Question

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) $\dfrac{3}{{\sqrt 2 }}$
B) $\sqrt {\dfrac{3}{2}}$
C) $\dfrac{{\sqrt 3 }}{2}$
D) $1$

Answer 1

Here, we will use the concept of moment of inertia. Firstly, we write the moment of inertia of a uniform cylinder of length L and radius R about its perpendicular bisector i.e. $I = \dfrac{{m{R^2}}}{4} + \dfrac{{m{L^2}}}{{12}}$ , after that the condition is given moment of inertia is minimum for this we differentiate the moment of inertia with respect to L then equate it to zero. Then we will get the desired value of the ratio of L/R.

Complete Step by step solution
Here, it is given that, Length of the cylinder = L
And radius of the cylinder = R
Now, we know that moment of inertia of the uniform cylinder about its perpendicular bisector is $I = \dfrac{{m{R^2}}}{4} + \dfrac{{m{L^2}}}{{12}}$ …………………………. (1)
As, we know that the volume of cylinder is $V = \pi {R^2}L$ , then we can write ${R^2} = \dfrac{V}{{\pi L}}$
Now, substitute the value of the R2 in equation (1), we get
$\Rightarrow I = \dfrac{m}{4}\left( {{R^2} + \dfrac{{{L^2}}}{3}} \right)$
$\Rightarrow I = \dfrac{m}{4}\left( {\dfrac{V}{{\pi L}} + \dfrac{{{L^2}}}{3}} \right)$ …………………… (2)
Now, the condition is given in the question i.e. moment of inertia must be minimum., for this we know that the condition of minima we can differentiate the value of I with respect to L and equate it equal to zero.
For this, differentiate the equation (2) with respect to L, we get
$\Rightarrow \dfrac{{dI}}{{dL}} = \dfrac{m}{4}\left( {\dfrac{{ - V}}{{\pi {L^2}}} + \dfrac{{2L}}{3}} \right)$
For maxima or minima equate the above equation equal to zero, we get
$\Rightarrow \dfrac{m}{4}\left( {\dfrac{{ - V}}{{\pi {L^2}}} + \dfrac{{2L}}{3}} \right) = 0$
On rearranging the above equation, we get
$\Rightarrow \dfrac{V}{{\pi {L^2}}} = \dfrac{{2L}}{3}$
Again, using here the volume of cylinder is , put this value in above equation, we get
$\Rightarrow \dfrac{{{R^2}}}{L} = \dfrac{{2L}}{3}$
$\Rightarrow \dfrac{{{L^2}}}{{{R^2}}} = \dfrac{3}{2}$
Taking square root on both sides, we get
$\Rightarrow \sqrt {\dfrac{{{L^2}}}{{{R^2}}}} = \sqrt {\dfrac{3}{2}}$
$\Rightarrow \dfrac{L}{R} = \sqrt {\dfrac{3}{2}}$
Hence, option B is correct.

Note:
Here, one must get confused between the expressions of the moment of inertia of the uniform cylinder about its perpendicular bisector and the moment of inertia of the cylinder about the axis passing through the centre of the cylinder.
Moment of inertia of uniform cylinder about its perpendicular bisector is $I = \dfrac{{m{R^2}}}{4} + \dfrac{{m{L^2}}}{{12}}$ , this can be derived by using perpendicular axis theorem.
And the moment of inertia of the cylinder about the axis passing through the centre of the cylinder is $\dfrac{{m{R^2}}}{2}$ , this can be derived by using the centre of mass concept.