Problem 31 - Master Resource Book in JEE Main Physics

Question

Moment of inertia of the cylinder about an axis perpendicular to the axis of the cylinder and passing through the centre is $$ I=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}\right) $$ If \(\rho\) is volume density of the cylinder, then $$ M=\left(\pi R^{2} L\right) \rho=\text { constant } $$ \(\therefore\) $$ L=\frac{M}{\pi R^{2} \rho} $$ Put in Eq. (i) $$ I=M\left(\frac{R^{2}}{4}+\frac{M^{2}}{12 \pi^{2} R^{4} \rho^{4}}\right) $$ For \(/\) to be minimum, \(\frac{d l}{d R}=0\) $$ \begin{aligned} &\frac{d l}{d R}=M\left(\frac{R}{2}-\frac{M^{2}}{3 \pi^{2} R^{5}}\right)=0 \\ &\frac{R}{2}=\frac{M^{2}}{3 \pi^{2} \rho^{2} R^{5}} \quad \text { or } \quad R^{6}=\frac{2 M^{2}}{3 \pi^{2} \rho^{2}} \end{aligned} $$ Using Eq. (ii), \(\quad R^{6}=\frac{2 \pi^{2} R^{4} L^{2} \rho^{2}}{3 \pi^{2} \rho^{2}}\) Or \(R^{2}=\frac{2}{3} L^{2}\) or \(\frac{L^{2}}{R^{2}}=\frac{3}{2}\) Or \(\frac{L}{R}=\sqrt{3 / 2}\)

Step-by-Step Solution

Answer

The ratio \(\frac{L}{R}\) is \(\sqrt{\frac{3}{2}}\).

Step 1: Calculate Mass

Given the cylinder's mass formula: \[M = \pi R^2 L \rho\]Solve for length using mass, radius, and density:\[L = \frac{M}{\pi R^2 \rho}\]

Step 2: Substitute for Length in Moment of Inertia

Substitute the expression for L into the moment of inertia formula:\[I = M\left(\frac{R^2}{4} + \frac{M^2}{12 \pi^2 R^4 \rho^4}\right)\]

Step 3: Minimize the Moment of Inertia

Set the derivative of I with respect to R to zero for minimization:\[\frac{dI}{dR} = M\left(\frac{R}{2} - \frac{M^2}{3\pi^2 \rho^2 R^5}\right) = 0\]This simplifies to:\[\frac{R}{2} = \frac{M^2}{3 \pi^2 \rho^2 R^5}\]

Step 4: Solve for Relationship Between R and L

Rearrange the minimization condition to find:\[R^6 = \frac{2M^2}{3\pi^2 \rho^2}\]Use Eq. (ii) which is:\[R^6 = \frac{2\pi^2 R^4 L^2 \rho^2}{3\pi^2 \rho^2}\]Resulting in:\[R^2 = \frac{2}{3} L^2\]

Step 5: Final Ratio of Length to Radius

From the equation \[R^2 = \frac{2}{3} L^2\], take the square root to find:\[\frac{L}{R} = \sqrt{\frac{3}{2}}\]

Key Concepts

Cylinder

A cylinder is a three-dimensional geometric shape that is characterized by its circular base and straight sides. It can be visualized as a tube with two identical circular ends and a specific height, often termed as length in certain contexts. The properties of a cylinder heavily influence its physical characteristics, such as moment of inertia. The moment of inertia is related to how mass is distributed in the cylinder, and it dictates how the cylinder will respond to rotational forces, making it a crucial factor in many engineering and physics applications. When analyzing a cylinder, knowing its radius and length is essential, because these dimensions directly impact its volume and overall mass. Understanding these basics helps in computing other significant characteristics like volume density and moment of inertia.

Volume Density

Volume density, often denoted as \( \rho \), is a measure of how much mass is concentrated in a certain volume of space. For a cylinder, this can be calculated using the formula: \[ M = \pi R^2 L \rho \]This formula tells us that mass \(M\) is derived from the circular base area \(\pi R^2\), and the length \(L\), all multiplied by the volume density \(\rho\). Understanding volume density is important because it gives insight into the distribution of mass within the cylinder, impacting its stability and structural attributes. It helps in defining the mass-to-space relationship which is crucial for calculations involving moment of inertia. With a constant volume density, the mass distribution is uniform, making it easier to predict how the object will behave under various conditions.

Minimization

In calculus and physics, minimization involves finding the point at which a particular function takes its smallest value. In the context of this problem, the focus is on minimizing the moment of inertia \(I\) of the cylinder with respect to a changing radius \(R\). To achieve this, the derivative of the moment of inertia with respect to \(R\) (\(\frac{dI}{dR}\)) is set to zero. This implies finding where the rate of change stops increasing and begins to decrease. By solving the equation:\[ \frac{dI}{dR} = M\left(\frac{R}{2} - \frac{M^2}{3\pi^2 \rho^2 R^5}\right) = 0 \]we can find the optimal relationship between the radius \(R\) and the moment of inertia. Minimization in this context is useful not only for theoretical analysis but also for practical applications where minimizing resistance to rotation could be crucial, such as in wheel or rotor design. It involves sophisticated understanding of calculus to ensure computations are precise and relevant results are obtained.

Mass Formula

The mass formula for a cylinder provides critical information about how mass is related to its geometric properties. This formula is given by:\[ M = \pi R^2 L \rho \]where \(M\) is the mass, \(R\) is the radius, \(L\) is the length, and \(\rho\) is the volume density. This relationship signifies that the mass of a cylinder is contingent upon its radius, length, and density. Using the mass formula, one can manipulate the equation to solve for other unknowns, such as length or density, when the mass and one dimension are known. This becomes particularly useful when considering practical scenarios involving manufacturing or materials planning. The mass formula also interlinks with other physics concepts, such as the moment of inertia, providing a comprehensive framework for understanding how a change in one property impacts the others. Consequently, it serves as a foundational element in the study and application of physical properties of cylindrical objects.