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  • Two steel balls of the same weight 500 gms are attached at the ends of a stick of length 1.5m (negligible mass ); The stick is rotating about an axis that is 90 degrees to its length with

Two steel balls of the same weight 500 gms are attached at the ends of a stick of length 1.5m (negligible mass ); The stick is rotating about an axis that is 90 degrees to its length with a uniform angular speed. When the force required for rotation is minimum the perpendicular axis passes through a specific point; The point’s location is 

Option: 1

0.75 m from any of the mass


Option: 2

0.6 m from the first mass


Option: 3

0.9 m from the second mass

 


Option: 4

0.87 m  from the second mass


Answers (1)

best_answer

Let us assume the distance from a mass to that specific point is “x” m

And another mass is at a distance of (1.5-x) m

Inertia for the two masses is I_{1} \ and \ I_{2}

Total inertia I=I_{1}+ I_{2}

Therefore Total inertia will be 

\\ I=(0.5)x^{2}+0.5\left ( 1.5-x \right )^{2}\\ I=0.5x^{2}+0.5\left ( 2.25-3x+x^{2} \right ) \\ I=x^{2}-1.5x+1.125

For work done, we know the equation 
W=\frac{1}{2}I\omega ^{2}

W=\frac{1}{2}\left [ x^{2}-1.5x +1.125\right ]\omega^{2}

For minimum force required 

\\ \frac{dW}{dx}=0 \\\Rightarrow 2x-1.5 =0 \\ \\ \Rightarrow x=0.75

 

Posted by

Ajit Kumar Dubey

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