Question

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

• A. 0.98 m
• B. 0.49 m
• C. 1.4 m
• D. Data insufficient

Answer 1

Answer: (A)

0.98 m

Let the point be at a distance of x m from.the mass of 0.3 kg. Then moment of inertia of the system
$l=m_1{r}_1^2+m_2{r}_2^2$
$l=0.3x^2+0.7(1.4-x{)}^2$
For minimum work moment of inertia of the system should be minimum hence, we have
$\frac{dl}{dx}=0$
Using $\frac{d}{dx}{x}^n=n{x}^{n-1}$, we have
$\frac{dl}{dx}=0.3\times 2x-0.7\times (1.4-x)=0$
$\Rightarrow x=0.98m$