Rotational MotionHard
Question
Two points masses of 0.3 kg and 0.7 kg are fixed at the end of the rod of length 1.4 cm 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
Options
A.0.42 m from mass of 0.3 kg
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
Solution
Work done W = 
Iω2
If x is the distance of mass 0.3 kg from the centre of mass, we will have
I = (0.3)x2 + (.7)(1.4 - x)2
For work to be minimum, the moment of inertia (I) should be minimum, or
= 0
or 2(0.3x) - 2(0.7)(1.4 - x) = 0
or (0.3x) = (0.7)(1.4 - x)
⇒ x =
= 0.98
Iω2 If x is the distance of mass 0.3 kg from the centre of mass, we will have
I = (0.3)x2 + (.7)(1.4 - x)2
For work to be minimum, the moment of inertia (I) should be minimum, or
= 0or 2(0.3x) - 2(0.7)(1.4 - x) = 0
or (0.3x) = (0.7)(1.4 - x)
⇒ x =
= 0.98Create a free account to view solution
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