m=ρ(πR2l)...(1)
dm=ρπ(2R.dRl+R2dl)
0=ρπ(2RldR+R2dl)
2RldR=−R2dl...(2)
I=4MR2+12Ml2
I=12M(3R2+l2)
dI=12M(6R(dR)+2ldl)=0...(3)
Putting (1),(2) in (3)
6R(2Rl−R2dl)+2ldl=0
(l−3R2+2l)dl=0
−3R2+2l2=0
2l2=3R2
Rl=23
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?
Held on 2 Apr 2017 · Verified 6 Jul 2026.
23
23
23
1
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