Question

A rod of length $L$ is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let $T_1$​ and $T_2$​ be the tensions at the points $\frac{L}{4}$ and $\frac{3L}{4}$ away from the pivoted ends. Then,

• A. $T_1$$=$$T_2$
• B. $T_1$$>$$T_2$
• C. $T_1$$<$$T_2$
• D. None of these.

Answer 1

The correct option is B

$T_1$$>$$T_2$

Step 1: Given data

1. The length of the rod is $L$.
2. The tension of the rod at a distance $\frac{L}{4}$is $T_1$.
3. The tension of the rod at a distance $\frac{3L}{4}$ is $T_2$.

Step 2: Centripetal force due to a body

1. The centripetal force due to rotation of a body of mass $m$ is defined by the form, $F_c=m{\omega }^{2}r$, where, $r$ is the distance of the body from the center of rotation and $\omega$ is the angular velocity.

Step 3: Diagram

Step 4: Finding the tension on $\frac{3L}{4}$due to $\frac{L}{4}$

Now, considering the tension act on the length $\frac{3L}{4}$ due to the length $\frac{L}{4}$ is $T_1$. And from the concept of rotation, we know that this tension is equivalent to the centrifugal force acting on the length $\frac{3L}{4}$.

Let the mass of the road is $m$. So mass per unit length is $\frac{m}{L}$ and $dx$ be the small portion at a distance x from the point $\frac{L}{4}$ on the road. And the mass of the portion is $\frac{m}{L}dx$.

So, the tension on the mass length $dx$ is $dT=\frac{m}{L}{\omega }^{2}xdx$ (Using $F_c=m{\omega }^{2}r$ formula)

So, the tension on the length $\frac{3L}{4}$ is

$$\begin{gathered} T_1={\int }_{\frac{L}{4}}^L\frac{m}{L}{\omega }^{2}xdx \\ or{T}_1=\frac{m{\omega }^2}{L}{\left[\frac{x^2}{2}\right]}_{\frac{L}{4}}^L=\frac{m{\omega }^2}{L}\left[\frac{L^2}{2}-\frac{L^2}{32}\right]=\frac{m{\omega }^2}{L}.\frac{15L^2}{32} \\ or{T}_1=\frac{m{\omega }^2}{L}.\frac{15L^2}{32}.......................\left(1\right) \end{gathered}$$

Step 4: Finding the tension on $\frac{L}{4}$ due to $\frac{3L}{4}$

Again considering the tension act on the length $\frac{L}{4}$ due to the length $\frac{3L}{4}$is $T_2$. And from the concept of rotation, we know that this tension is equivalent to the centrifugal force acting on the length $\frac{L}{4}$.

Let the mass of the road is $m$. So mass per unit length is $\frac{m}{L}$ and $dx$ be the small portion at a distance x from the point $\frac{3L}{4}$ on the road. And the mass of the portion is $\frac{m}{L}dx$.

Now, the tension on the mass length $dx$ is $dT=\frac{m}{L}{\omega }^{2}xdx$.

So, the tension on the length $\frac{L}{4}$ is

$$\begin{gathered} T_2={\int }_{\frac{3L}{4}}^L\frac{m{\omega }^2}{L}xdx \\ or{T}_2=\frac{m{\omega }^2}{L}{\left[\frac{x^2}{2}\right]}_{\frac{3L}{4}}^L=\frac{m{\omega }^2}{L}\left[\frac{L^2}{2}-\frac{9L^2}{32}\right]=\frac{m{\omega }^2}{L}.\frac{7L^2}{32} \\ or{T}_2=\frac{m{\omega }^2}{L}.\frac{7L^2}{32}.......................\left(2\right) \end{gathered}$$

From equations 1 and 2 we get,

$\frac{m{\omega }^2}{L}.\frac{15L^2}{32}>$$\frac{m{\omega }^2}{L}\frac{7L^2}{32}$.

So, $T_1>T_2$

So. option (B) is correct.