(B) According to the work-energy theorem,the work done is given by $W = \frac{1}{2} I \omega^2$.Since the angular speed $\omega$ is constant,the work

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

(B) According to the work-energy theorem,the work done is given by $W = \frac{1}{2} I \omega^2$.Since the angular speed $\omega$ is constant,the work done is minimum when the moment of inertia $I$ is minimum.Let the axis of rotation pass through a point at a distance $x$ from the $0.3 \ kg$ mass. Then the distance from the $0.7 \ kg$ mass is $(1.4 - x)$.The moment of inertia $I$ is given by $I = 0.3x^2 + 0.7(1.4 - x)^2$.To find the minimum value of $I$,we differentiate with respect to $x$ and set it to zero:$\frac{dI}{dx} = 0.3(2x) + 0.7(2)(1.4 - x)(-1) = 0$$0.6x - 1.4(1.4 - x) = 0$$0.6x - 1.96 + 1.4x = 0$$2.0x = 1.96$$x = 0.98 \ m$.Thus,the axis should pass at a distance of $0.98 \ m$ from the $0.3 \ kg$ mass.

Class 11 Physics System of Particles and Rotational Motion Moment of Inertia and Radius of gyration

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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

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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 $0.7 \ kg$

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System of Particles and Rotational Motion

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Moment of Inertia and Radius of gyration

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