Question

A rod of length $l$ and mass is rotated in a horizontal plane about its one end with constant angular velocity $\omega$. The tension at the middle section of the rod is

• A. $m{\omega }^{2}l$
• B. $\frac{m{\omega }^{2}l}{2}$
• C. $\frac{3}{4}m{\omega }^{2}l$
• D. $\frac{3}{8}m{\omega }^{2}l$

Answer 1

The correct option is D $\frac{3}{8}m{\omega }^{2}l$

Tension at the middle section of the rod =

Force required to rotate the other half of the rod

Linear mass density of the rod = $\lambda$ = $\frac{m}{l}$

Force required to rotate a mass $dm$ around the axis = $dm{\omega }^{2}r$

Now $dm$ = $\lambda dx$

Therefore , force required to rotate the other half of the rod

= $\int dm{\omega }^{2}x$

= ${\int }_{\frac{l}{2}}^l\frac{m}{l}{\omega }^{2}xdx$

= $\frac{3}{8}m\omega l^2$

Hence , the tension at the middle section of the rold is $\frac{3}{8}m\omega l^2$.