5 Marks Questions

Question 15 Marks

Identical springs of steel and copper are equally stretched. On which, more work will have to be done?

Answer

Key concept: Work Done in stretching a Wire or Spring. In stretching a wire work is done against internal restoring forces. This work is stored in the wire as elastic potential energy or strain energy. If a force F acts along the length L of the wire of cross - section A and stretches it by x, then$\text{Y}=\frac{\text{stress}}{\text{strain}}=\frac{\text{F/A}}{\text{x/L}}=\frac{\text{FL}}{\text{Ax}}\Rightarrow\text{F}=\frac{\text{YA}}{\text{L}}\text{x}$
So the work done for an additional small increase dx in length,$\text{dW}=\text{Fdx}=\frac{\text{YA}}{\text{L}}\text{x}.\text{dx}$
Hence the total work done in increasing the length by I,$\text{w}=\int^1_0\text{dw}=\int^1_0\text{fdx}=\int^1_0\frac{\text{YA}}{\text{L}}\text{ xdx}=\frac{1}{2}\frac{\text{YA}}{\text{L}}\text{ l}^2$
This work done is stored in the wire. According to this problem work done in stretching a wire is given by$\text{W}=\frac{\text{YAl}^2}{2\text{L}}$
As springs of steel and copper are equally stretched. Therefore, for same force (F),$\text{W}\propto\Delta\text{l}\ ...(\text{i})$
As both springs are identical, so A and L are same.$\Delta\text{l}\propto\frac{\text{l}}{\text{Y}}\ ...(\text{ii})$
From eqs. (i) and (ii), we get $\text{W}\propto\frac{\text{l}}{\text{Y}}$$\therefore \frac{\text{W}_\text{steel}}{\text{W}_\text{copper}}=\frac{\text{Y}_\text{copper}}{\text{Y}_\text{steel}}<(\text{As, Y}_\text{steel}>\text{Y}_\text{copper})$
$\Rightarrow\text{W}_\text{steel}>\text{W}_\text{copper}$
Hence, more work will be done in case of spring made of copper.

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Question 25 Marks

A truck is pulling a car out of a ditch by means of a steel cable that is $9.1m$ long and has a radius of $5mm$. When the car just begins to move, the tension in the cable is $800N$. How much has the cable stretched?
(Young’s modulus for steel is $2 \times 10^{11}N m^{–2}$.)

Answer

Length of cable, $L = 9.1m r = 5mm = 5 \times 10^{-3}m$, $\text{A}=\pi\text{r}^2$ Tension in cable $= F = 800N Y = 2 \times 10^{11}N m^{-2}$$\Delta\text{L}=\frac{\text{FL}}{\text{AY}}=\frac{800\times9.10}{3.14\times10^{-3}\times10^{-3}\times5\times5\times2\times10^{11}}$
$\Delta\text{L}=\frac{800\times910\times10^{-11+6}}{314\times5\times5\times2}$
$=\frac{728}{157}\times10^{-5\text{m}}=4.64\times10^{-5}\text{m}$
$\Delta\text{L}=4.64\times10^{-5}\text{m}.$

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Question 35 Marks

A steel rod ($Y = 2.0 \times 10^{11}N m^{-2}$, and $\alpha = 10^{-50} C^{-1}$) of length $1m$ and area of cross-section $1cm$ 2 is heated from $0°C$ to $200°C$, without being allowed to extend or bend. What is the tension produced in the rod?

Answer

$\therefore\text{L}_\text{t}=\text{L}_0(1+\alpha\Delta\text{t})$$\text{L}_\text{t}-\text{L}_0=\text{L}_0\alpha.\Delta\text{t}$
$\Delta\text{L}=1\times10^{-5}\times200=2\times10^{-3}$
$\text{Y}=\frac{\text{FL}_0}{\text{A}\Delta\text{L}}\ \ \ \text{L}_0=1\text{m}$
$\therefore\text{F}=\frac{\text{YA}\Delta\text{L}}{\text{L}_0}\ \ \ \text{A}=1\text{cm}^2=10^{-4}\text{ m}^2$
$\text{Y}=2\times10^{11}\text{Nm}^{-2}$
$\Delta\text{L}=2\times10^{-3}\text{m}$
$\text{F}=\frac{2\times10^{11}\times10^{-4}\times2\times10^{-3}}{1}$
$=4\times10^{11-7}=4\times10^4\text{N}.$

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Question 45 Marks

An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that temperature of each rod increases by $\Delta\text{T}$. Find change in the angle ABC. $[$Coeff. of linear expansion for 1 Cu is $\alpha_1$ Coeff. of linear expansion for 2Al is $\alpha_2]$

Answer

By trigonometry$\cos\theta=\frac{\text{l}_1^2+\text{l}_3^2-\text{l}_2^2}{2\text{l}_1\text{l}_3}$
$2\text{l}_1\text{l}_3\cos\theta=\text{l}_1^2+\text{l}_3^2-\text{l}_2^2$

Differentiating both sides$2\big[\text{d}(\text{l}_1\text{l}_3)\cdot\cos\theta+\text{l}_1\text{l}_3\text{d}(\cos\theta)\big]=2\text{l}_1\text{dl}_1+2\text{l}_3\text{dl}_3-2\text{l}_2\text{dl}_2$
$2\big[(\text{l}_1\text{dl}_3+\text{l}_3\text{dl}_1)\cos\theta-\text{l}_1\text{l}_3\sin\theta\text{ d}\theta\big]=2(\text{l}_1\text{dl}_1+\text{l}_3\text{dl}_3-\text{l}_2\text{dl}_2)$
$(\text{l}_1\text{dl}_3+\text{l}_3\text{dl}_1)\cos\theta-1_1\text{l}_3\sin\theta\text{ d}\theta=1_1\text{dl}_1+\text{l}_3\text{dl}_3-\text{l}_2\text{dl}_2\ \ \ (\text{i})$
$\text{L}_\text{t}=\text{L}_0(1+\alpha\Delta\text{t})$
$\text{L}_\text{t}-\text{L}_0(\text{L}_0\alpha\Delta\text{t})$
$\Delta\text{L}=\text{L}\alpha.\Delta\text{t})$
$\text{dl}_1=\text{l}_1\alpha_1\Delta\text{t}_2\text{ dl}_3=\text{l}_2\alpha_1\Delta\text{t}$
and $\text{dl}_2=\text{l}_2\alpha_2\Delta\text{t}$$\text{l}_1=\text{l}_2=\text{l}_3=\text{l}$
$\therefore\text{dl}_1=\text{l}\alpha_1\Delta\text{t}_2\text{ dl}_3=\text{l}\alpha_1\Delta\text{t}$
$\text{}$and $\text{dl}_2=\text{l}\alpha_2\Delta\text{t}$
Substitute their value in (i)$\cos\theta(\text{l}^2.\alpha_1\Delta\text{t}+\text{l}^2\alpha\Delta\text{t})-\text{l}^2\sin\theta\text{ d}\theta=\text{l}^2\alpha_1\Delta\text{t}+\text{l}^2\alpha_1\Delta\text{t}-\text{l}^2\alpha_2\Delta\text{t}$
$2\text{l}^2\alpha_1\Delta\text{t}\cos\theta-\text{ l}^2[\sin\theta.\text{d}\theta]=\text{l}^2[\alpha_1+\alpha_1-\alpha_2]\Delta\text{t}$
$\text{l}^2[2\alpha_1\Delta\text{t}\cos60^0-\sin60^0\text{d}\theta]=\text{l}^2[2\alpha_1-\alpha_2]\Delta\text{t}$
$2\alpha_1\Delta\text{t}\times\frac{1}{2}-2\alpha_1\Delta\text{t}+\alpha_2\Delta\text{t}=\frac{\sqrt{3}}{2}\text{d}\theta$
$\frac{\sqrt{3}}{2}\text{d}\theta=[\alpha_1-2\alpha_1+\alpha_2]\Delta\text{t}$
$\text{d}\theta=\frac{2(\alpha_2-\alpha_1)\Delta\text{t}}{\sqrt{3}}$
$[\therefore\Delta\text{t}=\Delta\text{T (given)}]$
$\text{d}\theta=\frac{2(\alpha_2-\alpha_1)\Delta\text{T}}{\sqrt{3}}$

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Question 55 Marks

In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{\text{Y}\pi\text{r}^4}{4\text{R}}.$ Y is the Young’s modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

Answer

By Pythagoras theorem: in right angled $\triangle\text{ABC}$ where point C is just outside the base of Trunk i.e. point C is at D$\text{R}^2=(\text{R}-\text{d})^2+\Big(\frac{\text{h}}{2}\Big)^2$
$\text{R}^2=\text{R}^2+\text{d}^2-2\text{Rd}+\frac{\text{h}^2}{4}$
$\therefore\text{d}<<<\text{R}$ $(\therefore d^2$ can be neglected$)$
$2\text{Rd}=\frac{\text{h}^2}{4}\text{ or }\text{d}=\frac{\text{h}^2}{8\text{R}} \ \ \ (\text{I})$
Let weight of Trunk per unit volume = $W_0$ The weight of trunk = Volume × $W_0 =(\pi\text{r}^2\text{h})\text{W}_0$ Torque by bending the trunk = Force × $\bot$dist.$\tau=\pi\text{r}^2\text{hW}_0\times\text{d}$
$\tau=\frac{\pi\text{r}^4\text{Y}}{4\text{R}}\text{(given)}$
$\therefore\pi\text{r}^2\text{hW}_0\times\frac{\text{h}^2}{8\text{R}}=\frac{\pi\text{r}^4\text{Y}}{4\text{R}}$
$\text{h}^3=\frac{\pi\text{r}^4\text{Y}\times8\text{R}}{4\text{R}\pi\text{r}^2\text{W}_0}=\frac{2\text{r}^2\text{Y}}{\text{W}_0}$
$\text{h}=\Big[\frac{2\text{Y}}{\text{W}_0}\Big]^{\frac{1}{3}}\text{r}^{\frac{2}{3}}$
Hence, h is the critical height given in this expression.

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Question 65 Marks

Consider a long steel bar under a tensile stress due to forces F acting at the edges along the length of the bar. Consider a plane making an angle $\Delta$ with the length. What are the tensile and shearing stresses on this plane?

For what angle is the tensile stress a maximum?
For what angle is the shearing stress a maximum?

Answer

According to the problem force F is applied along horizontal, so we resolve it in two perpendicular components - one is parallel to the inclined plane and other one is perpendicular to the inclined plane as shown in the diagram. Now, we can easily calculate the tensile and shearing stress. Here,$\text{F}_\bot=\text{F}\sin\theta,\text{F}_\parallel=\text{F}\cos\theta$
Let the cross - sectional area of the bar be A. Consider the equilibrium of the plane aa'. Here, $\text{F}_\bot$ produces tensile stress and $\text{F}_\parallel$ produces shear stress, on the plane aa'. Let the area of the face aa' be A, then$\therefore\ \sin\theta=\frac{\text{A}}{\text{A}'}\Rightarrow\text{A}'=\frac{\text{A}}{\sin\theta}$
Tensile stress on the plane aa'$\text{aa}'=\frac{\text{F}_\bot}{\text{A}'}=\frac{\text{F}\sin\theta}{\text{A}/\sin\theta}=\frac{\text{F}}{\text{A}}\sin^2\theta$
Shearing stress on the plane aa', Shearing stress $=\frac{\text{Parallel force}}{\text{Area}}$$=\frac{\text{F}_\parallel}{\text{A}'}=\frac{\text{F}\cos\theta}{\text{A}/\sin\theta}=\frac{\text{F}\sin\theta\cos\theta}{\text{A}}=\frac{\text{F}(2\sin\theta\cos\theta)}{2\text{A}}$
$=\frac{\text{F}\sin2\theta}{2\text{A}}$

For tensile stress to be maximum,

$\sin^2\theta=0\Rightarrow\sin\theta=0\Rightarrow\theta=\frac{\pi}{2}\text{ or }\theta=90^0$

For shearing stress to be maximum,

$\sin2\theta=1\Rightarrow2\theta=\frac{\pi}{2}\Rightarrow\theta=\frac{\pi}{4}\text{ or }\theta=45^0$

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Question 75 Marks

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by l. Another wire of the same material of length 2L and radius 2r, is pulled by a force 2f. Find the increase in length of this wire.

Answer

We have to apply Hooke’s law to compare the extension in each wire. According to the diagram which shows the situation. Now, Young’s modulus$\text{(Y)}=\frac{\text{f}}{\text{A}}\times\frac{\text{L}}{\text{l}}$
First case, length of wire = L, radius of wire = r Force applied = f, increase in length = l$\text{Y}_1=\frac{\frac{\text{f}}{\pi\text{r}^2}}{\frac{\text{l}}{\text{L}}}=\frac{\text{fL}}{\pi\text{r}^2\text{l}}\ \ ...\text{(i)}$
In second case, lengthof wire = 2L, radius of wire = 2r, force applied = 2f, increase in length = x (say)$\text{Y}_2=\frac{\frac{\text{2f}}{\pi\text{(2r})^2}}{\frac{\text{x}}{\text{L}}}=\frac{\text{fL}}{\pi\text{r}^2\text{x}}\ \ ...\text{(ii)}$

Both the wires are of same material, so Young's modulus will be same. From Eqs. (i) and (ii),$\frac{\text{f}}{\pi\text{r}^2}\times\frac{\text{L}}{\text{l}}=\frac{\text{f}}{\pi\text{r}^2}\times\frac{\text{L}}{\text{x}}$
Hence, x = l.

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Question 85 Marks

A stone of mass m is tied to an elastic string of negligble mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P.

Find the distance y from the top when the mass comes to rest for an instant, for the first time.
What is the maximum velocity attained by the stone in this drop?
What shall be the nature of the motion after the stone has reached its lowest point?

Answer

A stone is tied at P with string of length L. String is fixed with nail at ‘O’. Stone is lifted upto height L, so that string is stretched as shown in given fig. When stone fall under gravity. It tries to follow path PP’ but due to elastic string it will go a part of circular path P to Q. This is like a centrifugal force that stretches the string outward and increases its length $(\Delta\text{L}).$ So the change in Potential energy of stone at Q’ and p converts into mechanical energy in string of spring constant K. So P.E of stone = mechanical Energy of string.$\text{mgy}=\frac{1}{2}\text{K}(\text{y}-\text{L})^2$
$\text{mgy}=\frac{1}{2}\text{K}(\text{y}^2+\text{L}^2-2\text{yL})$
$2\text{mgy}=\text{K}[\text{y}^2+\text{L}^2-2\text{yL}]$
$2\text{mgy}=\text{K}\text{y}^2-2\text{K}\text{y}\text{L}+\text{KL}^2$
or $\text{K}\text{y}^2-2\text{K}\text{y}\text{L}-2\text{mgy}+\text{KL}^2=0$$\text{K}\text{y}^2-2(\text{K}\text{L}+\text{mg})\text{y}+\text{KL}^2=0$

Solving this equation by quadratic formula we get,

$\text{D}=\text{b}^2-4\text{ac}\ \ (\text{a}=\text{K}\text{ b}=-2(\text{KL+mg})\text{ c=KL}^2)$
$\text{D}=\big[-2(\text{KL+mg})\big]^2-4\text{(K)}\text{(KL)}^2$
$\text{D}=+4\big[(\text{KL})^2+\text{(mg)}^2+2\text{(KL)(mg)}\big]-4\text{K}^2\text{L}^2$
$\text{D}=4\big[\text{K}^2\text{L}^2+\text{m}^2\text{g}^2+2\text{KLmg}\big]-4\text{K}^2\text{L}^2$
$=4\text{K}^2\text{L}^2+4\text{m}^2\text{g}^2+8\text{KLmg}-4\text{K}^2\text{L}^2$
$\sqrt{\text{D}}=\sqrt{4\text{m}\text{g}[\text{m}\text{g}+2\text{KL}]}=2\sqrt{\text{m}\text{g}(\text{m}\text{g}+2\text{KL})}$
$\therefore\text{y}\frac{-\text{d}\pm\sqrt{\text{D}}}{2\text{a}}=\frac{+2\big(\text{KL}+\text{mg}\big)\pm2\sqrt{\text{mg(2}\text{KL}+\text{mg)}}}{2\text{K}}$
$\text{y}=\frac{2\big[\big(\text{KL}+\text{mg}\big)\big]\pm\sqrt{\text{mg(2}\text{KL}+\text{mg)}}}{2\text{K}}$
$\text{y}=\frac{\big(\text{KL}+\text{mg}\big)\pm\sqrt{\text{mg}(2\text{KL}+\text{mg})}}{\text{K}}$

At maximum velocity as its lowest point acceleration is zero.

$\therefore\ \text{F}=0$
So, the spring or string force Kx is blanced by gravitational force mg. so, these two forces will be equal and opposite.
$\therefore\ \text{mg}=\text{kx}\ ...(\text{i})$ where x is extension in string
Let v be the maximum velocity of stone at bottom of journey.
By law of conservation of energy,
KE of stone + PE gain by string = P.E. lost stone from p tp Q'
$\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{Kx}^2=\text{mg}(\text{L}+\text{x})$
$\text{mv}^2+\text{Kx}^2=2\text{mg}(\text{L}+\text{x})$
$\text{mv}^2=2\text{mgL}+2\text{mgx}-\text{Kx}^2$
$\text{mg}=\text{Kx}\ (\text{from i})$
$\text{x}=\frac{\text{mg}}{\text{K}}$
$\therefore\ \text{mv}^2=2\text{mgL}+2\text{mg}\cdot\frac{\text{mg}}{\text{k}}-\text{K}\frac{\text{m}^2\text{g}^2}{\text{K}^2}$
$=2\text{mgL}+\frac{2\text{m}^2\text{g}^2}{\text{K}}-\frac{\text{m}^2\text{g}^2}{\text{K}}$
$\text{mv}^2=\text{m}\Big[\text{2gL}+\frac{\text{mg}^2}{\text{K}}\Big]$
$\therefore\ \text{v}=\Big[2\text{gL}+\frac{\text{mg}^2}{\text{K}}\Big]^{\frac{1}{2}}$

At lowest point from figuare in part (a)

$\text{F}=\text{mg}\downarrow-\text{K}(\text{y}-\text{L})\uparrow\ (\text{by string})$
$\therefore\ \text{m}\frac{\text{d}^2\text{z}}{\text{dt}^2}=\text{mg}-\text{K}(\text{y}-\text{L})$
$\frac{\text{d}^2\text{z}}{\text{dt}^2}-\text{g}+\frac{\text{K}}{\text{m}}(\text{y}-\text{L})=0$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}+\frac{\text{K}}{\text{m}}\Big[(\text{y}-\text{L})-\frac{\text{mg}}{\text{K}}\Big]$
make a transformation of variables:
$\text{z}=\Big[(\text{y}-\text{L)}-\frac{\text{mg}}{\text{K}}\Big]$
then $\frac{\text{d}^2\text{z}}{\text{dt}}^2+\frac{\text{k}}{\text{m}}\text{z}=0$
it is differential equation of second order which represents S.H.M.
$\therefore\frac{\text{d}^2\text{z}}{\text{dt}^2}+\omega^2\text{z}=0$
Where w is angular frequency so $\omega=\sqrt{\frac{\text{K}}{\text{m}}}$
Solution of above differential equation is of type
$\text{z}=\text{A}\cos(\omega\text{t}+\theta)$
Where $\omega=\sqrt{\frac{\text{K}}{\text{m}}}\text{ and }\theta$ is phase difference.
$\text{z}=\Big(\text{L}+\frac{\text{m}}{\text{K}}\text{g}\Big)+\text{A}'\cos(\omega\text{t+}\theta)$
So the stone performs SHM with angular frequency $\omega$ about the point at y = 0
$|\text{z}_0|=\Big|-\Big(\text{L}+\frac{\text{mg}}{\text{K}}\Big)\Big|\ \ \ [\text{from (i)}]$
$\therefore\text{z}_0=\Big(\text{L}\frac{\text{mg}}{\text{K}}\Big)$

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Question 95 Marks

A steel wire of mass μ per unit length with a circular cross section has a radius of $0.1cm$. The wire is of length 10m when measured lying horizontal, and hangs from a hook on the wall. A mass of $25kg$ is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is $7860kg m^{–3}$ (Young’s modules $Y = 2 \times 10^{11} Nm^{–2})$.
If the yield strength of steel is $2.5 \times 10^8 Nm^{–2}$, what is the maximum weight that can be hung at the lower end of the wire?

Answer

Consider an element dx at a distance x from the load (x = 0).

If T (x) and T (x + dx) are tensions on the two cross sections a distance dx apart, then
T (x + dx) – T(x) = μgdx (where μ is the mass/ length)
$\Big(\frac{\text{dT}}{\text{dx}}\Big)\text{dx}=\mu\text{gdx}$
$\Rightarrow\text{T(x)}=\mu\text{gx+C}$
At $\text{x=0, T(O)=Mg}\Rightarrow\text{C}\Rightarrow\text{Mg}$
$\text{T(x)}=\mu\text{gx+Mg}$
Let the length dx at x increase by dr, then
$\frac{\text{T(x)}/\text{A}}{\text{dr/dx}}=\text{Y}$
or, $\frac{\text{dr}}{\text{dx}}=\frac{1}{\text{YA}}\text{T(x)}$
$\Rightarrow\text{r}=\frac{1}{\text{YA}}\int^\text{L}_0(\mu\text{gx+mg)dx}$
$=\frac{1}{\text{YA}}\Big[\frac{\mu\text{gx}^2}{2}+\text{mgx}\Big]_0^\text{L}$
$=\frac{1}{\text{YA}}\Big[\frac{\text{mgl}}{2}+\text{mgL}\Big]$
(m is the mass of the wire)
$\text{A}=\pi\times(10^{-3})^2\text{m}^2,\text{Y}=200\times10^9\text{Nm}^{-2}$
$\text{m}=\pi\times(10^{-3})^2\times10\times7860\text{kg}$
$\therefore\text{r}=\frac{1}{2\times10^{11}\times\pi\times10^{}-6}\Big[\frac{\pi\times786\times10^{-7}\times10\times10}{2}+25\times10\times10\Big]$
$=[196.5\times10^{-6}+3.98\times10^{-3}]\sim4\times10^{-3}\text{m}$

The maximum tension would be at x = L.

$\text{T}=\mu\text{gL+Mg =(m+M)g}$
The Yield force
$= 250\times10^6\times\pi×(10^{-3})2=250\times\pi\text{N}$=
At Yield
$\text{(m+M)g}=250\times\pi$
$\text{m}=\pi\times(10^{-3})^2\times10\times7860<<\text{M}\therefore\text{Mg}\sim250\times\pi$
Hence, $\text{M}=\frac{250\times\pi}{10}=25\times\pi\sim75\text{kg.}$

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Question 105 Marks

A steel rod of length 2l, cross sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young’s modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform.)

Answer

Consider in given figure an element (dr) of rod at a distance r from the centre.

Let T(r) and T (r + dr) are the tensions external force to rod extend at A and B ends of elemen (small) respectively. Centrifugal force on element dr due to tension difference = T(r + dr) - T(r) Centrifugal force = -dT (outward) Centripetal Force due to rotation on element $dr = dmrw^2$
$\therefore-\text{dt}=\text{dm }\omega^2\text{r}$ (Let $\mu$ =mass per unit length)
then $-\text{dT }\omega^2\text{r }(\text{dr.}\mu)$$-\text{dT }=\mu\text{w}^2\text{ r.}\text{dr}$
Integrating both sides$-\int\limits^{\text{T}}_0\text{dT}=\mu\text{w}^2\int\limits^1_\text{r}\text{rdr}$
Tension in rod at distance r from centre so limits will varies from r to l$\therefore-\text{T}(\text{r})=\mu\omega^2\Big[\frac{\text{r}^2}{2}\Big]^\text{l}_\text{r}=\frac{\mu\omega^2}{2}(\text{l}^2-\text{r}^2) \ \ \ (\text{I})$
Let the increase in length of dr element at distance r form centre is $\delta\text{r}$ then$\text{Y}=\frac{\text{strtess}}{\text{Strain}}=\frac{\text{T(r)/A}}{\delta\text{r/dr}}=\frac{\text{T(r)}}{\text{A}}.\frac{\text{dr}}{\delta\text{r}}$
$\frac{\delta\text{r}}{\text{dr}}=\frac{\text{T(r)}}{\text{AY}}=\frac{-\mu\text{w}^2}{2\text{AY}}(\text{l}^2-\text{r}^2)$
$\because$ Negative sign shows only that the direction of extension is opposite to restoring force
$\therefore\delta\text{r}=\frac{\mu\text{w}^2}{2\text{AY}}(\text{l}^2-\text{r}^2)\text{dr}$
$\int\limits_0^\delta\delta\text{r}=\int\limits_0^\text{l}\frac{\mu\text{w}^2}{2\text{AY}}(\text{l}^2-\text{r}^2)\text{dr}$ (for rod one from centre)
$\delta=\frac{\mu\text{w}^2}{2\text{AY}}\Big(\text{l}^3-\frac{\text{l}^3}{3}\Big)=\frac{\mu\text{w}^2}{2\text{AY}}\frac{2}{3}\text{l}^3$
$\delta=\frac{\mu\text{w}^2}{3\text{AY}}\text{l}^3$
$\therefore$ Total extension in rod both sides $=2\delta=\frac{2\mu\text{w}^2\text{l}^2}{3\text{AY}}$

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