Two point masses of 0.3 kg and 0.7 kg are fixed at ends of a rod of length 1.4 m &nbs

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6.System of Particles and Rotational Motion

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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

$0.4\ m$ from mass of $ 0.3\ kg$

B

$0.98\ m$ from mass of $0.3\ kg$

C

$0.70\ m$ from mass of $0.7\ kg$

D

$0.98\ m$ from mass of $0.7\ kg$

(IIT-1995) (AIIMS-2000)

Solution

$Work-energy$ theorem, $\quad W_{\text {all forces }}=\frac{1}{2} I w^{2}$

Given$:$ $w$ is constant

So required work done to be minimum implies that $I$ must be minimum.

Let the rotational axis passes through O. $I=0.3\left(x^{2}\right)+0.7(1.4-x)^{2}$

For $I$ to be minimum, $\quad \frac{d I}{d x}=0$

$\Longrightarrow 0.3 \times 2(x)-0.7 \times 2(1.4-x)=0$

$\Longrightarrow x=0.98 m$

Standard 11

Physics

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