Rotation Ques 120
- 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 $0.3 kg$
$(1995$, S)
(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 $0.7 kg$
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Answer:
Correct Answer: 120.(c)
Solution:
Formula:
- Work done, $W=\frac{1}{2} I \omega^{2}$
If $x$ is the distance of mass $0.3 kg$ from the centre of mass, we will have,
$$ I=(0.3) x^{2}+(0.7)(1.4-x)^{2} $$
For work to be minimum, the moment of inertia $(I)$ should be minimum, or $\frac{d I}{d x}=0$
$$ \begin{aligned} \text { or } & & 2(0.3 x)-2 & (0.7)(1.4-x) \\ \text { or } & & (0.3) x & =(0.7)(1.4-x) \\ \Rightarrow & & x & =\frac{(0.7)(1.4)}{0.3+0.7}=0.98 m \end{aligned} $$
BETA
