Rotation 1 Question 10

13. The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum?

(a) 32

(b) 1

(c) 32

(d) 32

(2017 Main)

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Answer:

Correct Answer: 13. (d)

Solution:

  1. MI of a solid cylinder about its perpendicular bisector of length is

I=Ml212+R24I=mR24+ml212=m24πρl+ml212[ρπr2l=m]

For I to be maximum,

dIdl=m24πρ1l2+ml6=0m24μπρ=ml36l3=3m2πρl=321/3mπρ1/3ρ=mπR2lR2=mπρlR2=mπρ231/3πρm1/3=mπρ2/3231/3R=mπρ1/3231/6

lR=32



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