Question

A uniform rod of mass 'm' and length 'L' is tied to a vertical shaft. It rotates in horizontal plane about the vertical axis at angular velocity $\omega$. How much horizontal force does the shaft excert on the rod?

• A. $m{w}^{2}L$
• B. $\frac{M{w}^{2}L}{2}$
• C. $\frac{m{w}^{2}L}{4}$
• D. $zero$

Answer 1

The correct option is B $\frac{M{w}^{2}L}{2}$

Let T be the force applied horizontally on the rod. The acceleration of centre of mass is ${\omega }^2\frac{L}{2}$, as it is moving in a circle of radius $\frac{L}{2}$ .
Using ${\overrightarrow{F}}_{ext}=M{\overrightarrow{a}}_{cm}\Rightarrow T=M{\omega }^2\left(\frac{L}{2}\right)$
Thus, the horizontal force exerted is $\frac{1}{2}M{\omega }^{2}L$