Question

A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let T₁ and T₂ be the tensions at the points L/4 and 3L/4 away from the pivoted ends.
(a) T₁ > T₂
(b) T₂ > T₁
(c) T₁ = T₂
(d) The relation between T₁ and T₂ depends on whether the rod rotates clockwise or anticlockwise.

Answer 1

(a) T₁ > T₂



Let the angular velocity of the rod be $\omega$.
Distance of the centre of mass of portion of the rod on the right side of L/4 from the pivoted end:
$r_1=\frac{L}{4}+\frac{1}{2}\left(\frac{3L}{4}\right)=\frac{5L}{8}$
Mass of the rod on the right side of L/4 from the pivoted end:
$m_1=\frac{3}{4}M$
At point L/4, we have:
$$\begin{gathered} T_1=m_1{\omega }^2{r}_1 \\ =\frac{3}{4}M{\omega }^2\frac{5}{8}L=\frac{15}{32}M{\omega }^{2}L \end{gathered}$$

Distance of the centre of mass of rod on the right side of 3L/4 from the pivoted end:
$r_1=\frac{1}{2}\left(\frac{L}{4}\right)+\frac{3L}{4}=\frac{7L}{8}$
Mass of the rod on the right side of L/4 from the pivoted end:
$m_1=\frac{1}{4}M$
At point 3L/4, we have:
$$\begin{gathered} T_2=m_2{\omega }^2{r}_2 \\ =\frac{1}{4}M{\omega }^2\frac{7}{8}L=\frac{7}{32}M{\omega }^{2}L \end{gathered}$$
∴ T₁ > T₂