Rotational Motion

JEE Main level, single correct, medium

88 of 280

Question 88 of 280

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

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Answer: (B) 32\sqrt{\dfrac{3}{2}}

m=ρ(πR2l)...(1)m=\rho \left(\pi {R}^{2}l\right)...(1)

dm=ρπ(2R.dRl+R2dl)dm=\rho \pi \left(2R.dRl+{R}^{2}dl\right)

0=ρπ(2RldR+R2dl)0=\rho \pi \left(2RldR+{R}^{2}dl\right)

2RldR=−R2dl...(2)2RldR=-{R}^{2}dl...(2)

I=MR24+Ml212I=\dfrac{M{R}^{2}}{4}+\dfrac{M{l}^{2}}{12}

I=M12(3R2+l2)I=\dfrac{M}{12}\left(3{R}^{2}+{l}^{2}\right)

dI=M12(6R(dR)+2ldl)=0...(3)dI=\dfrac{M}{12}\left(6R\left(dR\right)+2ldl\right)=0...(3)

Putting (1),(2)(1),(2) in (3)(3)

6R(−R2dl2Rl)+2ldl=06R\left(\dfrac{-{R}^{2}dl}{2Rl}\right)+2ldl=0

(−3R2l+2l)dl=0\left(\dfrac{-3{R}^{2}}{l}+2l\right)dl=0

−3R2+2l2=0-3{R}^{2}+2{l}^{2}=0

2l2=3R22{l}^{2}=3{R}^{2}

lR=32\dfrac{l}{R}=\sqrt{\dfrac{3}{2}}

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