Question

Two solid cylinders of the same density are found to have the same moment of inertia about their respective principal axes. The length of the second cylinder is 16 times that of the first cylinder. If the radius of the first cylinder is 4 cm, the radius of the second cylinder is \rule{1cm{0.15mm} cm. (in integer)}

Hint:

When comparing properties of two objects where some parameters are the same, it's often easiest to set up a ratio or an equality. Write the full formula for each object, set them equal, and cancel common terms. This quickly reveals the relationship between the parameters that are different.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
The problem involves comparing the moments of inertia of two solid cylinders. The moment of inertia depends on the mass and the radius of the cylinder. The mass, in turn, depends on the density, radius, and length. We are given that the densities and moments of inertia are the same, along with a relationship between their lengths and the radius of one cylinder. We need to find the radius of the other.
Step 2: Key Formula or Approach:
The moment of inertia \(I\) of a solid cylinder of mass \(M\) and radius \(R\) about its principal (longitudinal) axis is given by: \[ I = \frac{1}{2} M R^2 \] The mass \(M\) of a cylinder with density \(\rho\), radius \(R\), and length \(L\) is: \[ M = \text{density} \times \text{volume} = \rho \times (\pi R^2 L) \] We can substitute the expression for mass into the moment of inertia formula.
Step 3: Detailed Explanation:
Let the properties of the first cylinder be denoted by subscript 1 and the second cylinder by subscript 2.
Given: \(\rho_1 = \rho_2 = \rho\), \(I_1 = I_2\), \(L_2 = 16 L_1\), and \(R_1 = 4\) cm.
The moment of inertia for the first cylinder is: \[ I_1 = \frac{1}{2} M_1 R_1^2 = \frac{1}{2} (\rho \pi R_1^2 L_1) R_1^2 = \frac{1}{2} \rho \pi L_1 R_1^4 \] The moment of inertia for the second cylinder is: \[ I_2 = \frac{1}{2} M_2 R_2^2 = \frac{1}{2} (\rho \pi R_2^2 L_2) R_2^2 = \frac{1}{2} \rho \pi L_2 R_2^4 \] Since \(I_1 = I_2\), we can equate the two expressions: \[ \frac{1}{2} \rho \pi L_1 R_1^4 = \frac{1}{2} \rho \pi L_2 R_2^4 \] The common factor \(\frac{1}{2} \rho \pi\) cancels out, leaving: \[ L_1 R_1^4 = L_2 R_2^4 \] Now, substitute the given relation \(L_2 = 16 L_1\): \[ L_1 R_1^4 = (16 L_1) R_2^4 \] The term \(L_1\) cancels out: \[ R_1^4 = 16 R_2^4 \] We need to solve for \(R_2\): \[ R_2^4 = \frac{R_1^4}{16} \] Taking the fourth root of both sides: \[ R_2 = \sqrt[4]{\frac{R_1^4}{16}} = \frac{R_1}{\sqrt[4]{16}} = \frac{R_1}{2} \] Finally, substitute the given value \(R_1 = 4\) cm: \[ R_2 = \frac{4 \text{ cm}}{2} = 2 \text{ cm} \] Step 4: Final Answer:
The radius of the second cylinder is 2 cm.