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Question

Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L of negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point 'P' on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω0 is minimum is given by


A
x=m2Lm1+m2
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B
x=m1Lm1+m2
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C
x=m1m2L
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D
x=m2m1L
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Solution

The correct option is A x=m2Lm1+m2
Moment of inertia m1 about the axis I1=m1x2
Moment of inertia of m2 about the axis I2=m2(Lx)2
K.E. is rotational.
Total K.E. is E=12I1ω20+12I2ω20
=12ω20(m1x2+m2(Lx)2)
Work done is change in K.E.
To minimize E, differentiate wrt x and equate to zero.
m1xm2(Lx)=0
x=m2Lm1+m2
Alternatively, work done is minimum when the axis passes through the center of mass.
Center of mass is at m2Lm1+m2

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