Question

Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of

• A. 0.42 m from mass of 0.3 kg
• B. 0.70 m from mass of 0.7 kg
• C. 0.98 m from mass of 0.3 kg
• D. 0.98 m from mass of 0.7 kg

Answer 1

The correct option is C 0.98 m from mass of 0.3 kg


Let the axis of rotation
Pass through O.
$I=m{r}^2$ for point mass.
$\therefore I=I_1+I_2=0.3x^2+0.7(1.4-x{)}^2=0.3x^2+0.7(1.96+x^2-2.8x)=x^2+1.372-1.96x$
The work done for rotation of
The rod is stored as rotational kinetic energy,
$\frac{1}{2}$ ${\omega }^2$ of rod
Or $W=\frac{I{\omega }^2}{2}=\frac{1}{2}(x^2+1.372-1.96x){\omega }^2$
For work done to be minimum, $\frac{dW}{dx}=0$
$\therefore \frac{d}{dx}\left[\right(x^2+1.372-1.96x\left)\right]\frac{{\omega }^2}{2}=0$
Or 2x + 0 - 1.96 = 0
Or 2x = 1.96 or x = 0.98 m