Chapter 1 Physics PDF

# Unit 1 Properties of Matter ## Syllabus - Stress - Strain - Hooke's law - Elastic Behaviour of Material, Young's modulus by cantilever depression - Non-uniform bending, Uniform bending - Application - I - shaped girders, Torsional Pendulum - Couple per unit twist of a wire, Time period - Applicat...

# Unit 1 Properties of Matter ## Syllabus - Stress - Strain - Hooke's law - Elastic Behaviour of Material, Young's modulus by cantilever depression - Non-uniform bending, Uniform bending - Application - I - shaped girders, Torsional Pendulum - Couple per unit twist of a wire, Time period - Application - Determination of Rigidity Modulus ## Contents | Section | Topic | Marks | |---|---|---| | 1.1 | Introduction | 4 | | 1.2 | Stress - Strain - Hooke's Law | 3 | | 1.3 | Elastic Behaviour of Material | 7 | | 1.4 | Bending of Beam | 7 | | 1.5 | Torsional Pendulum - Couple per Unit Twist of a Wire, Time Period | 7 | | 1.6 | Application - I-shaped Girders | 4 | | 1.7 | Application - Determination of Rigidity Modulus | 4 | | 1.8 | Solved Examples | | ## 1.1 Introduction This chapter is based on solids. Until now you have studied about rigid bodies in earlier classes. - **What is a rigid body?** - A material body in which the distance between any two particles / atoms remains unchanged. - Distance between any two particles / atoms / molecules that lies on the rigid body never changes under the influence of external applied force. - That means body remains undeformed. - **Does perfectly rigid body exist?** - Answer to this question is No. In reality no such body exists: no body is perfectly rigid body. - **Deforming force for different bodies is different depending on the body.** - Let us see some definitions. - So we were talking about deforming force. When we apply a deforming force body gets deformed. What does this actually mean? What is the deforming force? ### Deforming force: - External force that changes the shape and size of the body. - ΔΙ is elongation. An external force that causes temporary or permanent deformation in a body. ### Restoring force: - The internal force developed inside the body due to change in shape and size of the body. - It tries to regain its original shape and size of the body. - It is also known as elastic force. - This force will help the body to regain its original shape and size. ### Elasticity: - It is the property of body by virtue of which it tries to regain its original shape and size when deforming force is removed. - The body possessing this property is called an elastic body. - Example: Steel, rubber, glass. ### Plastic body: - A body which does not regain its original shape and size after removal of deforming force. - Example: Chalk. ### Review Questions 1. Define and describe elasticity and plasticity in detail by giving suitable examples. 2. Define ductility, brittleness and elasticity. ## 1.2 Stress - Strain - Hooke's Law ### 1.2.1 Stress - When an external deforming force is applied on the body, a restoring force is developed due to deformation. - Hence, stress is restoring forces developed per unit area. $$Stress = \frac{Restoring Force}{Area}$$ - Under equilibrium, the magnitude of external deforming and internal restoring forces is equal. - At equilibrium, a = 0, because of very slow process $$F = F_{res}$$ - The magnitude of internal restoring force per unit area is termed as stress. $$Stress = \frac{F}{A}$$ - S.I. unit of stress is N/m2 - C.G.S. unit of stress is dyne/cm2 ### 1.2.2 Strain - Example: I have a wire. I will pull it from both ends, what will happen? - The wire will get stretched. Means its length will increase so I applied some kind of tension / force on wire say, deforming force, then the shape and size or say dimension of the wire changed. - Strain is defined as, $$Strain = \frac{Change in dimension}{Original dimension}$$ - It's basically a ratio. - The fractional change in dimension of a body is called as strain. - It has no units. - Strain is the effect that is produced because of deforming force. - There are three types of strain. 1. **Longitudinal strain:** - It is defined as change in length per unit original length. $$Longitudinal Strain = \frac{dl}{l}$$ $$Longitudinal Strain = \frac{Change in length}{Original length}$$ 2. **Volume strain:** - It is defined as changed in volume per unit original volume. $$Volume Strain = \frac{Change in Volume}{Original Volume}$$ 3. **Shear strain:** - It is defined as angular deformation (θ). $$tan \theta = \frac{\Delta{x}}{l} = Shear Strain$$ $$\Delta{x} \rightarrow 0 \text{ (very small)}$$ $$...tan \theta \approx \theta$$ $$\theta = \frac{\Delta{x}}{l}$$ ### 1.2.3 Hooke's Law - Now you know what is stress and strain. $$Stress \propto Strain$$ - Physicist Robert Hooke studied the relation between the extension produced in a wire and tension in wire. - "Within elastic limit of a material, the extension of an elastic body is directly proportional to applied force." - Modified by Thomas Young, true for all kinds of deformations. - "Within elastic limit of material, stress is proportional to the strain produced." $$Stress \propto Strain$$ $$Stress = Constant (E) \times Strain$$ - Hooke's law is valid for small deformation. - The constant E is called as modulus of elasticity, co-efficient of elasticity. - It depends on the nature of the material and not on the dimension of the substance. - S.I. unit is N/m2, dimension [M' L¯¹ T¯²] ### Review Question 1. Define Hooke’s law, stress, and strain. ## 1.3 Elastic Behaviour of Material ### 1.3.1 Types of Modulus of Elasticity - For a given material, there can be different types of modulus of elasticity depending on the types of stress and resulting strain produced.. - According to the nature of strain, there are three modulus of elasticity: 1. Young's modulus (Y) 2. Bulk modulus (K) 3. Modulus of rigidity (η) #### 1. Young's modulus (Y) - It is defined as the ratio of longitudinal stress to the longitudinal strain within elastic limit. $$Y = \frac{Longitudinal Stress}{Longitudinal Strain}$$ - When a change of length takes place the strain is known as longitudinal strain. - It is measured by the change in length per unit length. If L is the length of a wire and an increase l in length is caused by a force then, $$Strain = \frac{l}{L}$$ $$Y = \frac{Stress}{Strain}$$ $$Y = \frac{F/a}{l/L}$$ $$Y = \frac{FL}{al}$$ $$Y = \frac{FL}{al} = \frac{mgL}{\pi r^2 l}$$ #### 2. Bulk modulus: - It is defined as the ratio of stress to the volumetric strain $$Bulk Modulus (K) = \frac{Stress}{Volume Strain}$$ - Reciprocal of bulk modulus is called compressibility. - When a force is applied normally to the surface of a body and a change in volume takes place, the strain is known as volumetric strain. - It is measured by the change in volume per unit volume. - If v is changed in volume and V is original volume. - The bulk modulus K is given by $$K = \frac{v/V}{\Delta{v}}$$ $$K = \frac{\Delta{v}}{V}$$ $$K = \frac{P}{\Delta{v}/V}$$ - When a very small pressure dp is applied, the change in volume being very small is represented by dV, then $$-K = \frac{dp}{dV/v}$$ - The negative sign indicates that an increase in applied pressure causes a decrease in volume. #### 3. Modulus of rigidity: - It is defined as the ratio of the tangential stress to shearing strain. - To find the value of the shearing stains, consider a solid cube, shown in Fig. 1.3.2. - Let a tangential force F be applied to the upper face of cube of each edge L with lower face fixed. - The layers parallel to the upper face of the cube are displaced along F, through distance l w.r.t. lower fixed face, with point A shifting to A', B to B'. $$Shearing Strain = \frac{AA'}{AD} = \frac{tan \phi}{l}$$ - As Φ is small tan Φ = Φ - Modulus of rigidity, $$η = \frac{F/A}{\phi} = \frac{F}{A\phi}$$ - It is also known as coefficient of transverse elasticity. ### 1.3.2 Poisson’s Ratio - When a wire is stretched, its length increases but diameter decreases. - Thus, the linear strain, i.e. strain along the direction of applied load (or length of the wire) is always accompanied by the lateral strain or the strain along the lateral or perpendicular to the direction of applied load. - The Poisson’s ratio is defined as the ratio of lateral strain to the longitudinal strain. $$(σ) = \frac{Lateral Strain}{Longitudinal Strain}$$ Let: - The length of wire = L - Diameter of wire = D - Increase in length = l - Corresponding decrease in diameter = d $$... The Poisson's ratio = σ = \frac{-d/D}{+l/L}$$ $$σ = \frac{β}{α}$$ where, - α → Longitudinal strain per unit stress - β→ Lateral strain per unit stress. σ is dimensionless quantity and its value lies between 0.5 and – 1. ### 1.3.3 Elastic Limit - Elastic limit is the maximum stress within which the body exhibits the property of elasticity. - Below the elastic limit, the body regains its original position or shape or size when the deforming force is removed. - Beyond the elastic limit, the body does not regain completely its original position or shape or size even though external force is withdrawn. In solids, if the stress is gradually increased, the strain also increases in accordance with the Hooke's law, unit a point at which linear relation between strain and stress ceases out. - When a wire is loaded beyond the elastic limit, Hooke's law is no longer obeyed and the extension produced in the wire is not proportional to the stress. - The extension (strain) is more than the corresponding increase in stress. At this stage the particles of material go further apart. Now, if the load is withdrawn, the particles do not regain their original positions. - Point B in Fig. 1.3.4 represents yield point. - As the stress is further increased, a maximum value of stress is reached at which wire starts to elongation without further increase in stress called neck point C as shown in Fig. 1.3.4. - At point D, wire breaks. The maximum value of stress corresponding to point D is called breaking stress. ### 1.3.4 Work done in stretching a Wire - Strain Energy - Whenever a body is deformed by application of external forces the body gets strained. - The work done is stored in the body in the form of energy, known as energy of strain. [Work is done against the internal restoring force acting between the particles of wire] - Let us calculate the work done by stretching a wire, producing longitudinal strain. - Consider an elastic wire fixed at one end and other end is being stretched by an external force F. Let: - L- Length of wire - A → Area of cross - section of wire - Y→ Young's modulus of material of the wire - l→ Change in length of wire - The Young's modulus of elasticity $$Y = \frac{Longitudinal Stress}{Longitudinal Strain}$$ $$Y = \frac{F/A}{l/L}$$ $$Y = \frac{FL}{al}$$ $$Y = \frac{YAI}{L}$$ $$F = \frac{YAI}{L}$$ - Work done in producing a stretching dl = F. dl = dW $$dW = \frac{YAI}{L}dl$$ - Hence work done to produce a stretching of the wire from 0 to l $$W = \int_{0}^{l}{ \frac{YA}{L}dl}$$ $$W = \int_{0}^{l}{ \frac{YA}{L}dl}$$ $$W = [\frac{YAl}{L}]_{0}^{l}$$ $$W = \frac{YAI}{2L}(l^2 - 0)$$ $$W = \frac{YAI}{2L} \times l^2$$ $$W = \frac{1}{2} \times FXl$$ $$W = \frac{1}{2} \times Stretching force \times Extension$$ - Work done is stored as elastic potential energy in the wire which is represented by U. $$U = \frac{1}{2} \times YAI \times l$$ $$U = \frac{1}{2} \times Stress \times Strain \times Volume$$ $$Stress = Y$$ $$Stress = Y \times Strain$$ $$Elastic energy density = \frac{1}{2} \times Stress \times Strain$$ ## Angle of twist (θ) and Angle of shear (Φ) : - Consider a cylindrical rod / wire of length l and radius r fixed at lower end. - Suppose a twisting couple is applied to the free upper end shown by arrow head, in a direction perpendicular to the length of the cylinder. - Each circular cross-section is rotated about the axis of rod OO' through a certain angle. - The angle through which the cross-section is rotated about axis of cylinder is called angle of twist (θ) of thot cross-section. - It is proportional to the distance of cross section from fixed end of cylinder. - Thus, θ is zero at fixed end and maximum at the free end. - Due to twist at free end, let OB takes a new position OB'. - AT free end, Arc BB' = rθ $$...Arc BB' = r\phi .... (1.3.1)$$ - Now, due to twist at free end, a line AB on cylindrical layer of rod parallel to axis OO' takes new position AB'. The angle between AB and AB' is called as angle of shear (Φ) - It is proportional to the distance of layer from axis of rod. - Thus, Φ is minimum and zero at the axis of rod and maximum at the outermost cylindrical surface of rod. - At the outer surface, Arc BB' = l Φ $$... (1.3.2)$$ - From equations (1.3.1) and (1.3.2) rθ = l Φ (Torque is a twisting or turning force that tends to cause rotation around an axis.) $$\phi = \frac{r\theta}{l}$$ $$Thus, Angle of shear = \frac{Radius of rod}{Length of rod} \times Angle of twist$$ ### Review Questions 1. Explain the terms: (i) Young's modulus, (ii) Bulk modulus, (iii) Shear modulus and (iv) Poisson's ratio. Show any one relation between different moduli of elasticity. 2. Draw: Stress - Strain diagram with necessary notation. Explain : Elastic limit and upper yield point in detail. 3. Describe Stress-Strain diagram in detail. 4. Discuss in detail the different types of elasticity. List different factors affecting elasticity. 5. Draw stress versus strain graph with appropriate notations. Explain elastic limit and upper yield point in detail. 6. Define Poisson's ratio. Write its expression. 7. Discuss stress-strain diagram. 8. Derive an expression for twisting couple in a wire. 9. Explain the term bulk modulus, modulus of rigidity and yield point. 10. Draw Stress versus Strain diagram with necessary notation and discuss it. ## 1.4 Bending of Beam **Beam:** - A beam is defined as a structure of uniform cross-section, whose length is much larger as compared to its breadth and thickness. - So a rod of uniform cross-section (rectangular, circular, elliptical, etc.) of much greater length as compared to its thickness is called as beam. - Beams are used in the construction of bridges and infrastructure where heavy loads are to be supported. They are mostly used in the structure of multistoried buildings. - A beam can be bend in two ways : 1. When a beam is clamped horizontally at one end and loaded at the other, called as cantilever. 2. The beam is supported at its two ends and loaded at the middle. ### Some definitions: - Consider a beam clamped rigidly at one end and loaded at the free end. 1. **Longitudinal filaments:** - A beam may be considered to be made up of thin parallel layers (longitudinal). Each layer is considered to be made up of number of longitudinal filament. 2. **Neutral surface:** - During the bending of the beam the layer of the material in the upper half are extended and those in the lower half are compressed. There is one layer PQNS in the middle, whose length remained unchanged. This layer is called the neutral surface. - The neutral surface is perpendicular to the plane of bending and passes through the centre of area of cross - section of the beam. 3. **Neutral axis:** - The intersection of neutral surface and plane of bending is called the neutral axis. 4. **Plane of bending:** - The plane containing the neutral axis and the centre of curvature of the neutral axis. - The plane in which bending of beam takes place is called (plane of) bending of beam. ### 1.4.1 Bending Moments #### I) External Bending Moment (EBM) : - **Couple** → Pair of equal (parallel) and opposite (direction) forces. - Let a beam ABCD be fixed at the end AD and loaded (within elastic limit) with a load W at the free end BC so as to bend a little and remaining in equilibrium in the position shown in Fig. 1.4.3. - Consider a section PBCP' of the beam bent, cut by a plane PP' arbitrary right angles to its length and its plane of bending. - Clearly, the load W acting vertically downwards at its free end give rise to an equal and opposite reactional force W acting vertically upward at P. - These two equal and opposite forces form a couple, tending to bend the section clockwise. - This couple applied due to load W is called external bending moment. #### II) Internal bending moment : - Since the section is in equilibrium, there must be another equal and opposite couple called internal bending moment comes into play in the section due to tensile and compressive stresses set up in its upper and lower halves respectively. - For the filament above the neutral axis NN' are in a state of tension, being elongated and it exert an inward pull on the flaments next to them (towards fixed end). - The filaments below the neutral axis shortened and in the state of compression, exert an outward pull on the filament next to them (toward the loaded end). - These inward and outward pulls thus form a pair of equal and opposite forces form a couple, tending to bend the section in anti-clockwise, indicated by arrowheads. - The moment of this couple is called internal bending moments or restoring bending moment. - In equilibrium position of the section, both EBM and IBM are equal and opposite and in general called as bending moment. ### 1.4.2 Expression for External Bending Moment - Consider a beam of elastic material clamped horizontally at one end AD and loaded at another end BC with mass M. - Let l be the length of the beam. - Consider a cross - section C' at a distance x from O. At equilibrium, the forces acting on part BC are 1. The weight Mg acting vertically downward at BC end. 2. The weight mg of part BC acting downward at the centre of gravity G of part BC. - Resultant force acting downward = (Mg + mg) - This force give rise to an equal but opposite reaction force at section C'. There forces form a couple. - The moment of couple of force is called external bending moment. - External bending moment: Mg. BP + mg GC = Mg (1 - x) + mg $\frac{l-x}{2}$ where - l → Length of beam of - x → Distance of C from O. ### 1.4.3 Expression for Internal Bending Moment - Consider a portion of beam bent into a circular arc. - Let R be the radius of curvature of neutral surface at C, O be the angle subtended by part BC of neutral surface. $$Length of BC = R\theta$$ - On bending, the length of filament B'C' at a distance z from neutral surface = (R + z) (θ) - Change in length of filament = B'C' – BC = (R + z) θ - Rθ = Rθ + zθ – Rθ - Change in length = zθ $$Strain = \frac{Change in length}{Original Length}$$ $$Strain = \frac{z\theta}{R\theta}$$ $$Strain = \frac{z}{R}$$ - Now, Young's Modulus, $$Y = \frac{Stress}{Strain}$$ $$Stress = Y \times Strain$$ $$Stress = Y \times \frac{z}{R}$$ - Let, 'da' be an elementary area perpendicular to plane of bending and length of beam at distance z above neutral axis, - Then force on area da = Stress × Area $$f = Y\frac{z}{R}da$$ - The force f above and below neutral surface form a couple of force. $$Moment of force = Y\frac{z}{R}da \times z = Y\frac{z^2}{R}da$$ - This moment of force is called as internal bending moment. - The total internal bending moment about neutral axis of beam is given by $$I_{BM} = \int{Y\frac{z^2}{R}da}$$ where, I = ∑ z² da is called geometric moment of inertia of cross - section of beam. ### Condition for equilibrium - For equilibrium, - Internal bending moment = External bending moment - $$Y\frac{I}{R} = Mg (1-x)+ mg\frac{(1-x)}{2}$$ - If beam is light, the weight mg of part of beam can be neglected. - $$Y\frac{I}{R} = Mg (1-x)$$ - $$\frac{1}{R} \rightarrow Rate of change of slope $$ ## 1.4.4 Young's Modulus by Cantilever Method - **Depression of Cantilever** - Consider a light beam clamped horizontally at one end and loaded with mass M at free end B. - Let, C (x, y) be the curvature of neutral axis at C. - L be length of beam. - At equilibrium, Internal bending moment = External bending moment $$Y\frac{I}{R} = Mg (1-x) ...(1.4.1)$$ (. Beam is light) where, - y → Young's modulus of beam - z → Geometric moment of inertia - For small depression, $$ \frac{1}{R} = \frac{d^2y}{dx^2}$$ - Equation (1.4.1) becomes, $$YIx\frac{d^2y}{dx^2} = Mg (l-x)$$ - Integrating w.r.t. x, $$ \int{YIx\frac{d^2y}{dx^2}dx} = \int{Mg (l-x)dx}$$ $$YIx\frac{dy}{dx} = Mg(lx - \frac{x^2}{2} ) + C_1$$ - At x = 0 $\frac{dy}{dx}$ = 0 $$ YI \times 0 \times 0 = Mg( l \times 0 - \frac{0^2}{2} ) + C_1$$ $$0 = 0 + C_1$$ $$C_1=0$$ - Equation becomes, $$YIx\frac{dy}{dx} = Mg(lx - \frac{x^2}{2})$$ - Again integrating, $$ \int{YIx\frac{dy}{dx}dx} = \int{Mg(lx - \frac{x^2}{2})dx}$$ $$Yly = Mg(\frac{lx^2}{2} - \frac{x^3}{6}) + C_2$$ - at x = 0, y = 0 $$ YI \times 0 = Mg(\frac{l \times 0^2}{2} - \frac{0^3}{6}) + C_2$$ $$C_2 = 0$$ - Equation becomes, $$Yly = Mg(\frac{lx^2}{2} - \frac{x^3}{6})$$ $$y = \frac{Mg}{YI}(\frac{lx^2}{2} - \frac{x^3}{6})$$ - This gives the depression free end of neutral axis at any point (x, y). - If 8 be the depression at free end, then on putting y = 8 at x = l, we get, $$ 8 = \frac{Mg}{YI} (\frac{l \times l^2}{2} - \frac{l^3}{6})$$ $$8 = \frac{Mg}{YI} ( \frac{l^3}{2} - \frac{l^3}{6})$$ $$8 = \frac{Mg}{YI}(\frac{l^3}{3})$$ $$8 = \frac{Mg l^3}{3YI}$$ - This is an expression for cantilever at free end. a) For beam of rectangular cross - section, I = $\frac{bd^3}{12}$ where b→ breadth, d→ thickness $$8 = \frac{3MgL}{3Ybd}( \frac{bd^3}{12})$$ $$8 = \frac{MgL^3}{4Ybd^3}$$ b) For beam of circular cross - section of radius r, I = $\frac{\pi r^4}{4}$ $$8 = \frac{3}{3Y \pi r^4} \times{Mg l^3} \times \frac{4}{\pi r^4}$$ $$8 = \frac{4Mg l^3}{3 \pi r^4 Y}$$ ## 1.4.5 Expression for Depression of Beam supported at Two Knife Edges and loaded at the Middle - Consider a beam of uniform cross - section supported at two ends and loaded at the middle. Let: - l→ Distance between knife edges - Mg → Load at the middle of beam - Now, the reaction Mg/2 at each knife edge is acting vertically upward. - Thus each of half beam is equivalent to inverted cantilever of length l/2 loaded by Mg/2. - Hence, depression of the beam at the middle is given by, $$ 8 = (\frac{Mg}{2})(\frac{l}{2})^3 \times \frac{1}{3YI}$$ $$8 = \frac{Mg l^3}{2 \times 8 \times 3YI}$$ $$8 = \frac{Mg l^3}{48YI}$$ a) For beam of rectangular cross - section, I = $\frac{bd^3}{12}$ $$8 = \frac{MgL^3}{48Ybd^3} \times \frac{bd^3}{12}$$ $$8 = \frac{MgL^3}{4Ybd^3}$$ b) For beam of circular cross - section, I = $\frac{\pi r^4}{4}$ $$8 = \frac{Mg l^3}{48Y \pi r^4} \times \frac{\pi r^4}{4}$$ $$8 = \frac{Mg l^3}{48Y \pi r^4}$$ $$8 = \frac{Mg l^3}{12 \pi r^4 Y}$$ - It is used to describe the distribution of geometry about an axis. ## 1.4.6 Non-uniform Bending - Let us consider a beam of length ‘l’ (distance between two knife edges) supported on the two knife edges A and B as shown in Fig. 1.4.7. The load of weight ‘W’ is suspended at the centre ‘C’. It is found that the beam bends and the maximum displacement is at the point ‘D’ where the load is given. - Due to the load (W) applied, at the middle of the beam the reaction W/2 is acted vertically upwards at each knife edges. The bending is called non-uniform bending. - The beam may be considered as two cantilevers, whose free end carries a load of W/2 and fixed at the point ‘D’. - Hence we can say that the elevation of A above D as the depression below ‘A’. We know the depression of a cantilever. $$y=\frac{wi}{3YI}$$ $$...(1.4.2)$$ - Therefore substituting the value l and 1/2 and was W/2 in the expression for the depression of the cantilever we have. $$y = \frac{(W/2)(l/2)^3}{3YI} $$ $$...(1.4.3)$$ OR $$y = \frac{WI^3}{48YI}$$ $$...(1.4.4)$$ ## 1.4.7 Uniform Bending - Elevation at the centre of the beam loaded at both the ends. - Let us consider a beam of negligible mass, supported symmetrically on the two knife edges ‘A’ and ‘B’ and shown in Fig. 1.4.8. Let the length between A and B is ‘l’. Let equal weight W be added to either end of the beam ‘C’ and ‘D’. - Let CA = BD - Due to load applied the beam bends from F and E into an arc of a circle and produces as elevation ‘x’ from position F and E. Let ‘W’ be the reaction produced at the point A and B acts vertically upward as shown in Fig. 1.4.8. - Consider a point ‘P’ on the cross section of the beam. Then the forces acting on the part PC of the beam are: 1. Force W at ‘C’ and 2. Reaction W at A as shown in the Fig. 1.4.9. - Let the distance PC = a, and PA = a, then the external bending moment about ‘P’ is M₁ = Wxa₁-W × a₂ - Here the clockwise moment is taken as negative and anticlockwise moment is taken as positive. - External bending moment about P can be written as M₁ = W x (a₁ - a₂) M₂ = W P a - We know the internal bending moment = $$Y\frac{I}{R}$$ under equilibrium condition. - External bending moment = Internal bending moment ### Review Questions 1. What is cantilever? Obtain the expression for depression at free end of thin beam clamped horizontally at one end and loaded at other end. 2. Derive an expression for the depression of the free end of cantilever. What are the applications of cantilevers ? 3. Derive an expression for depression of cantilever. 4. Derive the expression for depression of cantilever loaded at free end. ## 1.5 Torsional Pendulum - Couple per Unit Twist of a Wire, Time Period ### 1.5.1 Expression for the Torque required to Twist a Cylinder - To find the value of the twisting couple imagine a cylindrical rod / wire of length l and radius r fixed at lower end and twisted at free upper end. - The rod is assumed to be made up of large number of thin co-axial cylindrical shells. - Consider one elementary shell of radius x and thickness dx. $$τ= \frac{\pi \eta x^4}{2l} \theta $$ - This equation gives the torque required to twist a cylinder at its free end. where, $$τ = C.θ$$ $$C = \frac{\pi \eta r^4}{2l}$$ known as torsional constant ### 1.5.2 Torsional Constant - The torque required to twist a cylinder at its free end is given by $$τ= \frac{\pi \eta r^4}{2l} \theta $$ where, - η → Modulus of rigidity - r → Radius of cylinder - l → Length of cylinder - θ → Angle of twist at free end - The constant $$\frac{\pi \eta r^4}{2l}$$ is called a torsional constant C. $$C = \frac{\tau}{\theta}$$ - Thus, the torsional constant is defined as the torque to twist a cylinder through one radian at the free end. It is also called as modulus of torsion or torsional rigidity. ### 1.5.3 Torsional Pendulum and its Time Period #### Torsional pendulum: - A regular body suspended from a rigid support by means of a long and thin elastic wire is called as torsional pendulum. - When a body is given slight rotation by applying couple, the wire is twisted and equal but opposite restoring couple is developed in it due to elasticity. - Restoring torque generates restoring angular acceleration opposite to the twist. The body therefore returns to the mean position. But due to inertia, body rotates in opposite direction thereby setting restoring torque in wire again in opposite direction and body returns to mean position. - Thus, the body oscillates about the wire in a plane. Such oscillations are called torsional oscillations. - The time period of such oscillations is given by $$T = 2π\sqrt{\frac{I}{C}}$$ where, - I → M.I. of disc about wire - C→ Torsional constant ### 1.5.4 Derivation for Time Period of Torsional Pendulum - Consider a torsional pendulum. - Let θ be the instantaneous angular displacement by which the pendulum is set into oscillations. - Then the restoring torque set up in the wire is given by, $$T = -C\theta$$ where, - C→ Torsional constant - θ→ Angle of twist at any time t. - If I be the M.I. of the body about the wire, then, $$τ = I\frac{d^2\theta}{dt^2}$$ $$I\frac{d^2\theta}{dt^2} = -C\theta$$ $$I\frac{d^2\theta}{dt^2} + C\theta = 0$$ $$I\frac{d^2\theta}{dt^2} + C\theta = 0$$ - Thus $$\frac{d^2\theta}{dt^2}$$ ∝ θ, angular acceleration is directly proportional to the angular displacement from the equilibrium position and is directed towards that position. - Hence, the pendulum performs simple harmonic motion. - The periodic time of the motion is given by, $$T = 2π\sqrt{\frac{Magnitude of angular acceleration per unit angular displacement}{}}$$ $$[T = 2π√\frac{1}{ω}]$$ $$[\frac{d^2x}{dt^2}+ω^2x = 0]$$ Differential equation of S.H.O. $$T = 2π√\frac{1}{C/I}$$ $$T = 2π√\frac{I}{C}$$ - This is an expression for time period of torsional pendulum. ### 1.5.5 Application - Determination of Modulus of Rigidity by using Torsional Pendulum - Consider a regular body (disc) suspended by means of thin elastic wire constituting torsional pendulum. - The system is set into torsional oscillations and time period of oscillations T, is noted. $$T_1 = 2π\sqrt{\frac{I}{C}}$$ where, $$C = \frac{\pi \eta r^4}{2l}$$ $$CT_1^2 = (2π)^2 I$$ $$CT_1^2 = 4