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}$

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B

${T}_{1}$$>$${T}_{2}$

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C

${T}_{1}$$<$${T}_{2}$

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D

None of these.

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Solution

## The correct option is B ${T}_{1}$$>$${T}_{2}$Step 1: Given dataThe length of the rod is $L$.The tension of the rod at a distance $\frac{L}{4}$is ${T}_{1}$.The tension of the rod at a distance $\frac{3L}{4}$ is ${T}_{2}$.Step 2: Centripetal force due to a bodyThe 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: DiagramStep 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 ${T}_{1}={\int }_{\frac{L}{4}}^{L}\frac{m}{L}{\omega }^{2}xdx\phantom{\rule{0ex}{0ex}}or{T}_{1}=\frac{m{\omega }^{2}}{L}{\left[\frac{{x}^{2}}{2}\right]}_{\frac{L}{4}}^{L}=\frac{m{\omega }^{2}}{L}\left[\frac{{L}^{2}}{2}-\frac{{L}^{2}}{32}\right]=\frac{m{\omega }^{2}}{L}.\frac{15{L}^{2}}{32}\phantom{\rule{0ex}{0ex}}or{T}_{1}=\frac{m{\omega }^{2}}{L}.\frac{15{L}^{2}}{32}.......................\left(1\right)$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 ${T}_{2}={\int }_{\frac{3L}{4}}^{L}\frac{m{\omega }^{2}}{L}xdx\phantom{\rule{0ex}{0ex}}or{T}_{2}=\frac{m{\omega }^{2}}{L}{\left[\frac{{x}^{2}}{2}\right]}_{\frac{3L}{4}}^{L}=\frac{m{\omega }^{2}}{L}\left[\frac{{L}^{2}}{2}-\frac{9{L}^{2}}{32}\right]=\frac{m{\omega }^{2}}{L}.\frac{7{L}^{2}}{32}\phantom{\rule{0ex}{0ex}}or{T}_{2}=\frac{m{\omega }^{2}}{L}.\frac{7{L}^{2}}{32}.......................\left(2\right)$From equations 1 and 2 we get,$\frac{m{\omega }^{2}}{L}.\frac{15{L}^{2}}{32}>$$\frac{m{\omega }^{2}}{L}\frac{7{L}^{2}}{32}$.So, ${T}_{1}>{T}_{2}$So. option (B) is correct.

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