A steel rod of length $2 l$, 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)
Consider an element of width $d r$ at $r$ as shown in the diagram.
Let $T(r)$ and $T(r+d r)$ be the tensions at $r$ and $r+d r$ respectively.
Net centrifugal force on the element $=\omega^2 r d m$ (where $\omega$ is angular velocity of the rod)
$=\omega^2 r \mu d r$ $(\because \mu=$ mass/length $)$
$$\begin{aligned} &\begin{array}{rrr} \Rightarrow & T(r)-T(r+d r) & =\mu \omega^2 r d r \\ \Rightarrow & -d T & =\mu \omega^2 r d r \end{array}\\ &\text { [ } \because \text { Tension and centrifugal forces are opposite] } \end{aligned}$$
$$\begin{array}{rlr} \therefore \quad -\int_\limits{T=0}^T d T & =\int_\limits{r=l}^{r=r} \mu \omega^2 r d r & {[\because T=0 \text { at } r=l]} \\ T(r) & =\frac{\mu \omega^2}{2}\left(l^2-r^2\right) & \end{array}$$
Let the increase in length of the element $d r$ be $\Delta r$
So, Young's modulus $Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T(r) / A}{\frac{\Delta r}{d r}}$
$$\begin{array}{ll} \therefore & \frac{\Delta r}{d r}=\frac{T(r)}{A}=\frac{\mu \omega^2}{2 Y A}\left(l^2-r^2\right) \\ \therefore & \Delta r=\frac{1}{Y A} \frac{\mu \omega^2}{2}\left(l^2-r^2\right) d r \end{array}$$
$$\begin{aligned} \therefore \Delta=\text { change in length in right part } & =\frac{1}{Y A} \frac{\mu \omega^2}{2} \int_0^l\left(l^2-r^2\right) d r \\ & =\left(\frac{1}{Y A}\right) \frac{\mu \omega^2}{2}\left[l^3-\frac{l^3}{3}\right]=\frac{1}{3 Y A} \mu \omega^2 L^2 \end{aligned}$$
$$\therefore \quad \text { Total change in length }=2 \Delta=\frac{2}{3 Y A} \mu \omega^2 l^2$$
An equilateral triangle $A B C$ is formed by two Cu rods $A B$ and $B C$ and one Al rod. It is heated in such a way that temperature of each rod increases by $\Delta T$. Find change in the angle $A B C$. [Coeffecient of linear expansion for Cu is $\alpha_1$, coefficient of linear expansion for Al is $\alpha_2$ ]
Consider the diagram shown
Let $$l_1=A B, l_2=A C, l_3=B C$$
$$\therefore \quad \cos \theta=\frac{l_3^2+l_1^2-l_2^2}{2 l_3 l_1} \quad \text { (assume } \angle A B C=\theta \text { ) }$$
$$\begin{aligned} & \Rightarrow \quad 2 l_3 l_1 \cos \theta=l_3^2+l_1^2-l_2^2 \\ & \text { Differentiating } 2\left(l_3 d l_1+l_1 d l_3\right) \cos \theta-2 l_1 l_3 \sin \theta d \theta \end{aligned}$$
$$=2 l_3 d l_1+2 l_1 d l_1-2 l_2 d l_2$$
$$\begin{aligned} \text{Now,}\quad & d l_1=l_1 \alpha_1 \Delta t \quad \text { (where, } \Delta t=\text { change in temperature) } \\ & d l_2=l_2 \alpha_1 \Delta t \Rightarrow d l_3=l_3 \alpha_2 \Delta t \end{aligned}$$
$$\begin{aligned} &\text { and } &l_1=l_2=l_3=l \end{aligned}$$
$$\left(l^2 \alpha_1 \Delta t+l^2 \alpha_1 \Delta t\right) \cos \theta+l^2 \sin \theta d \theta=l^2 \alpha_1 \Delta t+l^2 \alpha_1 \Delta t-l^2 \alpha_2 \Delta t$$
$$\sin \theta d \theta=2 \alpha_1 \Delta t(1-\cos \theta)-\alpha_2 \Delta t$$
Putting $$\theta=60^\circ\quad$$ (for equilateral triangle)
$$\begin{aligned} d \theta \times \sin 60^{\circ} & =2 \alpha_1 \Delta t\left(1-\cos 60^{\circ}\right)-\alpha_2 \Delta t \\ & =2 \alpha_1 \Delta t \times \frac{1}{2}-\alpha_2 \Delta t=\left(\alpha_1-\alpha_2\right) \Delta t \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad d \theta & =\text { change in the angle } \angle A B C \\ & =\frac{\left(\alpha_1-\alpha_2\right) \Delta T}{\sin 60^{\circ}}=\frac{2\left(\alpha_1-\alpha_2\right) \Delta T}{\sqrt{3}} \quad(\because \Delta t=\Delta T \text { given }) \end{aligned}$$
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 $\frac{Y \pi r^4}{4 R} . Y$ 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.
Consider the diagram according to the question, the bending torque on the trunk of radius $r$ of the tree $=\frac{Y \pi r^4}{4 R}$
where $R$ is the radius of curvature of the bent surface.
When the tree is about to buckle $W d=\frac{Y \pi r^4}{4 R}$
If $R \gg h$, then the centre of gravity is at a height $l \approx \frac{1}{2} h$ from the ground.
$$\begin{aligned} & \text { From } \quad \triangle A B C R^2 \approx(R-d)^2+\left(\frac{1}{2} h\right)^2 \\ & \text { if } d \ll R, \quad R^2 \approx R^2-2 R d+\frac{1}{4} h^2 \\ & \therefore \quad d=\frac{h^2}{8 R} \end{aligned}$$
If $\omega_0$ is the weight/volume
$$\frac{Y \pi r^4}{4 R}=\omega_0\left(\pi r^2 h\right) \frac{h^2}{8 R}\quad$$ $[\because$ Torque is caused by the weight]
$$ \Rightarrow \quad h \approx\left(\frac{2 Y}{\omega_0}\right)^{1 / 3} r^{2 / 3}$$
$$\text { Hence, critical height }=h=\left(\frac{2 Y}{\omega_0}\right)^{1 / 3} r^{2 / 3}$$
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 $P$.
(a) Find the distance $y$ from the top when the mass comes to rest for an instant, for the first time.
(b) What is the maximum velocity attained by the stone in this drop?
(c) What shall be the nature of the motion after the stone has reached its lowest point?
Consider the diagram the stone is dropped from point $p$.
(a) Till the stone drops through a length $L$ it will be in free fall. After that the elasticity of the string will force it to a SHM. Let the stone come to rest instantaneously at $y$. The loss in PE of the stone is the PE stored in the stretched string.
$$m g y=\frac{1}{2} k(y-L)^2$$
$$\begin{aligned} &\begin{aligned} \Rightarrow \quad m g y & =\frac{1}{2} k y^2-k y L+\frac{1}{2} k L^2 \Rightarrow \frac{1}{2} k y^2-(k L+m g) y+\frac{1}{2} k L^2=0 \\ y & =\frac{(k L+m g) \pm \sqrt{(k L+m g)^2-k^2 L^2}}{k}=\frac{(k L+m g) \pm \sqrt{2 m g k L+m^2 g^2}}{k} \end{aligned}\\ &\text { Retain the positive sign. }\\ &\therefore \quad y=\frac{(k L+m g)+\sqrt{2 m g k L+m^2 g^2}}{k} \end{aligned}$$
In SHM, the maximum velocity is attained when the body passes, through the "equilibrium, position" i.e., when the instantaneous acceleration is zero. That is $m g-k x=0$, where $x$ is the extension from $L$.
$$\Rightarrow \quad m g=k x$$
Let the velocity be $v$. Then,
$$ \begin{aligned} &\begin{aligned} & \frac{1}{2} m v^2+\frac{1}{2} k x^2=m g(L+x) \quad \text { (from conservation of energy) }\\ & \frac{1}{2} m v^2=m g(L+x)-\frac{1}{2} k x^2 \end{aligned}\\ \end{aligned}$$
Now,
$$ \begin{aligned} m g & =k x \Rightarrow x=\frac{m g}{k} \\ \therefore \quad \frac{1}{2} m v^2 & =m g\left(L+\frac{m g}{k}\right)-\frac{1}{2} k \frac{m^2 g^2}{k^2}=m g L+\frac{m^2 g^2}{k}-\frac{1}{2} \frac{m^2 g^2}{k} \\ \frac{1}{2} m v^2 & =m g L+\frac{1}{2} \frac{m^2 g^2}{k} \\ \therefore \quad v^2 & =2 g L+m g^2 / k \\ v & =\left(2 g L+m g^2 / k\right)^{1 / 2} \end{aligned}$$
When stone is at the lowest position i.e., at instantaneous distance $Y$ from $P$, then equation of motion of the stone is
$$\frac{m d^2 y}{d t^2}=m g-k(y-L) \Rightarrow \frac{d^2 y}{d t^2}+\frac{k}{m}(y-L)-g=0$$
Make a transformation of variables, $z=\frac{k}{m}(y-L)-g$
$$\therefore \quad \frac{d^2 z}{d t^2}+\frac{k}{m} z=0$$
It is a differential equation of second order which represents SHM.
Comparing with equation $\frac{d^2 z}{d t^2}+\omega^2 z=0$
Angular frequency of harmonic motion $\omega=\sqrt{\frac{k}{m}}$
The solution of above equation will be of the type $z=A \cos (\omega t+\phi)$; where $\omega=\sqrt{\frac{k}{m}}$
$$y=\left(L+\frac{m g}{k}\right)+A^{\prime} \cos (\omega t+\phi)$$
Thus, the stone will perform SHM with angular frequency $\omega=\sqrt{k / m}$ about a point $y_0=L+\frac{m g}{k}$.