

Recall,

           Elastic potential energy is a stored energy which can be transformed into other forms of energy e.g. when stretch the rubber of a catapult and project a stone from it, the elastic potential energy stored in the rubber is transformed into the kinetic energy of the flying stone according to the law of conservation of energy i.e. potential energy is equal to kinetic energy of flying stone

            ½ Ke² = ½ mv²

Example 3: A stone of mass 20g is released from a catapult whose rubber has been stretched through 4cm. If the force constant of the rubber is 20Nm^-1, calculate the velocity with which the stone leave the catapult.

Solution

Given: m = 20g = 0.02kg; e = 4cm = 0.04m; K = 200Nm^-1

 Required: v=?

Analysis:

            P.E = K.E

            ½ Ke² = ½ mv²

            v = √(Ke²/m)

            v = √ (200 x 0.04²/0.02)         

            v = 4m/s

ASSIGNMENT

1. A metallic bar 50cm long has a uniform cross-sectional area of 4cm². If a tensile force of 35KN produces an extension of 0.25mm, calculate the value of Young Modulus.
2. The spiral of a spring balance is 75cm long when 5N acts on it and 30cm when the weight is 10N. what is the length of the spring when the weight is 3N. Assuming Hooke’s law is obeyed.
3. A stone of mass 30g is projected vertically upward with a catapult whose rubber cord has a force constant of 80N/m. If the tension in the cord at the point of release is 40N. Calculate;
4. The velocity of projection of the stone,
5. Maximum height attained by the stone.
6. Young Modulus for steel is 2 x 10¹²N/m². What extension will be produced in a steel wire of 1mm in diameter and 2.5m long by hanging a mass of 3kg on it.