The figure below shows three 0.0120 kg particles that have been glued to a rod o

Question

The figure below shows three 0.0120 kg particles that have been glued to a rod of length L = 6.90 cm and negligible mass. The assembly can rotate around a perpendicular axis through point O at the left end.

(a) Assume one particle (that is, 33% of the mass) is removed. By what percentage does the rotational inertia of the assembly around the rotation axis decrease when that removed particle is the innermost one?

(b) What is the percentage decrease when that removed particle is the outermost one?

Explanation / Answer

Let m be mass of each particle.

Moment of inertia before removing any particle=m(d2+(2d)2+(3d)2)=m*14md2

a)moment of inertia when innermost particle has been removed=m((2d)2+(3d)2)=13md2

Percent decrease in moment of inertia=1/14*100=7.14%

b)moment of inertia when outermost particle has been removed=m(d2+(2d)2)=5md2

Percent decrease in moment of inertia=9/14*100=64.3%

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