Question

A rod of length L is hinged at one end and it is rotated with a constant angular velocity in a horizontal plane. Let $T_1$ and $T_2$ be the tensions at the points L/4 and 3L/4 away from the hinged end.

• A. $T_1>T_2$
• B. $T_2>T_1$
• C. $T_1=T_2$
• D. The relation between $T_1$ and $T_2$ depends on whether the rod rotates clockwise or anticlockwise

Answer 1

Answer: (A)

$T_1>T_2$

Let $M$ be the mass of the rod, amd $\omega$ its angular velocity. The tension in an infinitesimal length of the rod can be written as:

$\left[SinceT\right(L)=0]$ $T\left(x\right)-T(x+dx)=\frac{M}{L}{\omega }^{2}xdx$

$-dT=\frac{M}{L}{\omega }^{2}xdx$

Integrating both sides:

${\int }_{T\left(L/4\right)}^{T\left(3L/4\right)}dT=-\frac{M}{L}{\omega }^2{\int }_{T\left(L/4\right)}^{T\left(3L/4\right)}xdx$

$⟹T_2-T_1=-\frac{M}{2L}{\omega }^2[x^2{]}_{T\left(L/4\right)}^{T\left(3L/4\right)}$

$⟹T_2-T_1=-\frac{M}{2L}{\omega }^2[\frac{9L^2}{16}-\frac{L^2}{16}]$

$⟹T_2-T_1=-\frac{M{\omega }^{2}L}{4}$

$\therefore T_1>T_2$